Measuring financial risk

Sebastiano Manzan

ECO 4051 | Spring 2017

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Why measure and manage risk?

I The risk-return tradeoff is at the core of any business and financial decision

I Modern companies and financial institutions are large enterprises with many(thousands?) of individuals making decisions in an uncertain environment

I How can we measure, aggregate, and manage (= control) the overall riskexposure of an organization?

I What type of risks are we trying to measure and control?

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Types of financial risk

I Market risk: losses due to movements in the level and/or volatility of assetprices

I directional: exposure to movements of the marketI non-directional: exposure to changes of volatility

I Credit risk: losses arising from defaults on bonds and loans; changes tocredit ratings or market perception of default (sub-prime, mortgage-backedsecurities etc)

I Liquidity risk:I Asset liquidity risk: arises when transactions cannot be executed at

prevailing market prices due to size of the position or thinness of themarket

I Funding liquidity risk: inability to meet margin calls on leveragedportfolios (LTCM)

I Operational Risk: due to human or technical mistakes, employees’ fraud,inadequacy of risk management procedures,inadequacy of pricing models(model risk)

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Cases of financial disasters

I In the last 20 years there have been many cases in which failure to measureand manage risk lead to huge losses or even bank failures

I Leeson/Barings: loss of $1.3 billion and that wiped out the capital ofBarings; shareholders lost everything, bondholders almost everything

I Orange County: loss of $1.6 billion

I LTCM: loss of $4 billion

I JP Morgan & London Whale: $6.2 billion

I 2008 Financial Crisis: $4 trillion?

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Leeson/Barings (1994-1995)

I Large long bet on the Japanese equity market through NIKKEI 225 indexfutures (notional $7 billion) and sold options betting on a stable market

I At the beginning of 1995 the Japanese equity market lost more than 15%and Leeson’s position started bleeding

I The total loss was of $1.3 billion that wiped out the equity capital of thebank which was sold to ING for £1

I Leeson was able to accumulate such a large directional bet for lack ofinternal controls

I Operational and market risks played a role in Barings disaster

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Derivatives and Risk Management

I Many of the financial disasters are due to the use of derivatives and leverage

I Derivatives are used to hedge but also speculate (directional bets) whileleverage is used to boost profits from limited capital

I The problem is not so much the use of derivatives and/or leverage, ratherthe uncontrolled risk-taking allowed by these positions

I Good risk management practice means measuring and managing these risk(not eliminating it!)

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Risk management policies and regulation

I The increasingly frequent financial disasters has called the attention of theprofession and regulators on the need for a more stringent risk managementpractice

I Professional initiatives:I Derivatives Policy Group

I management controlsI enhanced reportingI evaluation of risk versus capitalI measure market risk via Value at Risk (VaR) at 99% level and

2-week horizonI RiskMetrics from J.P. Morgan

I established a framework for VaR calculationI Regulators:

I Basel Accord on Banking Supervision (1988 and later amendments)

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Basel Accord I and II and III and . . .

I The Accords set rules on bank capital adequacy as a way tomonitor/control the amount of risk-taking

I Banks are allowed to use their internal models to determine the capitalcharge

I They should demonstrate to have an appropriate risk management system,conduct regular stress testing, having independent risk-control unit andexternal audits

I Each bank is subject to the Market Risk Charge (MRC) which iscalculated as:

MRC = k * max [yesterday’s VaR, 60-days average VaR]

where VaR is defined as maximum loss that we expect at a certainconfidence level and target horizon

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I The calculation of MRC is based on the following parameters:I Horizon: 10 trading days (2 weeks); allowed to scale up the 1-day VaR

using√10

I Confidence level: 99%I Estimation: at least 1 year of data and update of estimates at least

once a quarterI Correlation: can be accounted among and across broad categories

(bonds, equities, commodities, etc.)I Multiplicative Factor k: at least 3I Plus Factor: k can be increased by regulators to values higher than 3

if backtesting indicates inadequacy of the internal risk model

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What is Value-at-Risk (VaR)?

I Denote the profit/loss of a financial institution in day t + 1 byRt+1 = 100 ∗ ln(Wt+1/Wt), where Wt+1 is the portfolio value in day t + 1

I We define Value-at-Risk (VaR) as the maximum loss that it is expect at acertain confidence level and target horizon; using a 99% confidence level and1-day horizon VaR for day t + 1, VaRt+1, is defined as

P(Rt+1 > VaR0.99t+1 ) = 0.99

I VaR can also be expressed as the minimum loss that we expect at a certainconfidence level and target horizon, that is,

P(Rt+1 ≤ VaR0.99t+1 ) = 0.01

In statistical terms, VaR is the 0.01 quantile of the probability distribution ofRt+1

I For a portfolio value of Wt the $VaR is calculated as$VaR0.99

t+1 = Wt ∗ (exp(|VaR0.99t+1 |/100)− 1)

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I VaR at 100 ∗ (1− α)% represents the 100 ∗ α% quantile of the profit/lossdistribution

Probability Density Function (PDF)

α

VaR1−α µ

Cumulative Distribution Function (CDF)

VaR1−α µ

0α

1

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VaR assuming normality

I Assume that Rt+1 ∼ N(µ, σ2) with constant mean and variance

I Based on this assumption the P(Rt+1 ≤ VaR1−α) = α, with 1− α = 0.99 is

VaR0.99t+1 = µ− 2.33 ∗ σ

and, more generally,VaR1−α

t+1 = µ+ zα ∗ σ

where zα represents the quantile from the standard normal distribution atlevel α

I Example: we hold a portfolio that replicates the S&P 500 Index and weobtain historical data for the Index from Jan 03, 1990, to May 08, 2017. Rimplementation:

mu = mean(sp500daily)sigma = sd(sp500daily)var = mu + qnorm(0.01) * sigma ## qnorm(0.01) = -2.33

[1] -2.569

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Time-varying VaR

I The assumption that Rt+1 ∼ N(µ, σ2) restricts the mean and the varianceof the portfolio return to be constant over time

I This is an unrealistic assumption given the evidence that daily returns showsignificant variation of volatility over time

I We can thus assume that Rt+1 ∼ N(µt+1, σ2t+1) where:

I Conditional mean µt+1: an AR(1) process, i.e., µt+1 = φ0 + φ1RtI Conditional variance σt+1: MA, EMA, GARCH(1,1), or

GJR-GARCH(1,1)I Value-at-Risk is then calculated as:VaR0.99

t+1 = µt+1 − 2.33σt+1

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I Model: Rt+1 = µ+ σt+1εt+1

I σt+1 estimated by EMA(0.06)I εt+1 ∼ N(0, 1)

I For 1.074% of the 6891 days the return was smaller relative to VaR

library(TTR)mu <- mean(sp500daily)sigmaEMA <- EMA((sp500daily-mu)^2, ratio=0.06)^0.5var <- mu + qnorm(0.01) * sigmaEMA

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Expected Shortfall (ES)

I VaR represents the maximum (minimum) loss that is expected with99% (1%) probability

I However, it does not provide a measure of how large the losses are likely tobe in case an extreme event happens

I An alternative risk measure is Expected Shortfall (ES) that is defined as

ES1−αt+1 = E(Rt+1|Rt+1 ≤ VaR1−α

t+1 )

and represents the expected return conditional on the return being smallerthan 100*(1− α)% VaR

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I If we assume that returns are normally distributed, ES is given by

ES1−αt+1 = −σt+1

φ[VaR1−α

t+1 /σt+1]

α= −σt+1

φ [zα]α

where φ(·) represents the PDF of the standard normal distribution

I If we are using 1− α = 0.99 then we have that (for µ = 0):I VaR0.99

t+1 = −2.33σt+1

I Φ−1(VaR0.99t+1 /σt+1) = φ(−2.33) = 0.026

I ES0.99t+1 = −σt+1

φ(−2.33)0.01 = − 2.643σt+1

I Given the same level α, ES provides a more conservative risk measurerelative to VaR

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I Density and Distribution functions for the standard normal distributionx ∼ N(0, 1)

I Φ(−2.33) =I Φ(2.33) =I Φ−1(0.01) =I Φ−1(0.99) =

Probability Density Function

α φ(x)

φ(VaR1−α/σ)

VaR1−α/σ 0

Cumulative Density Function (CDF)

zα = Φ(−1)(α) x

0

α

Φ(x)

1

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sigma = 1alpha = seq(0.001, 0.05, by=0.001)ES = - dnorm(qnorm(alpha)) / alpha * sigmaVaR = qnorm(alpha) * sigma

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alpha

ES

,VaR

Measure

ES

VaR

Expected Shortfall and Value at Risk

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√K rule

I So far we discussed how to calculate VaR at the 1-day horizon but regulatorsask financial institutions to report their potential losses over a 10 dayhorizon

I Basel rules allow banks to scale-up the one-day risk measures by√10

I Where does this√10 rule come from?

I The return of holding the portfolio in the following K days is

Rt+1:t+k =K∑

k=1

Rt+k = Rt+1 + · · ·+ Rt+K

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I Assuming that the daily returns are independent and identically distributed(i.i.d.) with mean µ and variance σ2, then we have that the expected valueof the multi-period return is

E

(K∑

k=1

Rt+k

)=

K∑k=1

µ = Kµ

and its variance is

Var

(K∑

k=1

Rt+k

)=

K∑k=1

σ2 = Kσ2

I So that the standard deviation of the Rt+1:t+k is equal to√Kσ

I Expected return for unit of risk:E(Rt+1:t+K )/

√Var(Rt+1:t+K ) = Kµ√

Kσ=√K µσ

I VaR at 1% of the multi-period return Rt+1:t+K is given by

VaR1−αt+1:t+K = Kµ− 2.33

√Kσ

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I The√10 rule relies on the assumption that daily returns are independent

over time

I What happens if this assumption fails to hold?

I Assume that K = 2 so that Rt+1:t+2 = Rt+1 + Rt+2

I the variance of the multiperiod return Var(Rt+1 + Rt+2) is equal to

Var(Rt+1) + Var(Rt+2) + 2Cov(Rt+1,Rt+2)

= σ2 + σ2 + 2ρσ2 = 2σ2(1 + ρ)

I If ρ > 0 the variance (and thus VaR) of the multiperiod return is largerrelative to the case with no correlation since 2σ2(1 + ρ) > 2σ2

I Bottom line: assuming independence we under-estimate the risk in ourportfolio if returns are actually positively correlated

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VaR assuming non-normality

I Fat tails is the characteristic of daily financial returns to have events 3 (orlarger) standard deviations away from the mean more often relative to thenormal distribution

I Is time-varying volatility a possible explanation for the non-normality in thedata?

I This could be the case if returns are switching between two regimes of highand low volatility

I The sample skewness of the standardized residuals is -0.431 and the sampleexcess kurtosis is 2.261

I In this case, there is still some skewness and, more generally, there mightstill be some non-normality in the standardized residuals

I We can depart from the assumption of normality in two ways:I Cornish-Fisher approximationI t distribution

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Cornish-Fisher approximation

I The Cornish-Fisher (CF) approximation consists of a Taylor expansion ofthe normal density which has the effect of making the approximate density afunction of skewness and kurtosis

I VaR is calculated by VaR1−αt+1 = zCF

α σt+1 where the value zCFα is given by:

zCFα = zα +

SK6

((zα)2 − 1

)+

EK24

((zα)3 − 3zα

)+

SK2

36

(2(zα)5 − 5zα

)where SK and EK represent the skewness and excess kurtosis, respectively, and zαrepresents the α quantile

I If SK = EK = 0 then zCFα = zα = −2.33 (normal)

I For α = 0.01 we have that zα = −2.33 and zCFα is then equal to:

zCFα = −2.33 + 0.7 ∗ SK − 0.23 ∗ EK − 3.46 ∗ SK2

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I Effect of changing skewness and kurtosis on zCFα

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Skewness

z α

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z α

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A distribution with fat tailsI An alternative approach is to discard the normal distribution in favor of a

distribution with fat tails such as the t distribution with ddegrees-of-freedom

I If we assume that εt+1 ∼ td in Rt+1 = σt+1εt+1 we can estimate thevolatility parameters and d by ML

q1 <- ggplot(data.frame(x = c(-8, 8)), aes(x = x)) +stat_function(fun = dnorm, aes(colour = "N(0,1)")) +stat_function(fun = dt, args = list(df=4), aes(colour = "t(4)")) +stat_function(fun = dt, args = list(df=10), aes(colour = "t(10)")) +theme_bw() + labs(x = "Probability", y= NULL) + coord_cartesian(x=c(-3,3)) +scale_colour_manual("Distribution", values = c("seagreen3", "steelblue3", "tomato3"))

q2 <- q1 + coord_cartesian(x=c(-8, -3), y=c(0,0.03))

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Probability

Distribution

N(0,1)

t(10)

t(4)

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Probability

Distribution

N(0,1)

t(10)

t(4)

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I A simple way to obtain a rough estimate of the degrees of freedom isd = 6/EK + 4, where EK is the excess kurtosis of the returns

I This rule is based on the fact that the td distribution (for d > 4) has excesskurtosis equal to 6/(d − 4) (and it is infinite for 2 < d < 4)

I The standardized returns have sample excess kurtosis of 2.26 so that theestimate of d is equal to 6.65 which can be rounded to 7

I This value confirms the evidence that the returns standardized by thevolatility forecast still deviate from normality, in particular on the left tail

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Historical Simulation (HS)

I The risk models that we discussed are based on two building blocks:I mean and volatility models (constant or AR(p); EMA, GARCH(1,1),

or GJR-GARCH)I error distribution (standard normal, t)

I Historical Simulation (HS) is an alternative approach that calculates VaRat α level as the 1− α quantile of the most recent M days

I HS does not make any assumptions about the mean/volatility model and/orthe error distribution

I In this sense, it can be considered a non-parametric approach because of thelack of parametric assumptions

I A typical value for M is 250 days (one trading year)

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M1 = 250M2 = 1000alpha = 0.01hs1 <- rollapply(sp500daily, M1, quantile, probs=alpha, align="right")hs2 <- rollapply(sp500daily, M2, quantile, probs=alpha, align="right")

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Variable

HS1000

HS250

Historical Simulation

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I Zoom on the financial crisis period 2008-2009

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HS1000

HS250

Historical Simulation

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I Comparison of VaR based on EMA (with normal errors) and HS during thefinancial crisis

I EMA is clearly faster in adapting to changes in volatility

I Percentage of violations from January 03, 1990 to May 08, 2017:I HS: 1.3061%I EMA: 1.0013%

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EMA

HS250

Historical Simulation

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I Pros of HS:I No model required for the volatility dynamics and error distribution

I Easy and fast to calculate, no estimation required

I Cons of HS:I Choice of the estimation window M

I Transforming the 1-day VaR to 10-day VaR (practice: multiply by√10)

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Monte Carlo (MC) Simulation

I In the 1990s EMA and HS were the most widely used methods to calculate VaR

I Both for parametric and non-parametric methods we need to useapproximations to derive the distribution of the cumulative return

I This task is much simplified by considering simulation-based methods

I Simulation methods are based on the following idea: simulate data from theestimated risk model (volatility model + error distribution) to determine thedistribution of the cumulative return at horizon K

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I One application of MC methods is to simulate the future distribution ofreturns from 1 to K periods ahead

I We consider two dates:I January 22, 2007: period of low volatilityI September 29, 2008: period of high volatility

January 22, 2007

September 29, 2008

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I We are at the closing of day T and we want to forecast the distribution ofthe portfolio profit/loss from 1 to K periods ahead. A MC simulation iscomposed of the following steps:

1. Estimate the volatility model Rt+1 = σt+1εt+1 (e.g., GARCH(1,1) +εt+1 ∼ N(0, 1)) and produce a forecast σT+1

2. Generate S random draws for the error denoted εs,T+1 (for s = 1, · · · ,S)and S random returns for T + 1 as Rs,T+1 = σT+1εs,T+1

3. Produce a forecast σs,T+2 as σ2s,T+2 = ω + αR2

s,T+1 + βσ2T+1

4. Generate other S random draws for the error denoted εs,T+2 and S randomreturns for T + 2 as Rs,T+2 = σs,T+2εs,T+2

5. Produce a forecast σs,T+3 as σ2s,T+3 = ω + αR2

s,T+2 + βσ2s,T+2

6. Repeat 4-5 until the forecast horizon K

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Step 1: estimate GARCH model and produce the forecast for T+1

require(rugarch)spec = ugarchspec(variance.model=list(model="sGARCH",garchOrder=c(1,1)),

mean.model=list(armaOrder=c(0,0), include.mean=FALSE))fitgarch = ugarchfit(spec = spec, data = window(sp500daily, end=lowvol))

omega alpha1 beta10.005 0.054 0.942

2007-01-22T+1 0.483

fitgarch1 = ugarchfit(spec = spec, data = window(sp500daily, end=highvol))

omega alpha1 beta10.006 0.059 0.937

2008-09-29T+1 3.085

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Step 2-6 Iterate the simulation step until K days ahead

set.seed(9874)

S = 10000 # number of MC simulationsK = 250 # forecast horizon

# create the future dateslowvol <- as.Date("2007-01-22")futdates <- lowvol + 1:K

# create the matrices to store the simulated return and volatilityR = xts(matrix(sigma*rnorm(S), K, S, byrow=TRUE), order.by=futdates)Sigma = xts(matrix(sigma, K, S), order.by=futdates)

# iteration to calculate R and Sigma based on the previous dayfor (i in 2:K){

Sigma[i,] = (gcoef[’omega’] + gcoef[’alpha1’] * R[i-1,]^2 + gcoef[’beta1’] * Sigma[i-1,]^2)^0.5R[i,] = rnorm(S) * Sigma[i,]

}

I The assumption that the error distribution is N(0, 1) can be easily changedby replacing rnorm(S) with, e.g., rt(S, df=6) to simulate from t6

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I At each horizon k plot the quantiles (across S) at 0.10, · · ·, 0.90I For the simulated volatility we also plot the average (across S)

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MC cumulative returns

I The cumulative return is defined as Rt+1:t+k =∑k

j=1 RT+jI MC VaR is represented by the 1− α quantile of the simulated returns at each

horizon k

Rcum <- cumsum(R)R1cum <- cumsum(R1)Rcumq <- xts(t(apply(Rcum, 1, quantile,

probs=c(0.01,0.05,seq(0.10,0.90,0.10),0.95,0.99))), order.by=futdates)R1cumq <- xts(t(apply(R1cum, 1, quantile,

probs=c(0.01,0.05,seq(0.10,0.90,0.10),0.95,0.99))), order.by=futdates1)

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VaR and ES

I The simulated return values can also be used to calculate ES by averaging(at each k) the returns values that are smaller or equal to VaR

VaR = xts(apply(Rcum, 1, quantile, probs=0.01), order.by=futdates)VaRmat = matrix(VaR, K, S, byrow=FALSE)Rviol = Rcum * (Rcum < VaRmat)ES = xts(rowSums(Rviol) / rowSums(Rviol!=0), order.by=futdates)

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VaR ES

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VaR ES

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Filtered Historical Simulation (FHS)

I A drawback of HS is that it adapts very slowly to changes in asset volatility

I An alternative that has been proposed is Filtered Historical Simulation(FHS) which combines features of parametric modeling, MC, and HS

I FHS assumes:

1. A time-series volatility model for returns, e.g., Rt+1 = σt+1εt+1,where σt+1 is assumed to follow a GARCH model

2. No error distribution is assumed; instead, the simulated errors εs,t+kare random draws from the standardized returns Rt+1/σt+1

3. 1 and 2 are used in a MC simulation as shown above

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I The only difference between MC and FHS consists of the distributionassumed to draw the simulated errors: a parametric distribution in the caseof MC (standard normal or td) or the empirical distribution of thestandardized returns (Rt+1/σt+1)

I This implies that the resampled εs,t+k maintain the features of the data,such as skewness and kurtosis

I In R this is implemented by drawing the εs,t+k by sample(std.resid, S,replace=TRUE) instead of rnorm(S) (or rt(S, df))

I In the next plots we compare the expected one-day volatility and the 99%VaR based on the k-period cumulative return conditional on January 22,2007 and September 29, 2008

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# standardized residualsstd.resid = as.numeric(residuals(fitgarch, standardize=TRUE))std.resid1 = as.numeric(residuals(fitgarch1, standardize=TRUE))

for (i in 2:K){

Sigma.fhs[i,] = (gcoef[’omega’] + gcoef[’alpha1’] * R.fhs[i-1,]^2 +gcoef[’beta1’] * Sigma.fhs[i-1,]^2)^0.5

R.fhs[i,] = sample(std.resid, S, replace=TRUE) * Sigma.fhs[i,]Sigma.fhs1[i,] = (gcoef1[’omega’] + gcoef1[’alpha1’] * R.fhs1[i-1,]^2 +

gcoef1[’beta1’] * Sigma.fhs1[i-1,]^2)^0.5R.fhs1[i,] = sample(std.resid1, S, replace=TRUE) * Sigma.fhs1[i,]

}

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Portfolio VaR

I In practice, the portfolio of large financial institutions is composed of severalassets and its return can be express as

Rp,t =J∑

j=1

wj,tRj,t

where:I Rp,t represents the portfolio return in day tI wj,t is the weight of asset j in day t (and there is a total of J assets)I Rj,t is the return of asset j in day t

I Let’s assume that the bank holds only 2 assets (i.e., J = 2)

I The expected portfolio return is given by: E(Rp,t) = µp,t = w1,tµ1 + w2,tµ2

I The portfolio variance is given by:Var(Rp,t) = σ2

p,t = w21,tσ

21 + w2

2,tσ22 + 2w1,tw2,tρ12σ1σ2

which is a function of the individual variances and the correlation betweenthe two assets, ρ12

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I The portfolio Value-at-Risk is then given by

VaR1−αp,t = µp,t + zασp,t

= w1,tµ1 + w2,tµ2 + zα√

w21,tσ

21 + w2

2,tσ22 + 2w1,tw2,tρ12σ1σ2

I If we assume that µ1 = µ2 = 0, the VaR1−αp,t formula can be expressed as

follows:

VaR1−αp,t = +zα

√w2

1,tσ21 + w2

2,tσ22 + 2w1,tw2,tρ12σ1σ2

= −√

z2αw2

1,tσ21 + z2

αw22,tσ

22 + 2 ∗ z2

αw1,tw2,tρ12σ1σ2

= −√

(VaR1−α1,t )2 + (VaR1−α

2,t )2 + 2 ∗ ρ12VaR1−α1,t VaR1−α

2,t

and:

1. if ρ12 = 1: VaR1−α1,t + VaR1−α

2,t2. if ρ12 = −1: −|VaR1−α

1,t −VaR1−α2,t |

3. if −1 < ρ12 < 1:VaR1−α

1,t + VaR1−α2,t < VaR1−α

p,t < −|VaR1−α1,t −VaR1−α

2,t |

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I If we assume that the mean and the variance vary over time then VaR1−αp,t is:

VaR1−αp,t = w1,tµ1,t+w2,tµ2,t+zα

√w2

1,tσ21,t + w2

2,tσ22,t + 2w1,tw2,tρ12,tσ1,tσ2,t

where:I µ1,t = β0 + β1R1,t−1 and µ2,t = β0 + β1R2,t−1I σ2

1,t , σ22,t MA, EMA, and a GARCH

I ρ12,t ??

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Modeling correlations

I A simple approach to modeling correlations consists of using MA and EMAsmoothing as in the case of modeling volatility

I Denote the returns of asset 1 by R1,t and of asset 2 by R2,t

I The MA(M) estimate of the covariance of the two assets is:

σ12,t+1 =1M

M∑m=1

R1,t−m+1R2,t−m+1

I The correlation is then given by: ρ12,t+1 = σ12,t+1/ (σ1,t+1 ∗ σ2,t+1)

I If the portfolio is composed of J assets there are J ∗ (J − 1)/2 correlations toestimate

I An alternative approach is to use EMA smoothing which can beimplemented using the recursive formula discussed earlier:

σ12,t+1 = (1− λ)σ12,t + λR1,tR2,t

and the correlation can be obtained as earlier by dividing with the standarddeviation forecasts.

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Application in R

I We hold a portfolio that is invested for a fraction w1 in a gold ETF (ticker:GLD) and the remaining fraction 1− w1 in the S&P 500 ETF (ticker: SPY)

I The data are from January 02, 1970, January 02, 1970 until January 03,1970, January 02, 1970 at the daily frequency

I I will forecast volatility and correlation using the EMA approach withλ = 0.94

data <- getSymbols(c("GLD", "SPY"), from="2005-01-01")R <- 100 * merge(ClCl(GLD), ClCl(SPY))names(R) <- c("GLD","SPY")# EMA for the product of returnsprod <- R[,1] * R[,2]cov <- EMA(prod, ratio=0.06)# EMA for the squared returnssigma <- do.call(merge, lapply(R^2, FUN = function(x) EMA(x, ratio=0.06)))^0.5names(sigma) <- names(R)# correlation given by covariance divided by product of the st. dev.corr <- cov / (sigma[,1] * sigma[,2])

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autoplot(corr) + geom_hline(yintercept=0, color="tomato3", linetype="dashed") +geom_hline(yintercept = as.numeric(cor(R[,1], R[,2], use="pairwise.complete"), color="seagreen4", linetype="dashed")) +theme_bw() + labs(x=NULL, y=NULL)

−0.5

0.0

0.5

2005 2010 2015

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I VaR for a portfolio investing w1 in GLD and 1− w1 in SPY

w1 = 0.5 # weight of asset 1w2 = 1 - w1 # weight of asset 2VaR = -2.33 * ( (w1*sigma[,1])^2 + (w2*sigma[,2])^2 +

2*w1*w2*corr*sigma[,1]*sigma[,2] )^0.5autoplot(VaR) + geom_hline(yintercept=0, color="tomato3", linetype="dashed") +

theme_bw() + labs(x=NULL, y=NULL)

−8

−6

−4

−2

0

2005 2010 2015

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I A comparison of portfolio VaR with VaR for a portfolio that is fully investedin GLD or SPY

VaRGLD = -2.33 * sigma[,1]VaRSPY = -2.33 * sigma[,2]mydata <- merge(VaR, VaRSPY, VaRGLD)autoplot(mydata, facets=FALSE) + theme_bw() +

geom_hline(yintercept=0, color="tomato2", linetype="dashed")

−10

−5

0

2005 2010 2015

plot_group

GLD

GLD.1

SPY

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Backtesting VaR

I What makes a risk model a good model?

I We need to identify some characteristics which should hold if our model isthe true one and test them on the data

I Backtesting means testing the goodness of our risk model in terms of threeproperties of the violations process, Vt+1, defined as

V 1−αt+1 = I (Rt+1 ≤ VaR1−α

t+1 )

where I (A) takes value 1 if the statement A is true, and 0 otherwiseI These properties are:

I Unconditional Coverage: we expect that 100 ∗ (1− α)% VaR isviolated 100*α% of the days

I Independence: violations should be independent over timeI Conditional Coverage: independent of what happened the previous

day, we should always have a 1% chance of having a violation

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I For the S&P 500 example above with volatility predicted using EMA andα = 0.01 we have:

T1 = sum(V)TT = length(V)alphahat = T1 / TTround(alphahat,3)

[1] 0.011

I The fraction of days with violations is 1.079% which is different from theexpected value of α = 1%

I Significantly different? We want to evaluate this hypothesis statistically bytesting the null that H0 : α̂ = α

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Unconditional Coverage (UC) test

I The violation r.v. V 1−αt+1 (for t = 0, · · · ,T − 1) is a binomial random

variable with probability α, that is,

P(V 1−αt+1 = 1) = α

andP(V 1−α

t+1 = 0) = 1− α

I Example: let’s say that we have only three days and in the second day therewas a violation, but not in the other days

I V1 = 0,V2 = 1,V3 = 0 and the probability of this event is

P(V1 = 0,V2 = 1,V3 = 0) = (1− α)1 ∗ α0︸ ︷︷ ︸P(V1=0)

∗ (1− α)0 ∗ α1︸ ︷︷ ︸P(V2=1)

(1− α)1 ∗ α0︸ ︷︷ ︸P(V3=0)

so thatP(V1 = 0,V2 = 1,V3 = 0) = (1− α)2α1

I More generally, if we have a sample of T days with T1 days with a violation(V 1−α

t+1 = 1) and T0 without a violation (V 1−αt+1 = 0), the joint probability of

the sequence of violation is given by

L(α,T1,T) = αT1 (1− α)T0

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I The UC test is performed by comparing the likelihood above calculated atthe theoretical probability 0.01 with the likelihood calculated at theestimate of α, α̂ = T1/T

I The statistic and distribution of the test for Unconditional Coverage(UC) are

UC = −2 ln(L(α,T1,T)L(α̂,T1,T)

)∼ χ2

1

where α is the level at which VaR was calculated, α̂ = T1/T and χ21 denotes

the chi-square distribution with 1 degree-of-freedom.

I The critical values at 1, 5, and 10% are 6.63, 3.84, and 2.71, respectively

I The null hypothesis H0 : α̂ = α is rejected if UC is larger than the criticalvalue

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I In practice, the test statistic can be calculated as follows:

−2[T1 ln

(α

α̂

)+ T0 ln

(1− α1− α̂

)]In the example discussed above we have α̂ = 0.01, T1 = 74, and T is 6860.The test statistic is thus 0.42

I Since 0.42 is smaller than 3.84 we do not reject at 5% significance level thenull hypothesis H0 : α̂ = α and conclude that the risk model providesappropriate unconditional coverage

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Independence

I Another property that we would like the risk model to satisfy is thatviolations happen at random times

I This property does not hold when violations happen in consecutive days

I To test this hypothesis we calculate two quantities:

1. α1,1: the probability of a violation in day t given that a violationoccurred in day t − 1 (and α1,0 = 1− α1,1 probability of no violationin t given the violation the previous day)

2. α0,1: the probability of having a violation in day t given that noviolation occurred the previous day (and α0,0 = 1− α0,1 )

I They can be estimated from the data by calculating T1,1 and T0,1 thatrepresent the number of days in which a violation was preceded by aviolation and a no violation, respectively.

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I In R we can determine these quantities as follows:

T11 = sum((lag(V)==1) & (V==1))T01 = sum((lag(V)==0) & (V==1))

I The results are:I T0,1 = 74 and T0,0 = 6711I T1,1 = 0 and T1,0 = 74

I The conditional probabilities are then given by:I α̂0,1 = T0,1/(T0,1 + T0,0) = 0.011 and α0,0 = 1− α0,1I α̂1,1 = T1,1/(T1,0 + T1,1) = 0 and α1,0 = 1− α1,1

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I The likelihood of observing T0,1 and T1,1 sequences out of T is given by

L(α̂0,1, α̂1,1,T0,1,T1,1,T) = α̂T1,01,0 (1− α̂1,0)T1,1 α̂

T0,10,1 (1− α̂0,1)T0,0

I The IND test statistic is calculated as

IND = −2 ln(

L(α̂,T0,T)L(α̂0,1, α̂1,1,T0,1,T1,1,T)

)∼ χ2

1

where:I the numerator is the same as the denominator in the UC test and

represents the likelihood under independenceI The logarithm of the denominator is calculated as:

T1,0lnα1,0 + T1,1ln(1− α1,0) + T0,1 lnα0,1 + T0,0 ln(1− α0,1)

I The test statistic and distribution for the hypothesis of Independence(IND) in this case is 1.61 which is smaller relative to the critical value at5% and we conclude that we do not reject the null hypothesisH0 : α0,1 = α1,1 = α̂. The violations are independent.

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Conditional coverage

I In this case the hypothesis that we are interested in testing isH0 : α0,1 = α1,1 = α

I CC tests jointly independence of the violations and that coverage is alwaysequal to 1%

I The test statistic is calculated as CC = LRIND + LRUC ∼ χ22

I The critical value at 5% is 5.99 and the test statistic is 2.03 so that we donot reject the null hypothesis of correct conditional coverage.

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Summary on testing risk models

1. Unconditional coverage:I H0 : α = 0.01I UC = −2 ln

(L(0.01,T1,T)L(α̂,T1,T)

)I χ2

1

2. Independence:I H0 : α0,1 = α1,1 = α̂

I IND = −2 ln(

L(α̂,T0,T)L(α̂0,1,α̂1,1,T0,1,T1,1,T)

)I χ2

1

3. Conditional coverage:I H0 : α0,1 = α1,1 = 0.01I CC = UC + INDI χ2

2

I The End!

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