Chapter 4 Commonly Used Distributions - Princeton Universityjqfan/fan/classes/245/chap4.pdfChapter 4...

Chapter 4

Commonly Used Distributions

4.1 Binomial distribution §3.4

Bernoulli trial: the simplest kind of random trial (like coin tossing)

♠ Each trial has two outcomes: S or F.

♠ For each trial, P (S) = p .

♠ The trials are independent.

76

ORF 245: Common Distributions – J.Fan 77

Associated r.v.: let Yi be the outcome of ith trial. Its takes only

two values, {0, 1}, with pmf P (Yi = 1) = p and P (Yi = 0) = 1− p,

called Bernoulli dist. Then, E(Yi) = p and var(Yi) = pq.

Generic name: ♣Quality control: Defective/Good; ♠Clinical trial: Survival/death; disease/non-

disease; cure/not cure; effective/ineffective; ♣sampling survey: male/female; old/young; Coke/Pepsi;

♠online ads: click/not click ....

Example 4.1 Which of following are Bernoulli trials?

♠ Roll a die 100 times and observe whether getting even or odd spots.

♠ Observe daily temperature and see whether it is above or below


freezing temperature.

♠ Sample 1000 people and observe whether they belong to upper,

middle or low SES class.

Definition: Let X be the number of successes in a Bernoulli process

with n trials and probability of success p. Then, the dist of X called

a Binomial distribution, denoted by X ∼ Bin(n, p).

Fact: Let X ∼ Bin(n, p), then

• (pmf) b(x;n, p) = P (X = x) =(nx

)pxqn−x.

• (summary) EX = np and var(X) = npq.

• (sum of indep. r.v.) X = Y1 + · · · + Yn, FYi = a Bernoulli r.v.


0 2 4 6 8

0.00

0.05

0.10

0.15

0.20

0.25

x

y

Binomial with n= 8 p= 0.5

0 5 10 15 20 25

0.00

0.05

0.10

0.15

x

y


10 15 20 25 30 35 40

0.00

0.04

0.08

x

y


0 2 4 6 8

0.00

0.10

0.20

0.30

x

y


0 5 10 15 20 25

0.00

0.05

0.10

0.15

x

y


5 10 15 20 25 30

0.00

0.04

0.08

x

y


0 2 4 6 8

0.0

0.1

0.2

0.3

0.4

x

y


0 5 10 15 20 25

0.00

0.05

0.10

0.15

0.20

0.25

x

y


0 5 10 15 20

0.00

0.04

0.08

0.12

x

y


Figure 4.1: Bionomial distributions with different parameters.


Example 4.2 Genetics

According to Mendelian’s theory, a cross fertilization of related

species of red and white flowered plants produces offsprings of which

25% are red flowered plants. Crossing 5 pairs, what is the chance of

♣ no red flowered plant?

Let X = # of red flowered plants. Then, X ∼ Bin(5, 0.25).

P (X = 0) = q5 = 0.755 = 0.237.

♣ no more than 3 red flowered plants?

P (X ≤ 3) = 1− P (X ≥ 4) =

= 0.984.


�Use dbinom, pbinom, qbinom, rbinom for density (pmf), probability (cdf), quantile, and random numbers.

> dbinom(0,5,0.25) #pmf P(X = 0) or density for discrete r.v.

[1] 0.2373047

> pbinom(3,5,0.25) #cdf P(X <= 3)

[1] 0.984375

> dbinom(3,5,0.25) #pmf P(X = 3)

[1] 0.08789063

4.2 Hypergeometric and Negative Binomial Distributions §3.5

In a sampling survey, one typically draws w/o replacement.

1

M defect N-M Good

Total N

X defect n-X Good

Total n

n draws w/oreplacement

Figure 4.2: Illustration of sampling without replacement.

Hypergeometric dist:

(Use hyper in R: e.g.


dhyper, phyper, qhyper, rhyper)

P (X = x) =

(Mx

)(N−Mn−x

)(Nn

) ,

for max{0, n−N + M} ≤ x ≤ min(n,M).

Remarks

♠When sampling with replacement, the distribution is binomial.

♠When sampling a small fraction from a large population, like a poll,

the trials are nearly Bernoulli.

E.g. N = 40000 and n = 1000, # of families with income ≥$50K is 40%.

P (S2|S1) =15999

39999≈ P (S2). P (S1000|S1 · · ·S999) =

15001

39001≈ P (S1000).

Fact: EX = np, var(X) =N − nN − 1︸︷︷︸corr factor

npq, where p = MN .


Example 4.3 Capture-recapture to estimate population size

Suppose that there are N animals in a region. Capture 5 of them,

tag them and then release them back. Now among newly 10 captured

animals, X of them are tagged.

Probabilistic question: If N = 25, how likely is it that X = 2?

The distribution of X is “hyper” with N = 25, M = 5, n = 10.

P (X = 2) =

(52

)(208

)(2510

) = .385.

> dhyper(2,5, 20, 10) #pmf P(X = 2), N-M = 20

[1] 0.3853755

> phyper(2,5, 20, 10) #cdf P(X <= 2)

[1] 0.6988142

Statistical question: What is the population size if we observe 3


tagged animal? Thus, the realization of X , denoted by x, is 3. Since

EX = n 5N and an observed x is a reasonable estimate of EX , then

x = n5

Nor N̂ =

5n

x= 16.667 or 17.

Definition: In a Bernoulli trial w/ P (S) = p, let Y be the number of draws required to get r

successes and X = Y − r (so that it begins from 0). The distribution of X is called a negative

Binomial (Use nbinom in R, namely dnbinom, pnbinom, qnbinom, rnbinom):

P (X = x) =

(x + r − 1

r − 1

)prqx, x = 0, 1, 2, · · ·

Y︷︸︸︷Y1︷︸︸︷

FFS

Y2︷︸︸︷FFFFFFFS

Y3︷︸︸︷FFFS · · ·

Yr︷︸︸︷FFFFFS

When r = 1, Y follows a geometric distribution:

P (Y = y) = pqy−1, y = 1, 2, · · · .

Recall that EYi = 1/p, var(Yi) = q/p2. Then,

F Y = Y1 + · · · + Yr and X = Y − r.


F EY = r/p, var(Y ) = rq/p2.

F EX = EY − r = rq/p, var(X) = rq/p2.

4.3 Poisson Distribution §3.6

�Useful for modeling number of rare events (# of traffic accidents, #

of houses damaged by fire; # of customer arrivals to a service center).

When is the Poisson distribution (Use pois in R) reasonable?

• independence: # of events occurring in any time interval is

independent of # of events in any other non-overlaping interval;

• rare: almost impossible for two or more events simultaneously;

• constant rate: The average # of occurrences per time unit is


constant, denoted by λ.

0 5 10 15 20 25 30

0.0

0.2

0.4

0.6

(a)

dpoi

s(x,

0.5

)

Poission dist with lambda = 0.5

0 5 10 15 20 25 30

0.0

0.1

0.2

0.3

(b)

dpoi

s(x,

1)

Poission dist with lambda = 1

0 5 10 15 20 25 30

0.00

0.05

0.10

0.15

(c)

dpoi

s(x,

7)


0 5 10 15 20 25 30

0.00

0.04

0.08

(d)

dpoi

s(x,

15)


Figure 4.3: The Poisson Distribution (line diagram) for different values of λ

Let X = # of occurrences in a unit time interval. It can be shown

that the distribution Poisson(λ) is given by:

P (X = x) =e−λλx

x!, x = 0, 1, 2, · · · .


x= 0:30

plot(x, dpois(x,0.5), xlab="(a)", type="n") #lambda = 0.5

for(i in 0:30) lines(c(i,i), c(0,dpois(i, 0.5)),lwd=1.5, col="blue")

plot(x, dpois(x,15), xlab="(d)", type="h", col="blue", lwd=1.5)

title("Poission dist with lambda = 0.5")

Poisson approximation to the Binomial: When n is large and p is small, and n and 1/p are

of the same order of magnitude (e.g., n = 100, p = 0.01; n = 1000, p = 0.003), then:(n

x

)pxqn−x ≈ e−λλx

x!, with λ = np.

Example 4.4 Suppose that a manufacturing process produces 1% of defective products. Com-

pute P( # of defectives ≥ 2 in 200 products).

Note that Bin(200, 0.01) ≈ Poisson(2). Thus,

P (X ≥ 2) = 1− P (X = 0)− P (X = 1)

= 1− e−220

0!− e−22

1!= 0.594

Direct calculation shows that the probability is 0.595.


Expected value and variance: EX = λ and var(X) = λ.

(Think of the Poisson approximation to the Binomial).

4.4 The normal distribution §4.3

Definition: A cont rv X has a normal dist (”dnorm, pnorm, qnorm, rnorm ” for

density, probability (cdf), quantile, and random number generator) if its pdf is given by

f (x;µ, σ) =1√2πσ

exp

(−(x− µ)2

2σ2

), −∞ < x <∞.

denoted by X ∼ N(µ, σ2).

Facts: It can been shown that EX = µ and var(X) = σ2.

Standard normal distribution refers to the specific case in


Figure 4.4: Normal - bell-shaped - distributions with different means (center) and variances (spreadness).

which µ = 0 and σ = 1. Let

φ(z) =1√2πe−z

2/2 and Φ(z) =

∫ z

−∞φ(u)du

be the density and the cumulative distribution functions of the

standard normal distribution (z-curve).

Standardization: Suppose X ∼ N(µ, σ2). Then, Z = (X − µ)/σ

transforms X into standard units. It says how many SDs x is


Figure 4.5: The density and the cdf.

above or below the mean µ. Indeed, Z ∼ N(0, 1).

Computation of the area under the normal curve:

Putting endpoints into SUs and compute the area under z-curve.

P (a ≤ X ≤ b) = P

(a− µσ≤ Z ≤ b− µ

σ

)= Φ

(b− µσ

)− Φ

(a− µσ

)


Example 4.5 Finding areas under the normal curve.

(a) If X ∼ N(3, 22), compute P (1 ≤ X ≤ 6).

P (1 ≤ X ≤ 6) = =

= Φ

(6− 3

2

)− Φ

(1− 3

2

)= Φ(1.5)− Φ(−1)

= .9332− .1587 = 77.45%

(b) If X ∼ N(−5, 62), find P (X ≥ −2).

P (X ≥ −2) = =

= Φ(∞)− Φ

(−2 + 5

6

)= 1− Φ(0.5)

= 1− .6915 = 30.85%


(c) If X ∼ N(0, σ2), find P (−σ ≤ X ≤ σ).

P (−σ ≤ X ≤ σ) = = P (−1 ≤ Z ≤ 1)

= .8413− .1587 = 68.26%

(d) If X ∼ N(0, σ2), find P (−2σ ≤ X ≤ 2σ).

P (−2σ ≤ X ≤ 2σ) = =

= .9772− .0228 = 95.44%.

x

y

-3 -2 -1 0 1 2 3

0.0

0.1

0.2

0.3

0.4

Figure 4.6: Probabilities under the z-curve (standard normal curve).


Example 4.6 Suppose that the arrival time of a passenger to a

train station is normally distributed with mean 11:55am and SD

10 minutes. If the train is scheduled to leave at 12:03pm, what

is the chance that the passenger will catch the train?

Let X be the arrival time relative to 12:00pm. Then, X ∼

N(−5, 102). The probability is

P (X < 3) = = 78.81%.

Quantile of normal distribution: Let zα be the value such that

P (Z ≥ zα) = α, the upper α percentile. It can be found through

qnorm. For example,


Table 4.1: Upper quantile of normal distribution

α 0.0005 0.001 0.005 0.01 0.025 0.05 0.10

zα 3.291 3.090 2.576 2.326 1.96 1.645 1.282

4.5 Normal approximation to a data histogram

Normal approx: �Use prob histogram to approx data histogram:

Approximate a data histogram with average µ and SD σ by a normal

curveN(µ, σ2). valid when the data histogram follows a bell curve.

Sufficient summary: From n data values we can summarize into

two numbers: x̄ (mean) and s (SD). From these two numbers, we

can infer all useful information about the original data. Thus, for


bell-shaped data histograms, x̄ and s are sufficient statistics.

Example 4.7 Normal approximation

The blood pressures of the users aged 25 to 34 in a drug study averaged

out to 121mm with an SD of 12.5mm. The data histogram follows a

normal curve.

1. Approximately, what percentage of users have blood pressure be-


tween 96mm and 133.5mm?

P (96 ≤ X ≤ 133.5) = Φ

(133.5− 121

12.5

)− Φ

(96− 121

12.5

)=

= .8413− .0228 = 81.85%

2. If the top 10% of users need to be singled out for further investi-

gation, what is the cut-off point?

(a) Convert the percentile into the standard units (SU):

upper 10th percentile = 90th percentile = z.10 = 1.28 SU

(b) Convert the standard unit into the original unit:

1.28 SU ←→ = 137mm


Example 4.8 Normal approximation and percentile

The IQ in a particular population (as measured by a standard test) is known to be approximately

normally distributed with mean 100 and SD 15.

1. What percentage of the population has an IQ at least 125?

Let X ∼ N(100, 152). The percentage is given by:

P (X ≥ 125) = = 4.78%

2. What is the 95th percentile of the population?

95th percentile = 1.96 SU ←→ = 129.4

4.6 The exponential and distributions §4.4

Definition: A continuous rv X has an exponential distribution

(use exp in R) if its pdf is given by

f (x;λ) = λ exp(−λx), x ≥ 0.


Facts:

♠ F (x;λ) = P (X ≤ x) = 1− exp(−x/λ), x ≥ 0.

♠ EX = λ and var(X) = λ2.

♠memoryless property:

P (X > a + x|X > a) =P (X > a + x)

P (X > a)=

exp(−λ(a + x))

exp(−λa)= exp(−λx) = P (X > x).

Gamma distribution: Model lifetime without memoryless prop-

erties, income distributions, etc. (Use gamma in R)

f (x;α, λ) =1

λαΓ(α)xα−1 exp(−x/λ), for x ≥ 0

♠α > 0 —shape parameter, ♠λ > 0 —scale parameter.

♠α = 1 =⇒ exponential dist, i.e. Gamma(α, λ) = Exp(λ)


Figure 4.7: Left panel: Gamma densities with λ = 1 and α = 0.6, 1, 3. Right panel: Gamma densities with α = 3, λ = 0.5, 1, 2.

4.7 Probability (Q-Q) plots §4.6

Purpose: To check whether the data x1, · · · , xn follow a dist F .

Q-Q plot: Let x(1) ≤ x(2) ≤ · · · ≤ x(n) be sorted data, called the


order statistics. Then x(i) is roughly the (i − 0.5)/n (empirical)

percentile. The Q-Q plot shows

{(theoretical quantiles, empirical quantiles)}

=

{(F−1

(i− 0.5

n

), x(i)

)}Check if the plot looks reasonably straightline.

�Normal probability plot: display {(Φ−1((i− 0.5)/n), x(i))}.

� Extending to compare distributions of two data sets.

x = seq(-4, 4, 0.02) #generate plots

plot(x,dunif(x,-1,1),col="red",type="l",lwd=2) #uniform density

lines(x,dnorm(x), col="blue", lwd=2) #normal density

lines(x,dgamma(x+3,3,1), col="black",lwd=2) #shifted Gamma density

title("Uinf(-1,1), N(0,1), Gamma(3,1)-3")

x <- runif(400,-1,1) #uniform random numbers

qqnorm(x, col="blue", main="") #checking normality, taking default title away

qqline(x, lwd=2, col="red") #add a line connecting Q_1 and Q3


−4 −2 0 2 4

0.0

0.1

0.2

0.3

0.4

0.5

x

duni

f(x,

−1,

1)

Uinf(−1,1), N(0,1), Gamma(3,1)−3

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−3 −2 −1 0 1 2 3

−1.

0−

0.5

0.0

0.5

1.0

Theoretical Quantiles

Sam

ple

Qua

ntile

s

(a) unif(0,1): lighter

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−3 −2 −1 0 1 2 3

−3

−2

−1

01

2


Sam

ple

Qua

ntile

s

(b) N(0,1): about right

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−3 −2 −1 0 1 2 3

−2

02

46

8


Sam

ple

Qua

ntile

s

(c) gamma(3,1): heavier

Figure 4.8: Normal probability plot for 400 data points generated from: (a) uniform(-1, 1) distribution (light tail); (b) N(0,1)distribution (correct); (c) Gamma(3, 1)− 3 (heavier tail).

title("(a) unif(0,1): lighter")

x <- rnorm(400) #Normal random numbers

qqnorm(x, col="blue", main=""); qqline(x, lwd=2, col="red")

title("(b) N(0,1): about right")

x <- rgamma(400, 3, 1) -3 #Shifted normal numbers

qqnorm(x, col="blue", main=""); qqline(x, lwd=2, col="red")

title("(c) gamma(3,1): heavier")

Example 4.9 Tails of Financial Returns.

We now consider the financial returns of SP500 and IBM, as well as SP500 before and during the 2008 financial

crisis.


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●●●●●●●●●●●●●●●●●●●●

●●●●●●●●●●●●●

●●●●●●

● ●

●

●

−10 −5 0 5 10

−10

−5

05

1015

rSP500

rIB

M

(d) Returns of IBM against SP500

Figure 4.9: Comparisons of distributions of financial returns.

pdf("FinReturns.pdf", width=10, height=3, pointsize=8) #print to pdf file

par(mfrow = c(1,4), mar=c(4,4,4,1)+0.1, cex=0.8)

qqnorm(rSP500, col="blue", main="") #Checking normality

qqline(rSP500, lwd=2, col="red") #Add a line

title("(a) SP500: heavier") #Add a title

x = (1:length(rSP500)-0.5)/length(rSP500) #compute percents

qt4 = qt(x,3.5) #compute theoretical quantiles of t(3.5)

res=qqplot(qt4, rSP500, col="blue", main="") #draw the QQplot and save the results

x=res$x #extract x component of the results

y=res$y #extract y component


abline(lsfit(x,y), col="red", lwd=2) #draw a regression line through the data

title("(b) rSP500 against t(3.5)") #add title

res=qqplot(rSP500b, rSP500a, col="blue", main="") #compare dist of two data sets and save results

x=res$x

y=res$y

abline(lsfit(x,y), col="red", lwd=2) # draw a regression line

title("(c) SP500 before & after criss")

res=qqplot(rSP500, rIBM, col="blue", main="")

x=res$x

y=res$y

abline(lsfit(x,y), col="red", lwd=2)

title("(d) Returns of IBM against SP500")

Definition: The t-distribution (Use t in R) with degree of freedom

ν has the density

fν(x) = d−1ν

(1 +

x2

ν

)−(ν+1)/2


where dν = B(0.5, 0.5ν)√ν. It has mean 0 and variance ν/(ν − 2).

−5 0 5

0.0

0.1

0.2

0.3

0.4

dens

ity

t−distributions with different dfs

Figure 4.10: Densities of tν with ν = 1, 3, 5, 10 and ∞.

�A 5%-event under normal distri-

bution occurs in 14000 years (=

1/pnorm(−5)/252)and under t10-

distribution in 15 years and under t4.5

is 1.5 years (= 1/pt(−5, 4.5)/252).

Note: pnorm(−5) = 2.8665× 10−7. There are 252 trading days.

Chapter 4 Commonly Used Distributions - Princeton Universityjqfan/fan/classes/245/chap4.pdfChapter 4...

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