Correlation Coefficient -1 0 1 Negative No Positive Correlation Correlation Correlation.
Risk Management Topic Two { Default correlation and...
Transcript of Risk Management Topic Two { Default correlation and...
Risk Management
Topic Two – Default correlation and industrial correlation models
2.1 Mixture models for modeling default correlation
2.2 CreditRisk+
2.3 CreditMetrics and copula functions
1
2.1 Mixture models for modeling default correlation
Why are we concerned about dependence between default events and/or
credit quality changes?
• It affects the distribution of loan portfolio losses – critical in deter-
mining quantities or other risk measures used for allocating capital for
solvency purposes.
• Clustering phenomena: simultaneous defaults could affect the stability
of the financial system with profound effects on the entire economy.
The Occurrence of disproportionately many joint defaults is termed
the “extreme credit risk”.
2
Difficulties
• Parameter specification for default dependence models is often some-
what arbitrary.
Issues
• The “marginal” credit risk of each issuer in a pool is usually “well”
determined. Modeling various correlation structures that work with
the given marginal characteristics is the major challenge.
• Counterparty risk: The counterparty of a financial contract cannot
honor the contractual specification. For example, in a credit default
swap, the counterparty risk is high when the credit quality of the
protection seller is correlated with that of the underlying reference
securities.
3
Bernuolli mixture model
• The default indicator variable 1{default} equals 0 or 1, corresponding
to “no default”or “default” of an obligor, respectively. This is a
discrete Bernuolli random variable with only two possible values.
• We describe the loss of a portfolio from a loss statistics L = (L1, · · · , Lm)
with Bernuolli variables Li ∼ B(1;Pi). Here, B(m; p) denotes the bino-
mial distribution with m independent trials and stationary probability
of success p. More generally, we allow the loss probabilities (over a
given time horizon) to be a vector of random variables
P = (P1, · · · , Pm) ∼ F
for some distribution function F with support in [0,1]m.
• In a mixture model, the default probability Pi of obligor i is assumed
to depend on a set of common economic factors. This is how default
correlation among the obligors arise.
4
Conditional independence
Two events R and B are said to be conditionally independent given the
third event G if and only if
P [R ∩B|G] = P [R|G]P [B|G].
The last relation can be replaced by the following equivalent relation:
P [R|B ∩G] = P [R|G].
That is, given that G occurs, the conditional probability that R occurs is
unchanged by whether or not B occurs.
5
Two random variables X and Y are conditionally independent given an
event G if they are independent in their conditional probability distribution
given G. In terms of density functions:
fX(x|G)fY (y|G) = fX,Y (x, y|G),
or in terms of distribution functions:
FX(x|G)FY (y|G) = FX,Y (x, y|G).
We say that X and Y are statistically independent given G. In short hand,
we write
X ⊥ Y |G.
6
In the context of the mixture model, we assume that conditional on the
realization P = (P1, · · · , Pm) of the vector of loss probabilities (random)
P , the Bernuolli default indicator variables L1, · · · , Lm are independent.
That is, given the default probabilities, defaults of different obligors are
independent.
The (unconditional) joint distribution of Li’s is obtained by integrating
over the probability distribution of F :
P[L1 = ℓ1, · · · , Lm = ℓm] =∫[0,1]m
m∏i=1
Pℓii (1− Pi)
1−ℓi dF (P1, · · · , Pm),
where ℓi ∈ {0,1}. Here “0” denotes “no default” and “1” denotes “de-
fault”. For example,
P[L1 = 1, L2 = 0, L3 = 1] =∫[0,1]3
P1(1− P2)P3 dF (P1, P2, P3).
7
First and second order moments of Li The first and second order mo-
ments of the single loss indicator variable Li are computed as follows.
Conditional on the random probability P , we have
E[Li|P ] = 1× Pi +0× (1− Pi) = Pi,
By the tower rule in conditional expectation, we obtain
E[Li] = E[E[Li|P ]] = E[Pi].
The dependence between defaults stems from the dependence of the
default probabilities on a set of common random factors.
8
Recall: E[L2i |P ] = 12 × Pi +02 × (1− Pi) = Pi. Also,
E[LiLj] = E[E[LiLj|P ]] = E[PiPj].
By the conditional variance formula, we obtain
var(Li) = var(E[Li|P ]) + E[var(Li|P )]
= var(Pi) + E[E[L2i |P ]− E[Li|P ]2]
= var(Pi) + E[Pi(1− Pi)] = E[Pi](1− E[Pi]).
The covariance between a pair of losses
cov(Li, Lj) = E[LiLj]− E[Li]E[Lj].
9
Note that
E[LiLj] = P(Li = 1, Lj = 1)× 1× 1+ P(Li = 1, Lj = 0)× 1× 0
+ P(Li = 0, Lj = 1)× 0× 1+ P(Li = 0, Lj = 0)× 0× 0
= P(Li = 1, Lj = 1)
=∫ 1
0
∫ 1
0PiPj dF (Pi, Pj) by conditional independence
= E[PiPj].
Hence,
cov(Li, Lj) = E[PiPj]− E[Pi]E[Pj] = cov(Pi, Pj),
so that the default correlation in a Bernuolli mixture model is
corr(Li, Lj) =cov(Pi, Pj)√
E[Pi](1− E[Pi])√E[Pj](1− E[Pj])
.
The covariance between a pair of losses in the portfolio is fully captured
by the covariance structure of the multivariate distribution F of the vector
of random loss probabilities P .
10
Correlation of defaults of a pair of risky assets
Consider two obligors A and B and a fixed time horizon T .
Define 1{A} =
{1 A defaults before T0 otherwise
and similar definition for 1{B}.
pA = probability of default of A before T = E[1{A}]
pB = probability of default of B before T = E[1{B}]
pAB = joint default probability that A and B default before T
pA|B = probability that A defaults before T , given that B has
defaulted before T
pA|B =pAB
pB, pB|A =
pAB
pA, and recall
ρAB = linear correlation coefficient between default events
=E[1{A}1{B}]− E[1{A}]E[1{B}]
σ1{A}σ1{B}
=pAB − pApB√
pA(1− pA)pB(1− pB).
11
Note that pAB can be expressed in terms of pA, pB and ρAB. When there
are 2 obligors, we can compute the probabilities of all elementary events
by using ρAB. Since default probabilities are very small, the correlation
ρAB can have a much larger effect on the joint risk of a position. Observe
that
pAB = pApB + ρAB
√pA(1− pA)pB(1− pB)
pA|B = pA + ρAB
√pApB
(1− pA)(1− pB) and
pB|A = pB + ρAB
√pBpA
(1− pA)(1− pB).
Write ρAB = ρ, which is assumed to be not small; and let pA = pB = p ≪1, then
pAB ≈ p2 + ρp ≈ ρp
pA|B ≈ ρ.
The joint default probability and the conditional default probability are
dominated by the correlation coefficient ρ.
12
Limitations of pairwise correlation coefficient of default events
• With 3 obligors, we have 8 elementary events but only 7 restrictions (3
individual probabilities, 3 correlations and sum of probabilities). The
probability of the joint default of all 3 obligors cannot be determined
by the 3 pairs of correlation.
• For N obligors, we have N(N −1)/2 correlations, N individual default
probabilities. Yet we have 2N possible joint default events. The
correlation matrix only gives the bivariate marginal distributions while
the full distribution remains undetermined.
• For any pair of random variables X and Y , we have
cov(1−X,1− Y ) = cov(X,Y )
so that the linear correlation coefficient between survival events is the
same as that between default events. Hence, the covariance measure
is unable to capture the fact that risky assets exhibit greater tendency
to crash together than to boom together.
13
One-factor Bernuolli mixture model
Retail banking portfolios and portfolios of smaller banks are often quite
homogeneous. Assuming Li ∼ B(1; p) with a common single random
default probability p ∼ F , where F is a distribution function with support
in [0,1]. As the mixture distribution is dependent on the single distribution
F (p), this leads to the one-factor Bernuolli mixture model.
The joint distribution of the loss variables Li’s:
P[L1 = ℓ1, · · · , Lm = ℓm] =∫ 1
0pk(1− p)m−k dF (p)
where k =m∑
i=1
ℓi and ℓi ∈ {0,1}, i = 1,2, · · · ,m. Here, k counts the number
of defaults among m obligors in the portfolio.
14
• We write L as the random number of defaults. Since there are Cmk
possible combinations of defaults among the m obligors that lead to
a total of k defaults, the probability that exactly k defaults occur is
P[L = k] = Cmk
∫ 1
0pk(1− p)m−k dF (p).
This is the mixture of the binomial probabilities with the mixing dis-
tribution F .
• The uniform default probability of any obligor in the homogenous
portfolio is given by
p = P[Li = 1] = E[Li] =∫ 1
0p dF (p).
15
• Note that E[LiLj] = P[Li = 1, Lj = 1]. The uniform default correla-
tion of a pair of obligors is
ρ = corr(Li, Lj) =P[Li = 1, Lj = 1]− p2
p(1− p)
=
∫ 10 p2 dF (p)− p2
p(1− p)=
var(p)
p(1− p).
• Intuitively, with a higher var(p), we have a higher corr(Li, Lj).
• Recall that the variance of the Bernuolli variable that assumes value
either 0 or 1, and has parameter p is
E[X2]− E[X]2 = p− p2 = p(1− p).
Note that ρ = 1 if and only if var(p) = p(1 − p). In order that this
occurs, the common random default probability p must be Bernuolli
with parameter p.
16
Remarks
1. Since var(p) ≥ 0, so corr(Li, Lj) ≥ 0. The non-negativity of default
correlation is obvious since Li and Lj are dependent on the common
mixture variable p. In other words, we cannot implement negative
dependencies between the default events of obligors under this one-
factor binomial mixture model.
2. corr(Li, Lj) = 0 if and only if var(p) = 0, implying no randomness with
regard to p. In this case, p assumes the single value p. The random
portfolio loss L follows a binomial distribution with constant default
probability p. Correspondingly, the default events are independent.
That is, corr(Li, Lj) = 0 ⇒ independence of default events. The
converse statement is obvious since independence of Li and Lj implies
corr(Li, Lj) = 0. We then have corr(Li, Lj) = 0 ⇔ independence of
default events.
17
3. corr(Li, Lj) = 1 implies a “rigid” behavior of losses in the portfolio.
This corresponds to p = 1 with probability p and p = 0 with probability
1− p, where the distribution F of the random default probability p is
a Bernoulli distribution. Financially speaking, when an external event
occurs with probability p, all obligors in the portfolio default and the
total portfolio is lost. Otherwise, with probability 1 − p, all obligors
survive.
• A non-financial analogy is the death events of all passengers in an
aeroplane, either there is no crash with probability 1−p (all passengers
survive) or an accident occurs with probability p (all passengers die).
18
Fractional losses under the one-factor binomial mixture model
Define Dn =∑n
i=1Li, which is the total number of defaults in the port-
folio. We then have
E[Dn] =n∑
i=1
E[Li] = np.
Recall the following formulas:
var(Li) = E[Pi](1− E[Pi]) = p(1− p).
var(Dn) =n∑
i=1
var(Li) +n∑
i=1
n∑j = 1j = i
cov(Li, Lj).
Under the assumption of a common random default probability, we have
var(Dn) = np(1− p) + n(n− 1)(E[p2]− E[p]2).
19
Therefore, we have
var(Dn
n
)=
p(1− p)
n+
n(n− 1)
n2var(p) −→ var(p) as n → ∞.
When considering the fractional loss for n large, the only remaining vari-
ance is that of the distribution of p.
• One can obtain any default correlation in [0,1] through appropriate
choices of var(p) and E[p]. Note that the correlation of a pair default
of events depends only on the first and second order moments of F .
However, the distribution of Dn can be quite different for different
distribution F .
20
Large portfolio approximation
We perform the sampling of default events among a population of n
obligors that share a common random default probability p. By the law
of large numbers, the sample mean converges to the population mean.
In the current context, we have
Dn
n→ p as n → ∞
where p denotes the realized default probability. The conditional proba-
bility distribution of Dnn is given by
P[Dn
n≤ θ
∣∣∣∣∣p = p
]n↑∞−→
{0 if θ < p1 if θ ≥ p
.
21
The fractional loss distribution is obtained by integrating over the distri-
bution of p, that is,
P[Dn
n≤ θ
]n↑∞−→
∫ 1
01{θ≥p}f(p) dp
=∫ θ
0f(p) dp = F (θ).
As n → ∞, it is the probability distribution of the random default proba-
bility p that determines the fractional loss distribution.
22
Example – beta distribution for the common random default probability
A beta distribution for p gives us a flexible class of distributions in the
interval [0,1]. The beta distribution is characterized by the two positive
parameters α, β. The mean and variance of the distribution are
E[p] =α
α+ β, var(p) =
αβ
(α+ β)2(α+ β +1).
If we look at various combinations of the two parameters for which αα+β =
p for a given level of expected default probability p, the variance of the
distribution decreases as we increase α [see Hw 3, Qn 3(b)].
It is the common dependence on the mixture variable p that induces
the correlation in the default events. The more variability that is in the
mixture distribution, the stronger correlation of default events and more
weight there in the tails of the loss distribution.
23
Two beta distributions are shown, both with the common mean 0.1 but
with different variances. The density corresponds to α = 1, β = 9 (the
curve that always slopes downwards) observes a higher probability of see-
ing smaller losses and larger losses, which then gives a higher variance in
the default-event correlation.
24
Comparison with the case of independence of defaults
• The binomial distribution for independent defaults has a very thin
tail, thus not representing the possibility of a large number of defaults
realistically.
• Taking N = 100 obligors
Default Prob. (%) 1 2 3 4 5 6 7 8 9 1099.0% VaR Level 5 7 9 11 13 14 16 17 19 20
What is VaRα(X)? It is the maximum loss which is not exceeded with
a given probability (or confidence level α).
V aRα(X) = inf{x ≥ 0|P [X ≤ x] ≥ α}.
Take p = 5%, the probability with 13 defaults or less is at least
99%, that is, 99% confidence level. VaRα(X) is seen to increase with
default probability.
25
The distribution of the number of defaults among 50 issuers in the case
of a pure binomial model with default probability 0.1 and in cases with
beta distributions (α = 1, β = 9) as mixture distributions over the default
probability. The later case exhibits a higher probability of seeing a large
number of default losses (commonly known as the thick tail) and a small
number of losses also.
26
Moody’s binomial expansion method
• For a binomial distribution with independent obligors, the tail with
fewer (smaller value of n) obligors is “fatter” than the tail with many
(larger value of n) independent obligors. Actually, when n → ∞, the
variance of fractional loss decreases as O(1n
)since var
(Dnn
)= p(1−p)
n .
• Moody is aware that a pure binomial distribution with independent
defaults is unrealistic. They make the tails of distribution fatter by
assuming a smaller number of obligors (the diversity score). For
example, adjustment is made for industry concentration
The idea is to approximate the loss on a portfolio of n positively correlated
loans with the loss on a smaller number of independent loans with larger
face value.
27
Moody’s binomial expansion technique (BET)
The two parameters in a binomial experiment are number of trials n and
probability of success p.
• Diversity score, weighted average rating factor (default probability)
and binomial expansion technique.
• Generate the loss distribution.
To build a hypothetical pool of uncorrelated and homogeneous assets
that mimic the default behavior of the original pool of correlated and
inhomogeneous assets.
28
Criteria of an idealized comparison portfolio
The diversity score of a given pool of participations is the number n
of bonds in an idealized comparison portfolio that meets the following
criteria:
• Comparison portfolio and collateral pool have the same face value.
• Bonds in the comparison portfolio have equal face values.
• Comparison bonds are equally likely to default, and their defaults are
independent.
• Comparison bonds are of the same average default probability as that
of the participations of the collateral pool.
• According to some measure of risk (say variance of the loss principal),
the comparison portfolio has the same total risk as does the collateral
pool.
29
Binomial approximation using diversity scores
Seek the reduction of the problem of generating the distribution of mul-
tiple defaults to binomial distributions.
If the n loans each with equal face value are independent and they have
the same default probability, then the distribution of the portfolio loss is
a binomial distribution with n as the number of trials.
Let Fi be the face value of each bond, pi be the probability of default with-
in the relevant time horizon, and ρij be the linear correlation coefficient
of default events.
Assuming zero recovery, the loss variable Li (in dollar amount) associated
with bond i with face value Fi is given by
Li = Fi1{Di},
where Di is the default event of bond i.
30
With n bonds, the total notional principal of the portfolio isn∑
i=1
Fi. The
mean and variance of the loss principal P (in dollar amount) is
E[P ] = E[L1 + · · ·+ Ln] =n∑
i=1
FiE[1{Di}] =n∑
i=1
piFi
var(P ) =n∑
i=1
n∑j=1
E[LiLj]− E[Li]E[Lj]
=n∑
i=1
n∑j=1
FiFj(E[1{Di}1{Dj}]− E[1{Dj}]E[1{Dj}])
=n∑
i=1
n∑j=1
FiFjρij√pi(1− pi)pj(1− pj).
Here, ρij is specified rather than as a quantity calculated based on the
information on joint defaults.
31
• We construct an approximating portfolio consisting D independent
loans, each with the same face value F and the same default proba-
bility p.
To determine the binomial distribution of the loss amount of the com-
parison portfolio, we need to specify the constant default probability p,
number of obligors D and the common face value of the bonds. These
lead to the following system of equations:
n∑i=1
Fi = DF
E[P ] =n∑
i=1
piFi = DFp
var(P ) = F2Dp(1− p).
32
Note that
FD(1− p) =n∑
i=1
(1− pi)pi
and
D var(P ) = FDp[FD(1− p)].
Solving the equations, we obtain
p =
∑ni=1 piFi∑ni=1 Fi
D =
∑ni=1 piFi
∑ni=1(1− pi)Fi∑n
i=1∑n
j=1 FiFjρij√pi(1− pi)pj(1− pj)
F =n∑
i=1
Fi
/D.
Here, D is called the diversity score.
33
Contagion models
Contagion means that once a firm defaults, it may bring down other
firms with it. Define Yij to be an “infection” variable. Both Xi and Yijare Bernuolli variables assuming values either 0 or 1.
Xi is the default indicator of firm i due to its firm specific causes
Yij =
{1 if default of firm i brings down firm j0 if default of firm i does not bring down firm j
.
Assuming homogeneous property on the parameters:
E[Xi] = p and E[Yij] = q.
Also, Xi and Yij, for all i and j, are assumed to be independent.
34
The default indicator of firm i is
Zi = Xi + (1−Xi)
1−∏j =i
(1−XjYji)
.Note that Zi equals one either when there is a direct default of firm i
or if there is no direct default and∏j =i
(1 − XjYji) = 0. The latter case
occurs when at least one of the factor XjYji is 1, which happens when
firm j defaults and infects firm i. Define Dn = Z1 + · · · + Zn, Davis and
Lo (2001) find that [see Qn (5) in Hw 3]
E[Dn] = n[1− (1− p)(1− pq)n−1]
var(Dn) = n(n− 1)βpqn − (E[Dn])
2
cov(Zi, Zj) = βpqn − (E[Dn/n])
2,
where
βpqn = p2 +2p(1− p)[1− (1− q)(1− pq)n−2]
+ (1− p)2[1− 2(1− pq)n−2 + (1− 2pq + pq2)n−2].
35
1. When there is no infection, q = 0, so zero contagion gives a pure
binomial model. In this case, E[Dn] becomes np, which is the same
as that of the binomial distribution.
2. Increasing the contagion brings more mass to high and low default
numbers. To preserve the mean, we must compensate for an increase
in the infection parameter by decreasing the probability of direct de-
fault. Note that
E[Zi] = E
1− (1−Xi)∏j =i
(1−XjYji)
= 1− (1− p)(1− pq)n−1.
By equating the probability of no default with and without infection
effect, we find p(q) such that
[1− p(q)][1− p(q)q]n−1 = 1− p.
We have p(0) = p and p(q) < p for q > 0. While the mean is preserved,
the variance is increased.
36
Probability mass function of Dn
Let F (k;n, p, q) denote the probability mass function of Dn, where
F (k;n, p, q) = P[Dn = k],
then F (k;n, p, q) = Cnkα
pqnk, where
αpqnk = pk(1− p)n−k(1− q)k(n−k)
+k−1∑i=1
Cki p
i(1− p)n−i[1− (1− q)i]k−i(1− q)i(n−k).
• It is necessary to consider separately the special case where all of the
k defaulting bonds do not infect any other non-defaulting bonds in
the portfolio. This is captured by the first term in αpqnk.
37
Recall the following results:
• (1− q)i = probability that all i self-defaulting bonds do not
affect a particular non-defaulting bond• 1−(1−q)i = probability that at least one of the self-defaulting
bonds affect a particular non-defaulting bond
• The k bonds that are defaulting can be chosen in Cnk combinations.
Write αpgnk as the probability that out of the n bonds, k (≤ n) particular
bonds default. Consider the two cases:
(i) All these k bonds are self-defaulting and they do not infect any of the
remaining n− k bonds;
(ii) Of the k bonds defaulting, i of them are self-defaulting while k − i
of them are infected by the first of these i self-defaulting bonds,
i = 1,2, · · · , k − 1.
38
• With k bonds that are defaulting, i of them are self-defaulting and
k− i of them are infected by the first i of these self-defaulting bonds.
For i = 1,2, · · · , k − 1, consider
0 0 0 0 0 0 0 0 0 0 0 0
first k bonds default
i of these
defaulting bonds
are self-defaulting
k - i bonds
defaulted since
they are
infected by
the first i
self-defaulting bonds
n – k bonds remain non-defaulting
Probability of occurrence for given value of i is pi(1 − p)n−i[1 − (1 −q)i]k−i(1− q)i(n−k). There are Ck
i combinations to choose these i bonds
from k bonds, and subsequently, we sum from i = 1 to i = k − 1.
39
2.2 CreditRisk+
This is an industrial code that applies actuarial mathematics to calculate
the loss distribution of a portfolio
• It requires a limited amount of input data and assumptions. It uses
as basic input the same data as required by the Basel II internal rating
system.
A loan is understood as a Bernuolli random variable (default or no
default over a given time horizon). No profits or losses from rating
migrations are considered.
40
• It provides an analytic-based portfolio approach for rapid and unam-
biguous calculations of the loss distribution (no simulations are need-
ed). Efficient numerical techniques, like the Fast Fourier transform,
and analytic approximation method, like the saddle point approxima-
tion method, can be applied to hasten the computation.
– Unambiguity of the distribution is important (in particular, infor-
mation on the tail of the distribution) since simulation based port-
folio approaches usually fail to give agreeing answers (due to low
probabilities of default).
– Rapid calculations are helpful when comparative statistics (“what
if” analyzes) are performed: performing a large number of scenario
tests under different scenarios for the parameters.
The resulting portfolio loss distribution can be expressed as a sum of
independent compound negative binomial random variables.
41
Limitations
• CreditRisk+ takes the Poisson approximation to the Bernuolli type
default events, which implicitly allow multiple defaults. This is a rea-
sonable approximation, given that the expected default probabilities
are small. Though highly unlikely, possible paradoxes may be created,
say, calculated losses > sum of exposures since multiple defaults of
single obligor is allowed.
• Limited range of default correlation: produces only positive and usu-
ally moderate levels of default correlation among 2 obligors sharing a
common risk driver.
• Recovery rates are absorbed in the exposure upon default and they
are assumed to be independent of default events.
42
Fundamentals of CreditRisk+ – mixture Poisson distribution
• No financial modeling for the default event: The reason for default
of an obligor is not the subject of the model. Instead the default is
described as a purely random event, characterized by a probability of
default.
• Stochastic probability of default: The probability of default of an
obligor is not seen as a constant, but a randomly varying quantity,
driven by one or more (systematic) risk factors. This feature is used
to generate default correlation among the obligors. The distribution
of the default intensity is usually assumed to be a gamma distribution.
43
• Conditional independence: Given the risk factors, the default of oblig-
ors are independent.
• Only implicit correlation via the risk drivers: Default correlation be-
tween obligors are not explicit, but arise only implicitly due to the
common risk factors which drive the probability of defaults. A lin-
ear relationship between the systematic risk factors and the default
probabilities is assumed.
• Using exposure bands: Group the exposures in the portfolio into var-
ious bands. Replace each exposure amount LA of obligor A by the
nearest integer multiple of L, where L is the basic unit of exposure.
The distribution of losses is derived from the probability generating func-
tions, assuming that the losses (either as the count of defaulting obligors
or dollar amount) take discrete values.
44
Probability generating functions GK(z)
The probability generating function (pgf) of a discrete non-negative integer-
valued random variable K is a function of the auxiliary variable z such that
the probability that K = k is given by the coefficient of zk in the polyno-
mial expansion of the probability generating function. We have
GK(z) = E[zK] =∞∑
k=0
P [K = k]zk,
where z can be complex with |z| ≤ 1 (technical condition that guarantees
convergence of the infinite series).
The pgf of the Poisson random variable N with parameter α is
GN(z) = eα(z−1) = e−α∞∑
n=0
(αz)n
n!,
which observes the following probability mass function of N :
P [N = n] =e−ααn
n!.
45
For a Bernuolli random variable Y with probability p, the corresponding
pgf is
GY (z) = (1− p)z0 + pz = 1+ p(z − 1).
The pgf contains all the information to calculate the probability mass
function of the associated non-negative integer valued random variable
N , where
P [N = n] =1
n!G(n)N (0).
Conditional pgf: GN(z|·) = E[zN |·].
Given GN(z|x) where x has distribution F (x), then the (unconditional)
pgf is given by
GN(z) =∫
GN(z|x) dF (x).
46
More properties on pgf
• Let K1 and K2 be a pair of independent non-negative integer-valued
random variables. The pgf of the sum K1 +K2 is simply the product
of the two pgf’s since E[zK1+K2] = E[zK1]E[zK2]. This stems from
the property that for any given pair of independent random variables
X1 and X2, we have
E[f1(X1)f2(X2)] = E[f1(X1)]E[f2(X2)]
for any functions f1 and f2. Here, we choose f1 and f2 to be expo-
nential functions.
• GnY (z) = GY (zn) for any natural number n.
For example, take n = 3, we have
G3Y (z) = P [3Y = 0]z0 + P [3Y = 3]z3 + P [3Y = 6]z6 + · · ·= P [Y = 0]z0 + P [Y = 1]z3 + P [Y = 2]z6 + · · ·
=∞∑
k=0
P [Y = k]z3k.
47
Input data for the model
• For an obligor A in a loan portfolio, we designate PA to be the ex-
pected probability of default of A (say, from the output of a rating
process).
• Write vA as the potential (dollar) loss for A upon default. The ex-
pected (dollar) loss for A is given by
ϵA = PAvA.
• To work with discretized losses, we fix a loss unit L and choose a
positive integer vA as a rounded version of vA/L. To compensate for
the error due to rounding, we adjust the expected default probability
PA =ϵAvAL
=
(vAvAL
)PA.
We then work with PA and the integer-valued loss vA.
48
For example, suppose we take vA = $160,000 and PA = 0.2%, then
ϵA = $160,000× 0.2% = $320.
We take L = $100,000 so that the rounded exposure in unit of L is 2.
The adjusted expected default probability is
PA =$160,000× 0.2%
2× $100,000= 0.16%.
49
Example Rounding of exposure to units of L
Obligor A Exposure ($) Exposure Round-off Band j(potential loss (in $100,000) exposuregiven default) (in $100,000)
vA vA/L vA1 150,000 1.5 2 22 460,000 4.6 5 53 435,000 4.35∗ 5 54 370,000 3.7 4 45 190,000 1.9 2 26 480,000 4.8 5 5
Out of the 6 obligors, 2 to Band 2, 1 to Band 4, and 3 to Band 5.
∗ Rounding up to the nearest integer is adopted so that the round-off
exposure as an integer multiple of L always starts at 1 for the sake of
convenience.
50
Allocation of 350 obligors into 6 bands
vj number of obligors ϵj µj1 30 1.5 1.52 40 8 43 50 6 24 70 25.2 6.35 100 35 76 60 14.4 2.4
vj = common exposure in band j in units of L
ϵj = expected dollar loss in band j in units of L
µj = expected number of defaults in band j
We have ϵj = vjµj. Say, µ4 = 6.3 means the expected number of defaults
among 70 obligors in band 4 is 6.3 over a given time horizon, and the
potential dollar loss in band 4 is E4 = 4× 6.3 = 25.2 (in units of L).
51
Pgf of the number of default in a portfolio
Since a default event is considered as a Bernuolli event, the pgf for a
single obligor A is
FA(z) = (1− PA) + PAz.
For the whole portfolio of obligors, we have
F (z) =∞∑
n=0
P [n defaults]zn.
As a consequence of independence between default events, the probability
generating function for the whole portfolio is
F (z) =∏A
FA(z) =∏A
[1 + PA(z − 1)].
52
Taking logarithm on both sides, and observing
ln[1 + PA(z − 1)] ≈ PA(z − 1),
we have
lnF (z) ≈∑A
PA(z − 1).
As a Poisson approximation, we take
F (z) ≈ e∑
APA(z−1) = eµ(z−1) =∞∑
n=0
e−µµn
n!zn,
where µ =∑A
E[1A] =∑A
PA =m∑
j=1
µj = expected number of default
events over a given time horizon from the whole portfolio. Here, 1A is
the indicator default variable for obligor A.
53
If the probabilities of individual defaults are small, though not necessarily
equal, the probability of realizing n default events in the portfolio over a
given time horizon is
P[n defaults] =e−µµn
n!.
This is the well known Poisson distribution with single parameter µ.
• The distribution has only one parameter, namely, the expected num-
ber of defaults µ of the whole portfolio. The distribution does not
depend on the exposures or individual probabilities of default, provided
that they are uniformly small.
• Up to this point, we assume the default probabilities to be fixed con-
stants. However, there is no necessity for the obligors to have equal
probabilities of default.
54
Remark
Suppose the Poisson approximation is not taken, then the probability
generating function is given by
F (z) =n∏
i=1
[1 + Pi(z − 1)],
where Pi is the expected default probability of the ith obligor in a portfolio
of n obligors. It would be very cumbersome to find the mass probability
function of the random number of the defaults in the portfolio involving
products of Pi (default cases) and 1− Pj (non-default cases).
55
For example,
P [no of defaults = 1] =n∑
i=1
[Pi
n∏j = ij = 1
(1− Pj)],
P [no of defaults = 2] =n∑
i=1
n∑i =j
[PiPj
n∏k = ik = j
(1− Pk)].
Using the Poisson approximation, letting µ =∑n
i=1 Pi, we have
P [no of defaults = 1] = e−µµ,
P [no of defaults = 2] = e−µµ2
2 .
56
Pgf of the loss amount from the whole portfolio in units of L
Likelihood of suffering a given level of loss amount for band j is captured
by:
Probability generating function of the distribution of loss amounts for band j
= Gj(z) =∞∑
n=0
P [loss = nvj]zn =
∞∑n=0
P [n defaults]znvj
≈∞∑
n=0
e−µjµnj
n!znvj = e−µj
∞∑n=0
(µjzvj)n
n!= exp(−µj + µjz
vj).
The dollar loss amount in units of L in band j assumes integer values:
0, vj,2vj, · · · , corresponding to no default, one default, two defaults, etc.
in that band.
57
The default events in the bands are independent so that the pgf of the
loss amount of the whole portfolio with m bands is given by the product
of the pgf’s of the loss amount of individual band. That is,
G(z) =m∏
j=1
exp(−µj + µjzvj).
We then have
P [dollar loss ofnL] =1
n!
dnG(z)
dzn
∣∣∣∣∣z=0
.
Remark
Even without the independence assumption, we still obtain a closed for-
m loss distribution when correlated defaults are assumed (see the later
section on sector analysis).
58
Poisson approximation of the pgf of loss amount (in units of L)
From G(z) = exp
− m∑j=1
µj +m∑
j=1
µjzvj
, µ =m∑
j=1
µj,
if we express the variability of exposure bands within the portfolio via the
following polynomial
p(z) =
∑mj=1 µjz
vj
µ=
∑mj=1
ϵjvjzvj∑m
j=1ϵjvj
=m∑
j=1
P [V = vj]zvj ,
then
G(z) = eµ[p(z)−1].
Here,
P [V = vj] = µj/µ.
Given that an obligor defaults, P [V = vj] gives the probability that the
loss amount is vjL, or equivalently, the defaulting obligor falls in band j.
59
The functional form for G(z) expresses mathematically the compounding
of two sources of uncertainty arising from
(i) Poisson randomness of the incidence of default events, and
(ii) variability of exposure bands within the portfolio.
Indeed, G(z) is a compound Poisson distribution with
G(z) = Gµ(GV (z)),
where Gµ(z) = eµ(z−1) and GV (z) is the pgf of a random variable V taking
values in {v1, · · · , vm} with distribution:
GV (z) = p(z) = P [V = v1]zv1 + · · ·+ P [V = vm]zvm.
60
Numerical example
Consider the following portfolio with 5 obligors:
P1 = 0.2%, v1 = 3; P2 = 0.5%, v2 = 6; P3 = 0.4%, v3 = 7;
P4 = 0.1%, v4 = 8; P5 = 0.9%, v5 = 10.
We have the mean number of defaults
µ = 0.2%+ 0.5%+ 0.4%+ 0.1%+ 0.9% = 0.021,
while the pgf for the variability of exposure is given by
GV (z) =0.2
2.1z3 +
0.5
2.1z6 +
0.4
2.1z7 +
0.1
2.1z8 +
0.9
2.1z10.
The pgf of the compound Poisson distribution is
G(z) = Gµ(GV (z)) = e0.021[(0.22.1z
3+0.52.1z
6+0.42.1z
7+0.12.1z
8+0.92.1z
10)−1].
61
Correlation in defaults
• The observed default probabilities are volatile over time, even for
obligors having comparable credit quality. The variability of default
probabilities can be related to the underlying variability in a number
of background factors, like the state of the economy.
• Two obligors are sensitive to the same set of background factors (with
differing weights), so their default probabilities will move together.
These co-movements in default probabilities give rise to correlation
in defaults.
• CreditRisk+ considers default rates as continuous random variables
and incorporates the volatility of default rates in order to capture
the uncertainty in the level of default rates. It does not attempt to
model default correlation explicitly but captures the same concentra-
tion effects through the use of the default rate volatilities and sector
analysis.
62
The standard deviation of the Poisson distribution with mean µ is√µ.
Historical evidence of the standard deviation of default event frequencies
is invariably much larger thanõ. Thus, the assumption of fixed de-
fault rates cannot account for what is observed in the data. With the
correlation of defaults becomes higher, we expect to observe higher prob-
abilities of seeing both “small number of defaults” and “large number of
defaults”.
63
Sector analysis – independent sector risk factors
• We consider n different sectors S1, · · · , Sn, with the corresponding
positive random risk factor xk, k = 1, · · · , n, whose mean is µk and
standard deviation is σk. The sector risk factors are assumed to be
independent. Sectors could be constructed with respect to industries,
countries or rating classes. Increasing the number of sectors reduces
the correlation (the highest correlation is resulted if all obligors are
allocated to single sector).
• Let Xk denote the discrete non-negative integer random variable that
equals the number of defaults in sector k, whose distribution is a
Poisson mixture distribution with random default intensity xk. Each
(say, industrial) sector is driven by a single underlying risk factor xk of
the kth industry, which models the variability over time in the mean
total number of defaults over a given time horizon measured for that
sector. Here, E[xk] = µk =∑
A∈SkPA.
64
Single-factor setting
The whole portfolio is parameterized by the n factors, one factor for each
sector. In the one-factor setting, we assume that the random default
rate xA of an individual obligor A is influenced by single factor. Later,
we generalize to the scenario where xA depends on multiple sector risk
factors.
Distribution assumption of the risk factors
Unlike the usual Poisson distribution whose parameter is a constant (the
Poisson parameter equals the mean and also the variance of the Poisson
distribution), the random default intensity xk is assumed to be gamma
distributed. The gamma distribution is chosen since it is an analytical-
ly tractable two-parameter distribution. It is a skew distribution which
approximates the normal distribution when its mean is large.
65
Gamma distribution
The density function of the gamma-distributed xk with parameters α and
β is given by
P [x ≤ xk ≤ x+ dx] = f(x) dx =1
βαγ(α)e−x/βxα−1 dx,
where γ(α) = (α−1)! =∫ ∞
0e−xxα−1 dx. The mean and standard deviation
of xk are
µ = αβ and σ2 = αβ2.
For sector k, the parameters αk and βk are related to µk and σk by
αk = µ2k/σ2k and βk = σ2k/µk.
Once µk and σk are specified, the two parameters αk and βk in the gamma
distribution can be determined.
66
How to specify µk and σk?
• The underlying factor influences the sector through the total number
of defaults in that sector, which is modeled as a gamma-mixed Poisson
distribution with random default intensity xk whose mean is µk and
standard deviation is σk.
• We assume that the standard deviation σA (besides the expected
default probability PA) has been assigned for the default rate for each
obligor in the sector.
• We take the mean of factor xk by
µk =∑
A∈Sk
ϵAvA
=∑
A∈Sk
PA =∑
A∈Sk
E[1A],
where the summation extends over all obligors in sector k.
• We randomize the default probability of obligor A, and write it as xA.
Its mean and standard deviation are PA and σA, respectively.
67
Linear relation between xk and xA
For any obligor xA, we assume a linear relationship between the random-
ized default probability xA and random default intensity xk, where
xAxk
=PA
µk.
As a check for consistency, we take the sum:
∑A∈Sk
xA =
∑A∈Sk
PA
µkxk = xk;
and the respective means:
xA =PA
µkxk = PA.
68
Following the linear relation between xA and xk, where xA =ϵAvA
xkµk
, we
take σA =ϵAvA
σkµk
so that for any pair of obligors A1 and A2
σA1
xA1
=σA2
xA2
=σkxk
and∑A
σA =∑A
ϵAvA
σkµk
=
∑APA
µkσk = σk.
For the obligors in the same sector k, we observeσA1
µA1
=σA2
µA2
= · · · ; and
they are all equal toσkµk
so that
σkµk
=
∑A σA∑A PA
=∑
A∈Sk
wAσAPA
, where wA =PA
ΣAPA.
The ratio of σk to the mean µk is a weighted average of the ratio of
standard deviation to mean for each obligor, weighted with respect to
the expected default probabilities.
69
Numerical example
The obligors’ expected defaulted probabilities over a given time period in
3 different industrial sectors are listed below:
{1%,2%,3%}, {0.5%,1.5%,2.5%,3.5%},{0.3%,0.6%,0.9%,1.2%,1.5%,1.8%}.
We then have
µ1 = mean number of defaults in sector 1 = 0.06, µ2 = 0.08, µ3 = 0.063.
Let xi, i = 1,2,3, denote the random default intensity of sector i. Suppose
x1 assumes the realized value y, then
P [no of defaults in sector 1 = n] =e−yyn
n!.
We also have
x1 = µ1 = 0.03, x2 = µ2 = 0.08, x3 = µ3 = 0.063.
70
Let xji denote the random default probability of the jth obligor in the ith
sector. We have
x11 =1
6x1, x21 =
2
6x1 and x31 =
3
6x3;
x11 = 1%, x21 = 2% and x31 = 3%;
µ1 = x1 = x11 + x21 + x31.
For the standard deviations of these random variables, we adopt the linear
relations where
σ1x1
=σ11x11
=σ21x21
=σ31x31
or σ11 =1
6σ1, σ21 =
2
6σ1, σ31 =
3
6σ1,
so
σ1 = σ11 + σ21 + σ31 orσ1x1
=1
6
σ11x11
+2
6
σ21x21
+3
6
σ31x31
.
71
Sector default distribution
The random number of defaults in sector k is gamma-mixed Poisson with
mixture variable xk. Suppose the sector risk factors are assumed to be
independent, then the pgf for the whole portfolio F (z) can be written as
the product of the pgf of the individual sectors. This gives
F (z) =n∏
k=1
Fk(z).
The pgf for the distribution of default events conditional on xk = x is
given by
Fk(z)
∣∣∣∣∣xk=x
= ex(z−1).
The probability generating function for Xk, which equals the random
number of default events in one sector is the average of the conditional
probability generating function over all possible values assumed by xk.
72
Let fk(x) denote the probability density of xk, we have
P [n defaults] =∫ ∞
0P [n defaults|x]fk(x) dx
and∞∑
n=0
znP [n defaults|x] = Fk(z)|z=x = ex(z−1)
The pgf of xk is given by
Fk(z) =∞∑
n=0
P [n defaults]zn
=∫ ∞
0
∞∑n=0
znP [n defaults|x]fk(x) dx
=∫ ∞
0ex(z−1)fk(x) dx
73
=∫ ∞
0ex(z−1)e
−x/βkxαk−1
βαkk γ(αk)
dx [set y = (1
βk+1− z)x]
=1
βαk γ(αk)
∫ ∞
0
y1βk
+1− z
αk−1
e−y
/(1
βk+1− z
)dy
=γ(αk)
βαkk γ(αk)
(1βk
+1− z)αk
=1
βαkk
(1βk
+1− z)αk
=
[1
1+ βk(1− z)
]αk
=
(1− Pk
1− Pkz
)αk
,
where Pk =βk
1+ βk, 1− Pk =
1
1+ βk, 1− Pkz =
1+ βk(1− z)
1 + βk.
74
Recall the binomial expansion with negative exponent:
(1− Pkz)−αk = 1+ (−αk)(−Pkz) +
(−αk)(−αk − 1)
2!(−Pkz)
2
+(−αk)(−αk − 1)(−αk − 2)
3!(−Pkz)
3 + · · · .
Expanding Fk(z) in its Taylor series, we obtain
Fk(z) = (1− Pk)αk
1+∞∑
n=1
(n+ αk − 1
n
)Pnk z
n
so that
P [n defaults] =
(n+ αk − 1
n
)Pnk (1− Pk)
αk.
= probability mass function of (modified) negative binomial
distribution with parameters αk and Pk.
= probability of having n failures before the αthk success wi-
-th probability of success of each trial equals 1− Pk.
Note that αk in general is NOT an integer.
75
Summary
The portfolio has been divided into n sectors with the random default
intensity xk distributed according to the Gamma distribution
Γ(αk, βk), h = 1,2, . . . , n.
The pgf for the random number of default events from the whole portfolio
is
F (z) =n∏
k=1
Fk(z) =n∏
k=1
(1− Pk
1− Pkz
)αk
where αk = µ2k/σ2k , βk = σ2k/µk and Pk =
βk1+ βk
=σ2k
σ2k + µk.
Note that F (z) is not Negative Binomial but it is a product of the Negative
Binomial sector distributions.
76
Default losses with variable default rates
We have obtained the pgf for the number of default events. In order
to pass from default events to default losses, this distribution must be
compounded with the information about the distribution of exposures
within each sector.
Define the pgf of the distribution of loss amount by
G(z) =∞∑
n=0
P [aggregate losses = nL]zn.
Invoking the sector independence assumption, we have
G(z) =n∏
k=1
Gk(z),
where Gk(z) is the loss probability generating function for sector k,1 ≤k ≤ n.
77
We link default events with losses for each sector via the compound
distribution
Gk(z) = Fk(Pk(z))
by defining the polynomial Pk(z) that captures the distribution of expo-
sures of dollar losses of the obligors in the kth sector by
Pk(z) =
∑m(k)j=1
ϵ(k)j
v(k)j
zv(k)j
∑m(k)j=1
ϵ(k)j
v(k)j
=1
µk
m(k)∑j=1
ϵ(k)j
v(k)j
zv(k)j , 1 ≤ k ≤ n.
Here, m(k) is the number of obligors in the kth sector, and
µk =m(k)∑j=1
ϵ(k)j
v(k)j
,
which is the expected total number of defaults µk over the sector k.
78
Recall that
P [V (k) = v(k)j ] =
P(k)j
µk=
ϵ(k)j /v
(k)j
µk.
We may rewrite Pk(z) as a sum over individual obligors belonging to sector
k, where
Pk(z) =1
µk
∑A∈Sk
ϵAvA
zvA.
When we deal with random default intensity xk and random default prob-
ability xA, we take the assumptionxkxA
=µkPA
and integrate out the distri-
bution with respect to xk.
79
Using the linear relation on default probabilities: xA =ϵAvA
xkµk
, we have
exp
− ∑A∈Sk
xA +∑
A∈Sk
xAzvA
= e
∑A∈Sk
xA(zvA−1)
= exkµk
∑A∈Sk
ϵAvA
(zvA−1)= exk[Pk(z)−1],
which gives the probability generating function of the distribution of losses
where each obligor A has default rate xA.
In a similar manner as shown in the calculation of Fk(z), we have
Gk(z) =∞∑
n=0
∫ ∞
0P [loss of nL|xk]fk(xk) dxk
=∫ ∞
0exp
∑A∈Sk
xA(zvA − 1)
fk(xk) dxk
=∫ ∞
0exk[Pk(z)−1]fk(xk) dxk.
80
By performing a similar integration procedure as that of F (z) (see P.71),
except that z is replaced by Pk(z), we obtain the following closed form
expression for the pgf of the distribution of loss amount:
G(z) =n∏
k=1
Gk(z) =n∏
k=1
[1− Pk
1− PkPk(z)
]αk
=n∏
k=1
1− Pk
1− Pkµk
∑m(k)j=1
ϵ(k)j
v(k)j
zv(k)j
αk
,
where Pk =σ2k
σ2k + µkand Pk(z) is the pgf of the variability of exposure in
the kth sector.
Summary
In the single-factor setting, it is assumed that the portfolio is divided into
sectors, each of which is a subset of the set of obligors. In other words,
obligors fall into classes, each of which is driven by one factor. These
random factors (random default intensity) are mutually independent.
81
General sector analysis
It is not necessary that the default rate of an individual obligor depends
on only one of the factors. Recall that
G(z) =n∏
k=1
Gk(z) =n∏
k=1
∫ ∞
0exk[Pk(z)−1]fk(xk) dxk
=∫ ∞
0· · ·
∫ ∞
0e∑n
k=1 xk[Pk(z)−1]n∏
k=1
fk(xk) dx1 · · · dxn.
The probability generating function is integrated over the space of all
possible states represented by (x1 x2 · · · xn) and weighted by their asso-
ciated probability density functions.
82
Consider the exponent term
n∑k=1
xk[Pk(z)− 1] =n∑
k=1
∑A∈Sk
xkµk
ϵAvA
(zvA − 1)
=n∑
k=1
∑A
δAkxkµk
ϵAvA
(zvA − 1), where δAk =
{0 A ∈ k1 A ∈ k
.
Here,∑
A means summation over all obligors within the portfolio.
• The reduction to the single-factor setting is captured via the use of
δAk so that the two different methods of summation∑
A δAk and∑
A∈Sk
are equivalent.
83
Generalization
To allow each obligor to be influenced by more than one factor xk, we
replace the delta function δAk with an allocation of the obligors among
sectors by choosing for each obligor A
θAk :n∑
k=1
θAk = 1.
Here, θAk represents the extent to which the default probability of obligor
A is affected by the random factor xk underlying sector k.
For example, with n = 3, we may choose
θA1 =1
2, θA2 =
1
6and θA3 =
1
3, where θA1 + θA2 + θA3 = 1.
This would mean the relative dependence of the random default rate of
obligor A on the sector risk factors x1, x2 and x3 is in the ratio 3 : 1 : 2.
84
If obligor A is affected by only one sector xk, then we take θAk = δAk. In
general, the exponent term is given by
n∑k=1
xk[Pk(z)− 1] =n∑
k=1
∑A
θAkxkµk
ϵAvA
(zvA − 1)
=∑A
ϵAvA
n∑k=1
θAkxkµk
(zvA − 1).
• Each obligor contributes the following term to the pgf G(z)
exp(xA(zvA − 1)) where xA =
ϵAvA
n∑k=1
θAkxkµk
.
• The variability of the exposures of the obligors for sector k is captured
by
Pk(z) =1
µk
∑A
θAkϵAvA
zvA where µk =∑A
θAkϵAvA
.
• In a similar manner, we deduce that
σk =∑A
θAkσA.
85
2.3 CreditMetrics and Copula functions
Key features
• It is based on credit migration analysis, using the probability of moving
from one credit quality to another within a given horizon as input data.
• It models the full forward distribution of the values of any bond or loan
portfolio, where the changes in values are related to credit migration
only, while interest rates are assumed to evolve in a deterministic
fashion.
• Credit-VaR of a portfolio is derived as the percentile of the distribution
corresponding to the desired confidence level.
86
Reference
M. Crouhy, D. Galai, R. Mark, “A comparative analysis of current risk
models,” Journal of Banking and Finance, vol.24 (2000) p.59-117.
Challenging difficulties
• The portfolio distribution is far from being normal (actually it is highly
skewed and fat-tailed).
• The correlation in credit quality changes for all pairs of obligors are not
directly observable. The correlation coefficient in the joint probability
of asset returns is used as a proxy for the correlation in credit quality
changes.
87
Credit-VaR for a bond
1. Specify a rating system
Rating categories are combined with the probabilities of migrating
from one credit quality to another over the credit risk horizon. Say,
we adopt Moody’s or S&P’s or a proprietary rating system internal to
the bank.
Assumption: all obligors are credit-homogeneous within the same rat-
ing class.
2. Specify the forward discount curve at the risk horizon(s) for each
credit category, and the recovery rate.
Translate the above information into the forward distribution of the changes
in portfolio value consecutive to credit migration.
88
Specification of the transition matrix
• The transition probabilities are based on more than 20 years of history
of firms across all industries.
• Actual transition and default probabilities vary quite substantially over
the years, depending whether the economy is in recession, or in ex-
pansion.
• Many banks prefer to rely on their own statistics which relate more
closely to the composition of their loan and bond portfolios. They may
have to adjust historical values to be consistent with one’s assessment
of current environment.
89
Transition matrix, probabilities of credit rating migrating from one rat-ing quality to another, within one year.
InitialRating
Rating at year-end (%)
AAA AA A BBB BB B CCC Default
AAA 90.81 8.33 0.68 0.06 0.12 0 0 0
AA 0.70 90.65 7.79 0.64 0.06 0.14 0.02 0
A 0.09 2.27 91.05 5.52 0.74 0.26 0.01 0.06
BBB 0.02 0.33 5.95 86.93 5.30 1.17 1.12 0.18
BB 0.03 0.14 0.67 7.73 80.53 8.84 1.00 1.06
B 0 0.11 0.24 0.43 6.48 83.46 4.07 5.20
CCC 0.22 0 0.22 1.30 2.38 1.24 64.86 19.79
90
Specify the one-year forward discount curve
Category Year 1 Year 2 Year 3 Year 4
AAA 3.60 4.17 4.73 5.12
AA 3.65 4.22 4.78 5.17
A 3.72 4.32 4.93 5.32
BBB 4.10 4.67 5.25 5.63∗
BB 5.55 6.02 6.78 7.27
B 6.05 7.02 8.03 8.52
CCC 15.05 15.02 14.03 13.52
One year forward zero curves for each credit rating (%)
⋆ For example, the discount rate applied to the cash flow to be received
at the end of the fifth year from now (4 years from one year forward)
for a BBB-rated bond is 5.63%.
91
Spread of high (low) investment grade bonds increases (decreases) with
maturity.
92
Specify the forward pricing model
One-year forward price of the BBB bond if it remains to be in the BBB
rating class:
VBBB = 6+6
1.041+
6
(1.0467)2+
6
(1.0525)3+
106
(1.0563)4= 107.55
93
Specify the recovery rate
Seniority Class Mean (%) Standard Deviation (%)
Senior Secured 53.80 26.86
Senior Unsecured 51.13 25.45
Senior subordinated 38.52 23.81
Subordinated 32.74 20.18
Junior subordinated 17.09 10.90
Source: Carty & Lieberman [1996]
Recovery rates by seniority class (% of face value, i.e., “par”)
94
One-year forward values for a BBB bond
Year-end rating Value($)AAA 109.37
AA 109.19
A 108.66
BBB 107.55
BB 102.02
B 98.10
CCC 83.64
Default∗ 51.13
⋆ The bond is senior unsecured and the corresponding mean recovery
rate is 51.13%.
95
Derive the forward distribution of the changes in bond value
Year-end Probability Forward Change in
rating of state: price: V($) value: ∆V
p (%) ($)
AAA 0.02 109.37 1.82
AA 0.33 109.19 1.64
A 5.95 108.66 1.11
BBB 86.93 107.55 0
BB 5.30 102.02 -5.53
B 1.17 98.10 -9.45
CCC 0.12 83.64 -23.91
Default 0.18 51.13 -56.42
Source: CreditMetrics, J.P. Morgan
Distribution of the bond values, and changes in value of a BBB bond, in
one year. Note the long downside tail and the limited upside gain.
96
86.93
Frequency
5.955.30
Probabilityof State
(%)
1.17
.33
.18
.12
.02
......
Default CCC B BB BBB A AA AAA
51.13
-56.42
83.64
-23.91
98.10
-9.45
102.2
-5.53
107.55
0
109.37
1.82
...
...
Forward Price: V
Change in value: ∆V
Histogram of the 1-year forward prices and changes in value of a BBB bond.
97
Mean (∆V ) = m =∑i
Pi∆Vi = −0.46
Variance (∆V ) = σ2 =∑i
Pi(∆Vi −mean) = 8.95
First percentile of a normal distribution = m− 2.33σ = −7.43.
• Recall that 98% of observations lie between 2.33σ and −2.33σ from
the mean for normal distributions.
98
Joint migration probabilities (%) with zero correlation for 2 issuers
rated BB and A
Obligor #2 (A)
Obligor #1 AAA AA A BBB BB B CCC Default
(BB) 0.09 2.27 91.05 5.52 0.74 0.26 0.01 0.06
AAA 0.03 0.00 0.00 0.03 0.00 0.00 0.00 0.00 0.00
AA 0.14 0.00 0.00 0.13 0.01 0.00 0.00 0.00 0.00
A 0.67 0.00 0.02 0.61 0.40 0.00 0.00 0.00 0.00
BBB 7.73 0.01 0.28 7.04 0.43 0.06 0.02 0.00 0.00
BB 80.53 0.07 1.83 73.32 4.45 0.60 0.20 0.01 0.05
B 8.84 0.01 0.02 8.05 0.49 0.07 0.02 0.00 0.00
CCC 1.00 0.00 0.02 8.05 0.49 0.07 0.02 0.00 0.00
Default 1.06 0.00 0.02 0.97 0.06 0.01 0.00 0.00 0.00
Assuming zero correlation, each entry is given by the product of the
transition probabilities for each obligor. For example, the joint probability
that Obligor #1 remains to be BB and Obligor #2 remains to be A is
given by
73.32% = 80.53%× 91.05%.
99
Correlation between changes in credit quality
1. Correlation is expected to be higher for firms within the same industry
or in the same region.
2. Correlations vary with the relative state of the economy in the business
cycle. For example, when the economy is performing well, default
correlations go down. However, when the economy is deteriorating,
most firms are dragged down.
3. Default and migration probabilities should not stay stationary over
time.
CreditMetrics have chosen the equity price as a proxy for the asset value
of the firm that is not directly observed.
100
Derive the credit quality thresholds for each credit rating
101
• We slice the distribution of asset returns into bands in such a way that
is we draw randomly from this distribution, we reproduce exactly the
migration frequencies shown in the transition matrix. For example,
P [−1.23 < rBB < 1.37] = 80.53%
P [−∞ < rBB < −2.30] = 1.06%, etc.
• Zccc is the threshold point to trigger default; Zccc is commonly called
the “distance-to-default”.
• CreditMetrics estimates the correlations between the equity returns
of various obligors, then the model infers the correlations between
changes in credit quality directly from the joint distribution of equal-
ity returns. The use of equity returns as a proxy is based on the
assumption that firm’s activities are all equity financed.
• The normalized log-returns on both assets follow a joint normal dis-
tribution:
f(rBB, rA; ρ) =1
2π√1− ρ2
exp
(−r2BB − 2ρrBBrA + r2A
2(1− ρ2)
).
102
Transition probabilities and credit quality thresholds for BB and A ratedobligors
Rated-A obligor Rated-BB obligor
Rating in 1 year Prob(%) Z(σ) Prob(%) Z(σ)
AAA 0.09 3.12 0.03 3.43
AA 2.27 1.98 0.14 2.93
A 91.05 -1.51 0.67 2.39
BBB 5.52 -2.30 7.73 1.37
BB 0.74 -2.72 80.53 -1.23
B 0.26 -3.19 8.84 -2.03
CCC 0.01 -3.24 1.00 -2.30
Default 0.06 1.06
103
Calculation of the joint rating probabilities
P(−1.23 < rBB < 1.37,−1.51 < rA < 1.98)
=∫ 1.37
−1.23
∫ 1.98
−1.51f(rBB, rA; ρ) drBBdrA = 0.7365
Rating of second company (A)
Rating
of first
company AAA AA A BBB BB B CCC Def Total
(BB)
AAA 0.00 0.00 0.03 0.00 0.00 0.00 0.00 0.00 0.03
AA 0.00 0.01 0.13 0.00 0.00 0.00 0.00 0.00 0.14
A 0.00 0.04 0.61 0.01 0.00 0.00 0.00 0.00 0.67
BBB 0.02 0.35 7.10 0.20 0.02 0.01 0.00 0.00 7.73
BB 0.07 1.79 73.65 4.24 0.56 0.18 0.01 0.04 80.53
B 0.00 0.08 7.80 0.79 0.13 0.05 0.00 0.00 8.84
CCC 0.00 0.01 0.85 0.11 0.02 0.01 0.00 0.00 1.00
Def 0.00 0.01 0.90 0.13 0.02 0.01 0.00 0.00 1.06
Total 0.09 2.27 91.05 5.52 0.74 0.26 0.01 0.06 100
Joint rating probabilities (%) for BB and A rated obligors when
correlation between asset returns is 20%.
104
Correlation coefficient of joint defaults
corr(DEF1, DEF2) =P (DEF1, DEF2)− P1 · P2√
P1(1− P1) · P2(1− P2)
P (DEF1, DEF2) = P [r1 ≤ d12, r2 ≤ d22] = N2(d12, d
22; ρ)
Probability of joint defaults as a function of asset return correlation
105
Sample calculation
Take the correlation coefficient of the joint log-returns of the two assets
to be ρ = 20%,
P (DEF1, DEF2) = N2(d12, d
22; ρ) = N2(−3.24,−2.30; 0.20) = 0.000054.
P1 = 0.06% and P2 = 1.06%; so
corr(DEF1, DEF2)
=0.000054− 0.06%× 1.06%√
0.06%× 99.94%× 1.06%× 98.94%= 0.019 = 1.9%.
Note that ρ and corr(DEF1, DEF2) refer to two different types of corre-
lation.
106
Credit diversification
Implement a Monte Carlo simulation to generate the full distribution of
the portfolio values:
1. Derivation of the asset return thresholds for each rating categories.
2. Estimation of the correlation between each pair of obligor’s asset
returns.
3. Generation of asset return scenarios according to their joint normal
distribution (using the Cholesky decomposition). Each scenario is
characterized by n standardized asset returns, one for each obligor.
4. Given the spread curves which apply for each rating, the portfolio is
revalued.
5. Repeat the procedure a large number of times and plot the distribution
of the portfolio values.
107
Remark
Suppose the covariance matrix of the log-returns of the assets is the
symmetric positive definite matrix∑, we perform the Cholesky decompo-
sition:∑
= MMT . Let e be the vector of n independent standard normal
random variables, then e′ = Me gives the vector of n correlated standard
normal random variables with covariance structure as specified by∑.
Write e′ = (e′1 e′2 · · · e′n)T . As an illustration, let the first asset be A-
rated and the second asset be BB-rated. Suppose the sampled outcomes
are e′1 = −2.1, e′2 = 2.60, then the A-rated asset migrates to the rating
class BBB and the BB-rated asset migrates to the rating class A (based
on the thresholds shown in the table on P.100).
108
Calculation of economic capital
Economic capital stands as a cushion to absorb unexpected losses related
to credit events.
V (p) = value of the portfolio in the worst case scenario at the p% confi-
dence level
VF = forward value of the portfolio = V0(1 +RP )
where RP = promised return on the portfolio and V0 = current mark-to-
market value of the portfolio
E[V ] = expected value of the portfolio = V0(1 + E[R])
where E[R] = expected return on the portfolio
EL = expected loss = VF −E[V ]. The expected loss does not contribute
to the capital allocation. The capital charge comes only as a protection
against unexpected losses.
Capital = E[V ]− V (p)
109
Credit-VaR and calculation of economic capital.
110
One may query the following assumptions in CreditMetrics
1. Firms within the same rating class are assumed to have the same
default rate.
2. The actual default rate (migration probabilities) are set equal to the
historical default rate (migration frequencies).
3. Default is only defined in a statistical sense (non-firm specific) without
explicit reference to the process which leads to default.
111
Empirical studies show:
• Historical average default rate and transition probabilities can deviate
significantly from the actual rates.
• Substantial differences in default rates may exist within the same bond
rating class.
Some other credit risk models attempt to make improvements on
• Defaults rates vary with current economic and financial conditions of
the firm.
• Default rates change continuously (ratings are adjusted in a discrete
fashion).
• Microeconomic approach to default: a firm is in default when it cannot
meet its financial obligations.
112
Copula approach
Two aspects of modeling the default times of several obligors.
1. Default dynamics of a single obligor.
2. Models the dependence structure of defaults between the obligors.
• Knowing the joint distribution of random variables allows us to derive
the marginal distributions and the correlation structure among the
random variables but not vice versa.
• A copula function links univariate marginals to their full multivariate
distribution and introduce the dependence structure to generate the
joint probabilities.
113
Proposition
If a one-dimensional continuous random variable X has distribution func-
tion F , that is, F (x) = P [X ≤ x], then the distribution of the random
variable U = F (X) is a uniform distribution on [0,1].
Proof
P [U ≤ u] = P [F (X) ≤ u] = P [X ≤ F−1(u)] =∫ F−1(u)
−∞f(s) ds
where f(x) = F ′(x) is the density function of X.
Let y = F (s), then dy = f(s) ds and
P [U ≤ u] =∫ F (F−1(u))
F (−∞)dy =
∫ u
0dy.
Conversely, if U is a random variable with uniform distribution on [0,1],
then X = F−1(U) has the distribution function F .
114
Remark
To simulate an outcome of X, one may simulate an outcome u from a
uniform distribution, Then let the outcome of X be x = F−1(u).
For example, suppose X is the standard normal random variables with
distribution function
F (x) = N(x) =1√2π
∫ x
−∞e−t2/2dt.
The random variable U = N(X) has a uniform distribution on [0,1]. To
simulate X, we simulate an outcome u from the uniform distribution,
and x = N−1(u) is a simulated outcome of the standard normal random
variable X.
115
Definition of a copula
A n-dimensional copula is a distribution function on [0,1]n with standard
marginal distributions. In other words, there are uniform random variables
U1, U2, . . . , Un taking values in [0,1] such that the copula C is their joint
distribution function. A function C: [0,1]n → [0,1] is a copula if
C(u1, u2, . . . , un) = P [U1 ≤ u1, U2 ≤ u2, . . . , Un ≤ un],
where ui ∈ [0,1], i = 1,2, . . . , n.
It is seen that C has uniform marginal distributions. For all i ≤ n, ui ∈[0,1], we have
C(1, . . . ,1, ui,1, . . . ,1) = ui.
In the analysis of dependency with copula, the joint distribution can be
separated into two parts, namely, the marginal distribution functions of
the random variables (marginals) and the dependence structure between
the random variables which is described by the copula.
116
Construction of the multivariate distribution functions
Given the copula C and the set of univariate marginal distribution function
F1(x1), F2(x2), . . . , Fn(xn), the function which is defined using a copula C:
C(F1(x1), F2(x2), . . . , Fn(xn)) = F (x1, x2, . . . , xn)
results in a multivariate distribution function with univariate marginal
distributions specified as F1(x1), F2(x2), . . . , Fn(xn).
Proof
C(F1(x1), . . . , Fn(xn)) = P [U1 ≤ F1(x1), . . . , Un ≤ Fn(xn)]
= P [F−11 (U1) ≤ x1, . . . , F
−1n (Un) ≤ xn]
= P [X1 ≤ x1, . . . , Xn ≤ xn]
= F (x1, . . . , xn).
The marginal distribution of Xi = F−1i (Ui) is
C(F1(∞), . . . , Fi(xi), . . . , Fn(∞))
= P [X1 < ∞, . . . , Xi ≤ xi, . . . , Xn < ∞]
= P [Xi ≤ xi] = Fi(xi).
117
As a converse of the above result, any multivariate distribution function
F can be written in the form of a copula.
Sklar’s theorem
If F (x1, x2, . . . , xn) is a joint multivariate distribution function with univari-
ate marginal distribution functions F1(x1), . . . , Fn(xn), then there exists a
copula C(u1, u2, . . . , un) such that
F (x1, x2, . . . , xn) = C(F1(x1), F2(x2), . . . , Fn(xn)).
If each Fi is continuous, then C is unique.
Remark
Going through all copula functions gives us all the possible types of de-
pendence structures that are compatible with the given one-dimensional
marginal distributions. The above theorem is just an existence proof.
The actual finding of the copula for a given joint distribution can be very
cumbersome.
118
CreditMetrics and normal copula
• CreditMetrics uses the normal copula in its default correlation formula
even though it does not use the concept of a copula explicitly.
• CreditMetrics calculates the joint default probability of two credits A
and B using the following steps:
(i) Let qA and qB denote the one-year default probabilities for A and
B, respectively. Let ZA and ZB denote the credit index of A and
B, respectively, both are assumed to be standard normal random
variables.
qA = P [ZA ≤ zA] and qB = P [ZB ≤ zB].
(ii) Let ρ denote the asset correlation, the joint default probability for
credits A and B is given by
P [ZA ≤ zA, ZB ≤ zB] =∫ zA
−∞
∫ zB
−∞n2(x, y; ρ) dxdy = N2(zA, zB; ρ). (A)
119
Bivariate normal copula
C(u1, u2) = N2(N−1(u1), N
−1(u2); γ), − 1 ≤ γ ≤ 1.
Suppose we use a bivariate normal copula with a correlation parameter
γ, and denote the default times for credits A and B as TA and TB.
We observe that the one-year default probabilities are given by
qi = P [Ti ≤ 1] = Fi(1) = N(zi) and zi = N−1(qi) for i = A,B.
The one-year joint default probabilities are given by
P [TA ≤ 1, TB ≤ 1] = C(FA(1), FB(1)) = N2(N−1(FA(1)), N
−1(FB(1)); γ) (B)
where FA(1) = P [TA ≤ 1] and FB(1) = P [TB ≤ 1] are the distribution
functions for the default times TA and TB.
Eqs. (A) and (B) are equivalent if we have ρ = γ. Note that this
correlation parameter is not the correlation coefficient between the two
default times TA and TB. The two stochastic state variables, Ti and Zi,
i = A,B, provide different approaches to characterize a default event.
120
Simulation of the default times of a basket of obligors
Let Fi(t) denote the distribution function of the random default time Ti.
Assume that for each credit i in the portfolio, we have constructed a
credit curve so that Fi(t) can be calibrated.
Using a copula C, we obtain the joint distribution of the default times as
follows:
F (t1, t2, . . . , tn) = C(F1(t1), F2(t2), . . . , Fn(tn)).
Suppose we adopt the normal copula, where
F (t1, t2, . . . , tn) = Nn(N−1(F1(t1)), N
−1(F2(t2)), . . . , N−1(Fn(tn))).
The normal credit index variable Yi and Ti are related by Yi = N−1(Fi(Ti)),
i = 1, . . . , n. To simulate the correlated default times, we introduce the
set of standard normal random variables:
Y1 = N−1(F1(T1)), Y2 = N−1(F2(T2)), . . . , Yn = N−1(Fn(Tn)).
121
There is a one-to-one mapping between the random default time T and
the credit index Y .
Simulation scheme
• Simulate the set of normally distributed credit indexes Y1, Y2, . . . , Yn
from an n-dimensional standard normal distribution with correlation
coefficient matrix∑.
• Obtain the simulated sample of the random default times T1, T2, . . . , Tn
using Ti = F−1i (N(Yi)), i = 1,2, . . . , n.
With each simulation run, we generate the default times for all the credits
in the portfolio. A portfolio credit derivative is a financial instrument
whose payoff depends on the number of defaults in the portfolio by the
maturity date of the derivative. With the information on the arrival times
of defaults, we can value any credit derivative payoff structure written on
the portfolio.
122
Generating the set of correlated default times via the copula approach
(here we assume F1 = F2). Note that N(ri) is U(0,1).
Term structure of the cumulative default probability(DP): F(1)=0.0071, F(2)=0.0180, etc.year 1 2 3 4 5 6 7 8 9 10DP 0.0071 0.0180 0.0320 0.0484 0.0666 0.0859 0.1060 0.1264 0.1469 0.1672
123
Exponential model for dependent defaults
Reference
Kay Giesecke, “A simple exponential model for dependent defaults,” Jour-
nal of Fixed Income, vol.13 (3), p.74-83 (2003).
Poisson process Let λ be the mean rate of arrivals (intensity) of a Pois-
son process N(t). The probability mass function of the random number
of arrivals within the time interval [0, t] is given by
P [N(t) = n] =e−λt(λt)n
n!, n = 0,1,2, . . . .
• A firm’s default is driven by idiosyncratic as well as other region-
al, sectoral or economy-wide shocks, whose arrivals are modeled by
independent Poisson processes. Here, the Poisson approximation is
adopted again for nice analytic tractability.
• Default times are assumed to be jointly exponentially distributed. In
this case, the exponential copula arises naturally.
124
Bivariate version of the exponential models
Suppose there are Poisson processes N1, N2 and N with respective inten-
sities λ1, λ2 and λ. Here, λi is the idiosyncratic shock intensity of firm
i and λ is the intensity of a macro-economic shock affecting both firms
simultaneously. Define the default time τi of firm i by
τi = inf{t ≥ 0 : Ni(t) +N(t) > 0}, i = 1,2.
That is, a default occurs unexpectedly if either an idiosyncratic or a sys-
tematic shock strikes the firm for the first time. Assuming independence
of the two shocks, firm i defaults with intensity λi+λ so that its survival
function is
Si(t) = P [τi > t] = P [Ni(t) +N(t) = 0] = e−(λi+λ)t.
The joint survival probability is found to be
S(t1, t2) = P [τ1 > t1, τ2 > t2]
= P [N1(t1) = 0, N2(t2) = 0, N(t1 ∨ t2 = 0)]
= e−λ1t1−λ2t2−λ(t1∨t2) = e−(λ1+λ)t1−(λ2+λ)t2+λ(t1∧t2)
= S1(t1)S2(t2)min(eλt1, eλt2).
125
Survival copula
There exists a unique solution Cτ : [0,1]2 → [0,1], called the survival
copula of the default time vector (τ1, τ2) such that the joint distribution
of the survival probabilities can be represented by
S(t1, t2) = Cτ(S1(t1), S2(t2)).
The copula Cτ describes the complete non-linear default time dependence
structure.
Define ui = Si(ti) ∈ [0,1], and θi =λ
λi + λ, i = 1,2, we obtain
Cτ(u1, u2) = S(S−11 (u1), S
−12 (u2)) = min(u2u
1−θ11 , u1u
1−θ22 ).
The parameter vector θ = (θ1, θ2) controls the degree of dependence
between the default times. Note that θi is the ratio of the intensity of
the economy wide shock to the total intensity of all shocks that affect
firm i.
126
1. Firms default independently of each other (λ = 0 or λ1, λ2 → ∞)
θ1 = θ2 = 0, Cτθ (u1, u2) = u1u2 (product copula)
2. Firms’ defaults are perfectly correlated (firms default simultaneously,
λ → ∞ or λ1 = λ2 = 0)
θ1 = θ2 = 1 and Cτθ (u1, u2) = u1 ∧ u2.
It can be shown that
u1u2 ≤ Cτθ (u1, u2) ≤ u1 ∧ u2, (θ1, θ2) ∈ [0,1]2, u1 ∈ [0,1], u2 ∈ [0,1].
Under this exponential model, the default events can only be positively
correlated. The positive default correlation is driven by the common
macro-economic stock, which always enhances the chance of default of
both firms.
127
Default copula
Similarly, we define the default copula by
Kτ(P1(t1), P2(t2)) = P [τ1 ≤ t1, τ2 ≤ t2] = P (t1, t2),
where Pi(t) = P [τi ≤ t] = 1− Si(t), i = 1,2.
Since the survival function and default functions are related by
S(t1, t2) = 1− P1(t1)− P2(t2) + P (t1, t2).
Setting ui = Pi(ti) = 1 − Si(ti), i = 1,2, the survival copula and default
copula are related by
Kτ(u1, u2) = Cτ(1− u1,1− u2) + u1 + u2 − 1
= min((1− u2)(1− u1)1−θ1, (1− u1)(1− u2)
1−θ2)
+ u1 + u2 − 1.
128
Multivariate extension
Assume that there are n ≥ 2 firms. The default of an individual firm
is driven by its individual idiosyncratic shock as well as other sectoral,
industry, country-specific or economy-wide shocks.
Define m =n∑
k=1
nCk = 2n − 1. Also, define a matrix (aij)n×m, where
aij = 1 if shock j ∈ {1,2, · · · ,m} modeled through the Poisson process Nj
with intensity λj, leads to a default of firm i ∈ {1,2, · · · , n} and aij = 0
otherwise. Taking n = 3 and m = 7, we consider the following 3 × 7
matrix:
(aij) =
1 0 0 1 1 0 10 1 0 1 0 1 10 0 1 0 1 1 1
.
129
• The fourth column indicates that the fourth shock affects Firms 1
and 2 only.
• The seventh shock is the economy wide shock.
• Suppose the economy-wide shock events are excluded, one then set
ai7 = 0 for i = 1,2,3. This reduces to bivariate dependence only,
where one shock can affect at most 2 firms.
• Firm 1 defaults by time t if
N1(t) +N4(t) +N5(t) +N7(t) =7∑
k=1
a1kNk(t) > 0.
130
Joint survival function
Under this exponential model, the random default time of the ith firm is
given by
τi = inf
t ≥ 0 :m∑
k=1
aikNk(t) > 0
.
Assuming independence of the shocks, this indicates that firm i defaults
with intensitym∑
k=1
aikλk and the survival function of the ith firm is given
by
Si(t) = P [τi > t] = P [m∑
k=1
aikNk(t) = 0] = exp
− m∑k=1
aikλkt
.
The joint survival function is given by
S(t1, . . . , tn) = P [τ1 > t1, . . . , τn > tn].
These n firms all survive up to the maximum value of survival times
max(t1, . . . , tn), provided that all these m shocks do not affect any firm.
131
Example (assuming the arrivals of shocks to be independent events)
Taking n = 3, the joint survival function is
S(t1, t2, t3) = exp(−λ1t1 − λ2t2 − λ3t3
− λ4max(t1, t2)− λ5max(t1, t3)
− λ6max(t2, t3)− λ7max(t1, t2, t3)).
• The first term e−λ1t1 gives the probability that the first shock has not
arrived up to time t1.
• The fourth term e−λ4max(t1,t2) gives the probability that the fourth
shock has not arrived up to max(t1, t2), so both the first and second
firms have not been affected by the fourth shock up to max(t1, t2).
• The seventh term e−λ7max(t1,t2,t3) gives the probability that the sev-
enth shock has not arrived up to max(t1, t2, t3), so all firms have not
been affected by the seventh shock up to max(t1, t2, t3).
132
General case
Consider the kth shock, k = 1,2, . . . ,m, it influences firm i only when
aik = 1. The condition that the kth shock does not affect any firm up to
max(t1, t2, . . . , tn) is given by
Nk(a1kt1) = 0, . . . , Nk(anktn) = 0
or equivalent to
Nk(max(a1kt1, . . . , anktn)) = 0.
Since the m shocks are independent, the joint survival function is given
by
S(t1, . . . , tn) = P [N1(max(a11t1, . . . , an1tn)) = 0, . . . ,
Nm(max(a1mt1, . . . , anmtn)) = 0]
=m∏
k=1
exp(−λkmax(a1kt1, . . . , anktn))
= exp
− m∑k=1
λkmax(a1kt1, . . . , anktn)
.
133
Survival copula function
The exponential survival copula associated with the survival function
S(t1, . . . , tn) can be found via Cτ(u1, . . . , un) = S(S−11 (u1), . . . , S
−1n (un)).
Fixing some i, j ∈ {1,2, . . . , n} with i = j, the two-dimensional marginal
copula is given by
Cτ(ui, uj) = Cτ(1, . . . ,1, ui,1, . . . ,1, uj,1, . . . ,1)
= S(0, . . . ,0, ti,0, . . . ,0, tj,0, . . . ,0)
= exp
− m∑k=1
λkmax(aikti, ajktj)
.
The survival function for firm i and firm j are, respectively, given by
ui = Si(ti) = exp(−(m∑
k=1
aikλk)ti),
uj = Sj(tj) = exp(−(m∑
k=1
ajkλk)tj).
134
In terms of Si(ti) and Sj(tj), the marginal copula can be expressed as
Cτ(ui, uj) = exp(−m∑
k=1
aikλkti)exp(−m∑
k=1
ajkλktj)
min(exp(m∑
k=1
aikajkti), exp(m∑
k=1
aikajktj))
= Si(ti)Sj(tj)min(exp(m∑
k=1
aikajkti), exp(m∑
k=1
aikajktj)).
We define, analogously to the bivariate case, the dependence parameters
in the bivariate marginal copula by
θi =
∑mk=1 aikajkλk∑mk=1 aikλk
, θj =
∑mk=1 aikajkλk∑mk=1 ajkλk
.
Note that the kth shock affects both firm i and firm j, provided that
aikajk = 1. Here, θi is the ratio of the sum of intensities due to shocks
that affect both firms i and j to the sum of intensities due to shocks that
affect firm i only. We then obtain
Cτ(ui, uj) = min(uju1−θii , uiu
1−θjj ).
135
Simulation of correlated default arrival times
The following three-step algorithm generates the default arrival times
with an exponential dependence structure Cτ while allowing for arbitrary
marginal default time distributions.
1. Simulate an m-vector (t1, . . . , tm) of independent exponential shock
arrival times with given parameter vector (λ1, . . . , λm), where λk > 0.
This is done by drawing, for k ∈ {1,2, . . . ,m}, an independent standard
uniform random variate Uk and setting
tk = −1
λklnUk.
Indeed, P [tk > T ] = P [− lnUk/λk > T ] = P [Uk ≤ e−λkT ] = e−λkT .
2. Simulate an n-vector (T1, . . . , Tn) of joint exponential default times
by considering, for each firm i ∈ {1,2, . . . , n}, the minimum of the
relevant shock arrival times:
Ti = min{tk : 1 ≤ k ≤ m,aik = 1}.
Only those shocks that correspond to aik = 1 lead to default of firm
i.136
3. Generate a sample (v1, . . . , vn) from the (survival) default time copula
Cτ by setting, for i ∈ {1,2, . . . , n},
vi = Si(Ti) = exp
−Ti
m∑k=1
aikλk
.
Let the marginal distribution of the random default time Zi of firm i be
denoted by qi(Zi). In order to generate an n-vector Z = (Z1, . . . , Zn) of
correlated default arrival times with given marginal survival function qi,
we set
Zi = q−1i (vi), i = 1,2, . . . , n;
provided that the inverse q−1i exists.
The above algorithm then generates correlated default times Zi with the
given marginal distribution qi and exponential dependence structure. The
joint survival copula is given by
P [Z1 > t1, . . . , Zn > tn] = Cτ(q1(t1), . . . , qn(tn)).
137
Mathematical Appendices
Understanding “conditional independence”
Example Hemophilia is a hereditary disease. If a mother has
it, then with probability 12, any of her sons independently will inherit it.
Let H,H1 and H2 denote the events that the mother, the first son, and
the second son are hemophilic, respectively. Note that H1 and H2 are
conditionally independent given H. However H1 and H2 are not indepen-
dent. This is because if we know that one son is hemophilic, the mother is
hemophilic. With probability 1/2, the other son is also hemophilic. Note
that
P [H1|H] = P [H1|H ∩H2] =1
2
P [H2|H] = P [H2|H ∩H1] =1
2;
P [H1 ∩H2|H] =1
2×
1
2=
1
4.
138
Conditional Variance
Just as we have defined the conditional expectation of X given the value
of Y , we can also define the conditional variance of X, given that Y = y,
as follows:
var(X|Y ) = E[(X − E[X|Y ])2|Y ].
That is, var(X|Y ) is equal to the (conditional) expected square of the
difference between X and its (conditional) mean when the value of Y
is given. In other words, var(X|Y ) is exactly analogous to the usual
definition of variance, but now all expectations are conditional on the
fact that Y is known.
Formula
var(X) = E[var(X|Y )] + var(E[X|Y ]).
139
To obtain the unconditional variance of X, we add the expectation of
conditional variance of X given Y to the variance of expectation of X
given Y .
Proof
We start from
var(X|Y ) = E[X2|Y ]− (E[X|Y ])2
so
E[var(X|Y )] = E[E[X2|Y ]]− E[E[X|Y ])2]
= E[X2]− E[(E[X|Y ])2]. (i)
Treating E[X|Y ] as a random variable, and observing E[E[X|Y ]] = E[X],
var(E[X|Y ]) = E[(E[X|Y ])2]− (E[X])2. (ii)
By adding eqs. (i) and (ii), we arrive at the formula.
140
Negative binomial distribution
• The number X of Bernuolli trials required to obtain r successes on
the set {r, r+1, r+2, · · · }. Out of the (k−1) earlier trials, there have
been r− 1 successes, where k = r, r+1, r+2, · · · . The distribution is
given by
Pr[X = k] =
(k − 1r − 1
)pr(1− p)k−r, k = r, r +1, r +2, · · · .
• The number Y = X − r of failures before the rth success, supported
on the set {0,1,2, · · · }. The modified negative binomial distribution
is given by
Pr[Y = k] =
(k + r − 1
r − 1
)pr(1− p)k
=
(k + r − 1
k
)pr(1− p)k, k = 0,1,2, · · · .
141