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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
ORF 245: Common Distributions – J.Fan 78
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.
ORF 245: Common Distributions – J.Fan 79
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
Binomial with n= 25 p= 0.5
10 15 20 25 30 35 40
0.00
0.04
0.08
x
y
Binomial with n= 50 p= 0.5
0 2 4 6 8
0.00
0.10
0.20
0.30
x
y
Binomial with n= 8 p= 0.3
0 5 10 15 20 25
0.00
0.05
0.10
0.15
x
y
Binomial with n= 25 p= 0.3
5 10 15 20 25 30
0.00
0.04
0.08
x
y
Binomial with n= 60 p= 0.3
0 2 4 6 8
0.0
0.1
0.2
0.3
0.4
x
y
Binomial with n= 8 p= 0.1
0 5 10 15 20 25
0.00
0.05
0.10
0.15
0.20
0.25
x
y
Binomial with n= 25 p= 0.1
0 5 10 15 20
0.00
0.04
0.08
0.12
x
y
Binomial with n= 100 p= 0.1
Figure 4.1: Bionomial distributions with different parameters.
ORF 245: Common Distributions – J.Fan 80
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.
ORF 245: Common Distributions – J.Fan 81
�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.
ORF 245: Common Distributions – J.Fan 82
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 .
ORF 245: Common Distributions – J.Fan 83
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
ORF 245: Common Distributions – J.Fan 84
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.
ORF 245: Common Distributions – J.Fan 85
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
ORF 245: Common Distributions – J.Fan 86
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)
Poission dist with lambda = 7
0 5 10 15 20 25 30
0.00
0.04
0.08
(d)
dpoi
s(x,
15)
Poission dist with lambda = 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, · · · .
ORF 245: Common Distributions – J.Fan 87
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.
ORF 245: Common Distributions – J.Fan 88
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
ORF 245: Common Distributions – J.Fan 89
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
ORF 245: Common Distributions – J.Fan 90
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
)
ORF 245: Common Distributions – J.Fan 91
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%
ORF 245: Common Distributions – J.Fan 92
(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).
ORF 245: Common Distributions – J.Fan 93
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,
ORF 245: Common Distributions – J.Fan 94
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
ORF 245: Common Distributions – J.Fan 95
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-
ORF 245: Common Distributions – J.Fan 96
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
ORF 245: Common Distributions – J.Fan 97
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.
ORF 245: Common Distributions – J.Fan 98
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)
ORF 245: Common Distributions – J.Fan 99
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
ORF 245: Common Distributions – J.Fan 100
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
ORF 245: Common Distributions – J.Fan 101
−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
Theoretical Quantiles
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
Theoretical Quantiles
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.
ORF 245: Common Distributions – J.Fan 102
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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$y #extract y component
ORF 245: Common Distributions – J.Fan 103
abline(lsfit(x,y), col="red", lwd=2) #draw a regression line through the data
title("(b) rSP500 against t(3.5)") #add title
res=qqplot(rSP500b, rSP500a, col="blue", main="") #compare dist of two data sets and save results
x=res$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
ORF 245: Common Distributions – J.Fan 104
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.