Chapter 5

Special Discrete and ContinuousDistributions

5.1 The Bernoulli and Binomial distributions

Definition 5.1 An experiment is called a Bernoulli trial if the outcome can be classified aseither a “success” or “failure”.

In the Bernoulli trial, let p = P (success), and

X =

{1 if the outcome is a success0 if the outcome is a failure

Thenx 1 0

fX(x) p q

(1)

where q = 1− p.

• If X is the number of successes that occur in n independent Bernoulli trials, then X iscalled a binomial random variable with parameters (n, p), denoted by X ∼ B(n, p).X is also said to have binomial distribution with parameters n and p.

• Suppose X ∼ B(n, p). Then, the probability density function of X is

b(x; n, p) =

(n

x

)px(1− p)n−x, x = 0, 1, ..., n

• If X ∼ B(n, p), then

EX = np, Var(X) = npq, where q = 1− p

5-1

M(t) = EetX

=n∑

x=0

etx

(n

x

)pxqn−x

=n∑

x=0

(n

x

)(pet)xqn−x

= (pet + q)n

M ′(t) = n(pet + q)n−1pet

M ′′(t) = n(n− 1)(pet + q)n−2(pet)2 + n(pet + q)n−1pet

M ′(0) = np

M ′′(0) = n(n− 1)p2 + np = n2p2 + np(1− p) = (np)2 + npq

ThereforeE(X) = np, Var(X) = npq

Example 5.1 A man claims to have extrasensory perception (ESP). As a test, a fair coin isflipped 109 times, and he is asked to predict the outcome in advance. The man gets 7 out of 10correct. What is the probability that he would have done at least this well if he had no ESP?

Solution:Let X be the number of correct predictions. Then X ∼ B(10, .5).

P (X ≥ 7) =

(10

7

).510 +

(10

8

).510 +

(10

9

).510 +

(10

10

).510

≈ .1719

P (X ≥ 7) = 1− P (X ≤ 6) = 1− .8281 = .1719

Example 5.2 A multiple-choice test consists of 20 questions each with 4 possible answers ofwhich only 1 is correct. Suppose that a student knows the correct answers for 70% of questions,and the student randomly chooses an answer if he doesn’t know the correct answer.

(i) what is the probability that the student will get at least 16 correct answers?

5-2

(ii) Suppose that each correct answer is awarded 5 marks and each incorrect answer carries apenalty of 1 marks, what is the expected overall mark of the student?

Solution:Let X be the number of correct answers. Then X ∼ B(20, p), where p = 0.7+0.3× .25 = 0.775

P (X ≥ 16) = 1− P (X ≤ 15) = 1− .4798 = .5202Let Y be the overall mark. Then Y = 5×X−(20−X) = 6X−20 and E(Y ) = 6E(X)−20 =

6× .775− 20 = 73

5.2 Hypergeometric distribution

Example 5.3 An urn contains 40 white and 60 black balls. 10 balls are selected with re-placement. Let X be the white balls selected. What is the distribution of X

Solution: X ∼ B(10, .4)What if the balls were selected without replacement?

Definition 5.2 Suppose that n balls are “randomly drawn” (without replacement) from an urncontaining N balls, of which k are white and N − k are black. Let X be number of white ballsselected . Then

h(x; N,n, k) = P (X = x) =

(kx

)(N−kn−x

)(

Nn

) , x = 0, 1, ..., n

and X is called a hypergeometric r.v. with parameters (n,N, k), denoted by X ∼ H(n,N, k).

• If the n balls are chosen randomly with replacement, then X has binomial distributionB(n, k/N)

• The simplest way to view the distinction between the binomial distribution and thehypergeometric distribution lies in the way the balls are chosen.

Hypergeometric distribution – without replacementBinomial distribution – with replacement

• The mean and variance of the hypergeometric distribution are

µ = nk

N, σ2 =

N − n

N − 1× n

k

N(1− k

N)

5-3

• Binomial approximation to the hypergeometric distribution:

Intuitively, when N and k are large in relation to n, it shouldn’t make much differencewhether the selection is being done with or without replacement. If X is hypergeometric,then

Xapprox.∼ B(n, k/N)

Example 5.4 An unknown number, say N , of tigers inhabit a certain region. To obtain someinformation about N , ecologists first catch m tigers, mark them in some manner, and releasethem. After one year, they catch n tigers, among which there are i marked tigers. What is themost likely N?

Solution:Let X denote the number of marked tigers in the second capture. Then

P (X = i) =

(mi

)(N−mn−i

)(

Nn

) := Pi(N)

The desired answer is the value of N that maximizes Pi(N). Note that

Pi(N)

Pi(N − 1)=

(N −m)(N − n)

N(N −m− n + i)> 1

if and only if N ≤ mn

i. Thus, Pi(N) ↑ for N ∈ [1,mn/i], and Pi(N) ↓ for N ∈ (mn/i,∞).

Hence, the most likely N is [mn/i].For example, if m = 10, n = 10, and i = 2, then N = 50.

Example 5.5 What is the probability that an IRS auditor will catch only 2 income tax returnswith illegitimate deductions if she randomly selects 5 returns from among 15 returns, of whichnine contain illegitimate deductions?

Solution:

P (X = 2) =(92)(

63)

(155 )

= 2401001

≈ 0.24

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Example 5.6 Muffy is studying for a history exam covering the French Revolution that willconsist of 5 essay questions selected are random from a list of 10 the professor has handed outto the class in advance. Not exactly a Napoleon buff, Muffy would like to avoid researchingall ten questions but still be reasonably assured of getting a fairly good grade. Specifically, shewants to have at least an 85% chance of getting at least 4 of the 5 questions right. Will it besufficient if she studies 8 of the 10 questions?

Solution:

P (X ≥ 4) =

(84

)(21

)(105

) +

(85

)(20

)(105

) =196

252≈ 0.78

5.3 Geometric and negative binomial distributions

Independent trials,P (success) = p, P (failure) = 1− p, 0 < p < 1,

are performed until a success is obtained. Let X be the number of trials required. Then

g(x; p) = P (X = x) = (1− p)x−1p, x = 1, 2, ...

and X is called a geometric r.v. with parameter p.

• If X has geometric distribution with parameter p, then

EX =1

p, Var(X) =

1− p

p2

andP (X > k) = (1− p)k, k = 0, 1, 2, ...

M(t) =∞∑

n=1

etnpqn−1

= p et∞∑

n=0

(qet)n

=pet

1− qet

Then find M ′(0) and M ′′(0).

5-5

Negative Binomial Random Variable (option)

Definition 5.3 Independent trials,

P (success) = p, P (failure) = 1− p, 0 < p < 1,

are performed until k successes are obtained. Let X be the number of trials required. Then

b∗(x; k, p) = P (X = x) =

(x− 1

k − 1

)pk(1− p)x−k,

x = k, k + 1, ... and X is called a negative binomial r.v. with parameters (k, p), denoted byX ∼ NB(k, p).

• A geometric r.v. is just a negative binomial r.v. with parameters (1, p).

• If X has a negative binomial distribution with parameters (k, p), then

EX =k

p, Var(X) =

k(1− p)

p2

Example 5.7 If the probability is 0.75 that a person will believe a rumor about the transgres-sions of a certain politician, find the probability that the eighth person to hear the rumor willbe the fifth to believe it.

Solution:

P =

(7

4

).755.253 ≈ .1298

Example 5.8 An oil drilling company ventures into various locations, and its success or failureis independent from one location to another. Suppose the probability of a success at any specificlocation is 0.25.

(a) What is the probability that a driller drills 10 locations and finds 1 success?

(b) The driller feels that he will go bankrupt if he drills 10 times before the first success occurs.What are the driller’s prospects for bankruptcy?

(c) The driller feels that he will “hit it big” if the second success occurs on or before the sixthattempt. What is the probability that the driller will “hit it big”?

5-6

Solution:Q.5.94, 5.95 on p.168

(a) Let X be the number of successes in 10 drills. X ∼ B(10, 0.25), P (X = 1) = 0.1877(b) Let X be the number of drills needed to have the first success. Then P (X > 10) =

(1− 0.25)10 = 0.056313(c) Let X be the number of drills needed to have the second success. Then with p = 0.25,

P (X ≤ 6) = P (X = 2) + P (X = 3) + ... + P (X = 6) = p2 + 2p2q + 3p2q2 + 4p2q3 + 5p2q4 =0.466064

5.4 The Poisson distribution

Definition 5.4 X is called a Poisson random variable with parameter λ, denoted by X ∼Poisson(λ), if the probability density function of X is given by

p(x; λ) =e−λ λx

x!, x = 0, 1, 2, ...

where λ > 0.

• If X ∼ P (λ), thenEX = λ, Var(X) = λ

Proof.

M(t) =∞∑

x=0

etx λxe−λ

x!

= e−λ∞∑

x=0

(λet)x

x!

= eλ(et−1)

M ′(t) = λet eλ(et−1)

M ′′(t) = λet eλ(et−1) + λ2e2t eλ(et−1)

Poisson approximation to the binomial distribution

5-7

Suppose X ∼ B(n, p). Then ∀ k = 0, 1, ..., n

f(k) =

(n

k

)pk(1− p)n−k

=n(n− 1) · · · (n− k + 1)

k!pk(1− p)n−k

=n(n− 1) · · · (n− k + 1)

nk k!(np)k (1− np/n)n−k

=n(n− 1) · · · (n− k + 1)

nk k!(np)k (1− np/n)n

(1− np/n)k

Suppose as n →∞, np → λ > 0. Then ∀ k = 0, 1, 2, ...

f(k) −→ e−λ λk

k!

Hence, for large n and small p (≤ 0.1) such that np is moderate,

Xapprox∼ P (λ) with λ = np

Thus, the Poisson r.v. is a good model for the number of “rare” events occurring. The Poissondistribution provides a good approximation to binomial probabilities when n ≥ 20 and p ≤ 0.05.

Examples of “rare” events:

(1) The number of misprints on a page of a book.

(2) The number of wrong telephone numbers that are dialed in a day.

(3) The number of accidents occurring in a city in a given day.

(4) The number of earthquakes occurring during some fixed time span.

(5) The number of wars per year

Example 5.9 The monthly worldwide average number of airplane crashes of commercial air-lines is 3.5. What is the probability that there will be

(i) at least 2 such crashes in the next month;

(ii) at least 6 months in 1999 that have at least 2 crashes each.

5-8

Solution:Let X be the number of crashes in the next month. Since each flight has a small probability ofcrashing, it seems reasonable to suppose that X ∼ P (λ), with λ = EX = 3.5.

(i) P (X ≥ 2) = 1− P (X < 2)

= 1− P (X = 0)− P (X = 1)

= 1− e−3.5 − 3.5 e−3.5 = 0.8641

(ii) Y ∼ B(12, .8641)

P (Y ≥ 6) = 1− P (Y ≤ 5) = 1− 0.0004 = 0.9996

Example 5.10 Average number of typographical errors per page is 2.5. What is the probabilitythat

(i) a given page has no error;

(ii) at least 5 out of 100 given pages have no errors;

(iii) 2 given pages have k errors?

Solution:

(i) Let X be the number of errors on a given page. Then

X ∼ P (λ), λ = 2.5

P (X = 0) = e−λ = e−2.5 = 0.0821

(ii) Let Y be the number of pages with no errors among the 100 given pages. We have

Y ∼ B(100, p), where p = P (X = 0) = 0.0821

Hence,

P (Y ≥ 5) =100∑

k=5

(100

k

)pk(1− p)n−k

= 1−4∑

k=0

(100

k

)pk(1− p)n−k

5-9

We can also use Poisson approximation. So

Yapprox.∼ P (λ)

with λ = 100× 0.0821 = 8.21 and

P (Y ≥ 5) = 1−4∑

k=0

e−8.218.21k

k!

= 0.9118

(iii) Average number of typographical errors in the 2 given pages is 2.5 × 2. Let Z be thenumber of errors on the 2 given pages. Then

Z ∼ P (λ), λ = 5

P (Z = k) =e−55k

k!

Poisson Process:

Consider the following random variables:

(i) N(t) – be the number of people arrived in a Bank in the time interval (0, t].

(ii) N(t) – be the number of telephone calls received by an office in the time interval (0, t].

Poisson Process with rate λ:Let N(t) be the number of outcomes occurring in (0, t]. If

(1) The number of outcomes occurring in one time interval is independent of the number thatoccurs in any other disjoint time interval;

(2) P (N(t) = 1) = λt + o(t) as t → 0;

(3) P (N(t) ≥ 2) = o(t) as t → 0;

then {N(t), t ≥ 0} is called a Poisson process. Moreover, N(t) ∼ P (λt).

5-10

5.5 The uniform distribution

Definition 5.5 X is called a uniform r.v. over (a, b), denoted by X ∼ U(a, b), if the p.d.f.of X is given by

f(x) =

{ 1

b− aif a < x < b

0 otherwise

• If X ∼ U(a, b), then

E(X) =a + b

2, Var(X) =

(b− a)2

12

Example 5.11 You arrive at a bus stop at 8 o’clock, knowing that the bus will arrive at sometime uniformly distributed between 8 and 8 : 30a.m.

(i) What is the probability that you will have to wait longer than 10 minutes?

(ii) If at 8 : 15 the bus has not yet arrived, what is the probability that you will have to waitat least an additional 10 minutes?

Solution:Let X be your waiting time. Then

X ∼ U(0, 30)

(i) P (X ≥ 10) =

∫ 30

10

1

30dx =

2

3

(ii) P (X ≥ 25|X ≥ 15) =P (X ≥ 25, X ≥ 15)

P (X ≥ 15)

=P (X ≥ 25)

P (X ≥ 15)=

5/30

15/30=

1

3

5.6 The normal distribution

Definition 5.6 X is called a normal random variable with parameters µ and σ2, denoted byX ∼ N(µ, σ2), if the p.d.f. of X is given by

f(x) =1√2π σ

e−(x−µ)2

2σ2 for −∞ < x < ∞

where σ > 0. X is referred to a standard normal random variable if µ = 0 and σ = 1.

5-11

The normal distribution is the most important distribution of probability theory and statis-tics. It was first found by the French mathematician Abraham de Moivre in 1733 and wasused by him to approximate probabilities associated with binomial random variables when thebinomial parameter n is large. However, de Moivre never found the integral expression. In 1808the American surveyor Rober Adrain (1775-1855) published a remarkable paper, in which hesuggested the normal density as the proper distribution of errors of measurements. One yearlater, Karl Friederich Gauss (1777-1855) did the same thing. At that time the United Stateswas far from the leading mathematical circles, and Adrain’s paper passed unnoticed until a fewyears ago. And since Gauss was the leading mathematician of his times, the normal distributionis also called the Gaussian distribution.

Why are the normal distributions important in statistics?

1. Good descriptions for some distributions of real data, such as scores on test taken bymany people, repeated careful measurements of the same quantity, characteristicsof biological populations

2. Good approximations to the results of many kinds of chance outcomes, the central limittheorem

3. Important roles in statistical inference

Theorem 1 If X ∼ N(µ, σ2), then

E(X) = µ, Var(X) = σ2

• If X ∼ N(µ, σ2), then

Z =X − µ

σ∼ N(0, 1)

• If Z ∼ N(0, 1), thenX = µ + σ Z ∼ N(µ, σ2)

• It is conventional to denote the d.f. of a standard normal random variable by Φ(x). Thisis,

Φ(x) =1√2π

∫ x

−∞e−y2/2 dy

• Φ(−x) = 1− Φ(x) for −∞ < x < ∞• P (Z < −x) = P (Z > x)

5-12

Example 5.12 Let Z ∼ N(0, 1). Find

P (Z > 1.14), P (Z > −0.36), P (−0.46 < Z < −0.09), P (−0.58 < Z < 1.12)

Solution:

(a) P (Z > 1.14) = 0.5− .3729 = .1271

(b) P (Z > −.36) = P (Z < .36) = .5 + .1406 = .6406

(c) P (−.46 < Z < −.09) = P (0.09 < Z < 0.46) = .1772− .0359 = .1413

(d) P (−.58 < Z < 1.12) = .3686 + .2190 = .5876

Example 5.13 Let X ∼ N(µ, σ2). Find

(i) P (|X − µ| ≤ σ);

(ii) P (|X − µ| ≤ 2σ);

(iii) P (|X − µ| ≤ 3σ).

Solution:(i) 0.6827, (ii) 0.9545, (iii) 0.9973

Example 5.14 If X ∼ N(3, 9), find P (|X| > 6).

Solution:

P (|X| > 6) = P (X > 6) + P (X < −6)

= P

(X − 3

3>

6− 3

3

)+ P

(X − 3

3<−6− 3

3

)

= P (Z > 1) + P (Z < −3)

= P (Z > 1) + P (Z > 3)

= (.5− .3413) + (.5− .4987) = 0.1587 + 0.0013 = 0.16

5-13

Example 5.15 The scores of final examination in a statistics class are (approximately) nor-mally distributed with mean µ = 76 and standard deviation σ = 10.

(a) What percent of students in that class have final exam scores above 90?

(b) What score must a student achieve on the final exam in order to fall in the top 5% in theclass?

Solution:(a) 0.0808, (b) 92.45, z= 1.645

Find a value given a probability for N(0, 1) :Use Table A.4

Example 5.1 Use Table A.4 to find the value z of a standard normal variable that satisfieseach of the following conditions.

(a) the point z with 10% of the observations falling below it

(b) the point z with 5% of the observations falling above it

Solution:(a) z = −1.285 (b) z = 1.645

Find a value given a probability for N(µ, σ)

1. Formulate the problem in terms of N(0, 1) and use Table A.4 to find z

2. Use z-score transformation

z =x− µ

σor x = µ + zσ

to find x

5-14

Example 5.2 The scores of a reference population on the Wechsler Intelligence Scale for Chil-dren (WISC) are normally distributed with µ = 100 and σ = 15. What score must a childachieve on the WISC in order to fall in the top 5% of the population? In the top 1%?

Solution:(a) z = 1.645, x = µ + zσ = 100 + 1.645× 15 = 124.75

(b) z = 2.325, x = 100 + 2.325× 15 = 134.875

Example 5.3 It is known that 15% of all homeowners pay a monthly mortgage of more than$2400 and that the standard deviation of the monthly mortgage payment of all homeown-ers is $350. Suppose that the monthly mortgage payment of all homeowners have a normaldistribution. What is the mean monthly mortgage paid by all homeowners?

Solution:σ = 350, z = 1.036, x = 2400. µ = x− σz = 2400− 1.036× 350 = 2400− 362.6 = 2037.4

Example 5.4 Let X ∼ N(µ, σ2). Find the first quartile Q1 and third quartile Q3. Also findP (X > µ + 1.5(Q3 −Q1)) and P (X < µ− 1.5(Q3 −Q1)).

Solution:Q1 = µ−0.6745σ, Q3 = µ+0.6745σ, P (X > µ+1.5(Q3−Q1)) = P (X > µ+2.0235σ) = 0.0215

5.7 The Normal Approximation to the Binomial Distri-

bution

The normal distribution provides a close approximation to the binomial distribution when n isvery large, say, np(1− p) ≥ 10.

The Demoivre-Laplace Central Limit Theorem:If X := Xn denotes the number of successes that occur when n independent trials, each

resulting in a success with probability p, are performed, then for any a < b,

P

(a ≤ X − np√

npq≤ b

)−→ Φ(b)− Φ(a)

5-15

as n →∞, where q = 1− p. Equivalently,

P

(a ≤ X − EX√

Var(X)≤ b

)−→ P (a ≤ Z ≤ b) = Φ(b)− Φ(a)

or

B(n, p) ≈ N(np, npq)

• If X ∼ B(n, p), then

{X = k} = {k − 0.5 < X < k + 0.5}{X ≥ k} = {X > k − 0.5}{X ≤ k} = {X < k + 0.5}

for any integer k. Hence,

P (X = k) = P (k − 0.5 < X < k + 0.5)

≈ P (k − 0.5 < N(np, npq) < k + 0.5)

P (X ≥ k) = P (X > k − 0.5)

≈ P (N(np, npq) > k − 0.5)

P (X ≤ k) = P (X < k + 0.5)

≈ P (N(np, npq) < k + 0.5)

Example 5.16 Let X be the number of heads occurring in flipping a fair coin 40 times. Findthe probability that X = 20

Solution:X ∼ (40, 0.5), EX = 20, Var(X) = 10

5-16

P (X = 20) = P (19.5 < X < 20.5)

= P

(19.5− 20√

10<

X − EX√Var(X)

<20.5− 20√

10

)

≈ P (−0.16 < Z < 0.16)

= 1− P (Z > 0.16)− P (Z < −0.16)

= 1− 2P (Z > 0.16)

= 1− 2× 0.4364 = 0.1272

The exact result is

P (X = 20) =

(40

20

)(1

2

)40

= 0.1254

Example 5.17 The lifetimes of interactive computer chips are normally distributed with pa-rameters µ = 1.4 × 106 hours and σ = 3 × 105 hours. What is the probability that a batch of100 chips will contain at least 90 whose lifetimes are less than 1.8× 106?

Solution:Let Y be the lifetime of a chip, and X be the number of chips whose lifetimes are less than1.8× 106. Then

Y ∼ N(µ, σ2)

X ∼ B(100, P (Y < 1.8× 106))

We have

P (Y < 1.8× 106) = P

(Y − µ

σ<

1.8× 106 − µ

σ

)

= P (Z < 1.33) = 1− P (Z ≥ 1.33)

= 1− 0.0918 = 0.9082

Thus,

P (X ≥ 90) = P (X > 89.5)

= P

(X − 90.82√

90.82× 0.0918>

89.5− 90.82√90.82× 0.0918

)

≈ P (Z > −0.46)

= 1− P (Z > 0.46)

= 1− 0.3228 = 0.6772

5-17

5.8 The exponential distribution

Definition 5.7 If the p.d.f. of X is given by

f(x) =

1

βe−x/β if x ≥ 0

0 if x < 0

where β > 0, then X is called an exponential r.v. with parameter β, denoted by X ∼ Exp(β).

• If X ∼ Exp(β), then

F (x) =

{1− e−x/β if x ≥ 00 if x < 0

EX = β, Var(X) = β2

Proof. For x ≥ 0,

F (x) =∫ x

−∞f(t) dt =

∫ x

0

1β

e−t/β dt

= −e−t/β

∣∣∣∣x

0

= 1− e−x/β

LetΓ(n) =

∫ ∞

0

yn−1e−ydy

ThenΓ(n) = (n− 1)!

By using the above equation,

E(X) =∫ ∞

0

x1β

e−x/β dx = β

∫ ∞

0

(x/β) e−x/β d(x/β)

= β

∫ ∞

0

y e−y dy

= β

E(X2) =∫ ∞

0

x2 1β

e−x/β dx

= β2

∫ ∞

0

y2 e−y dy

= 2β2

and henceVar(X) = E(X2)− (EX)2 = β2

Relationship to the Poisson random variables

5-18

Suppose that “rare” events occur at a rate of λ events per unit time. Let

N(t) = The number of events occurring in[0, t]

Then N(t) ∼ P (λt).Let X be the waiting time to the first event (amount of time that lapses until the first eventoccurs).

FX(t) = 0 for t < 0

For t ≥ 0,

FX(t) = 1− P (X > t) = 1− P (N(t) = 0)

= 1− e−λt

HenceX ∼ Exp(1/λ)

• Since λ = average number of events per unit time,1

λ= average length of time to 1st event

This agrees with E(X) =1

λ.

Therefore, exponential distribution is a good model for the amount of time until somespecific event occurs. For instance, the amount of time (starting from now) until an earthquakeoccurs, or until a new war breaks out, or until a telephone call you receive turns out to be awrong number are all random variables that tend in practice to have exponential distributions.

Example 5.18 Suppose that a system contains a certain type of component whose time inyears to failure is given by the r.v. X, distributed exponentially with mean time 5 to failure. If5 of these components are installed in different systems, what is the probability that at least 2are still functioning at the end of 8 years?

Solution:X ∼ Exp(β) with β = 5.

P (X > 8) =

∫ ∞

8

1

βe−x/β dx

= e−8/β = e−8/5 ≈ 0.2019

5-19

Let Y be the number of components functioning after 8 years. Then

Y ∼ B(5, 0.2019)

Thus,

P (Y ≥ 2) = 1− P (Y < 2)

= 1− P (Y = 0)− P (Y = 1)

= 1− (1− 0.2019)5 − 5(0.2019)(1− 0.2019)4

≈ 0.2666

5.9 The Gamma distribution and Chi-square distribu-

tion (Option)

Definition 5.8 X has a gamma distribution with parameters (α, β), denoted by X ∼Γ(α, β), if its p.d.f. is given by

f(x) =

{1

βαΓ(α)xα−1 e−x/β if x > 0

0 if x < 0

where α > 0, β > 0, and Γ(α), called the gamma function, is defined as

Γ(α) =

∫ ∞

0

yα−1e−y dy

• Γ(α) = (α− 1)Γ(α− 1)

• The gamma distribution with parameters (n, β) often arises as the distribution of amountof time one has to wait until a total of n specific events has occurred.

• Let X be a gamma r.v. with parameters (α, β). Then

E(X) = α β, Var(X) = α β2

A very important special case of the gamma distribution is obtained by letting α = v/2 andβ = 2. The result is called the chi-square distribution with v degrees of freedom.

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Definition 5.9 X has a chi-square distribution with v degrees of freedom if the pdf is given by

f(x) =

{ 12v/2Γ(v/2)

xv/2−1e−x/2 x > 0,

0 elsewhere

The chi-square distribution plays a vital role in statistical inference. It has considerableapplication in both methodology and theory.

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