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Lecture 24 Agenda 1. Examples from Normal Distribution 2. Beta distribution 3. Moment Generating Function Example Mainly two kind of examples are done for normal distribution. Example 1 Suppose that men’s neck sizes are approximately normally distributed with a mean of 16.2 inches and variance of 0.81 square inch. Find the probability that the neck size of a randomly chosen man lies between 13.5 and 18.9 inhes. Let X = Men’s neck size in inches. Then X N (μ = 16.22 =0.81). P (13.5 X 18.9) = P (X 18.9) - P (X 13.5) = Φ 18.9 - 16.2 0.9 - Φ 13.5 - 16.2 0.9 = Φ (3) - Φ(-3) = 0.997 Example 2 Suppose scores in an exam follow normal distribution with mean 80 and standard deviation 5. What’s the minimum score that you should get to be in the top 10% ? Let X be the score of a randomly chosen student and suppose you have to score minimum x to be in the top 10%. X N (μ = 802 =5 2 ) Then P (X x)=0.9. We also know P (X x)=Φ ( x-μ σ ) ( x-80 5 ) . Hence Φ ( x-80 5 ) =0.9. Now from computers I can find out that Φ -1 (0.9) = 1.2815. Hence x-80 5 =1.2815, i.e. x = 86.40776. 1
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Example - Department of Statisticsabhisheksaha/sta4321/lect24.pdfLecture 24 Agenda 1.Examples from Normal Distribution 2.Beta distribution 3.Moment Generating Function Example Mainly

Mar 22, 2018

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Page 1: Example - Department of Statisticsabhisheksaha/sta4321/lect24.pdfLecture 24 Agenda 1.Examples from Normal Distribution 2.Beta distribution 3.Moment Generating Function Example Mainly

Lecture 24

Agenda

1. Examples from Normal Distribution

2. Beta distribution

3. Moment Generating Function

Example

Mainly two kind of examples are done for normal distribution.

Example 1

Suppose that men’s neck sizes are approximately normally distributed

with a mean of 16.2 inches and variance of 0.81 square inch. Find

the probability that the neck size of a randomly chosen man lies

between 13.5 and 18.9 inhes.

Let X = Men’s neck size in inches. Then X ∼ N(µ = 16.2, σ2 = 0.81).

P (13.5 ≤ X ≤ 18.9) = P (X ≤ 18.9)− P (X ≤ 13.5)

= Φ

(18.9− 16.2

0.9

)− Φ

(13.5− 16.2

0.9

)= Φ (3)− Φ (−3) = 0.997

Example 2

Suppose scores in an exam follow normal distribution with mean 80and standard deviation 5. What’s the minimum score that you should

get to be in the top 10% ?

Let X be the score of a randomly chosen student and suppose you haveto score minimum x to be in the top 10%.

X ∼ N(µ = 80, σ2 = 52)

Then P (X ≤ x) = 0.9.We also know P (X ≤ x) = Φ

(x−µσ

)= Φ

(x−805

). Hence Φ

(x−805

)= 0.9. Now

from computers I can find out that Φ−1(0.9) = 1.2815. Hence x−805

= 1.2815,i.e. x = 86.40776.

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Page 2: Example - Department of Statisticsabhisheksaha/sta4321/lect24.pdfLecture 24 Agenda 1.Examples from Normal Distribution 2.Beta distribution 3.Moment Generating Function Example Mainly

Beta Distribution

Every continuous distribution which we have encountered except the uni-form distribution, takes values over an infinite interval like (0,∞) or R. Thebeta distribution is an alternative model for random variables which can beconstrained in the interval (0, 1). In general if X can be constrained in theinterval (a, b) then Y = X−a

b−a can be constrained in (0, 1) and thus by study-ing properties of Y we can study properties of X.

Definition 1. A random variable X is said to follow the Beta distributionwith parameters (α, β) for some α > 0 and β > 0 if,

Range(X) = (0, 1)

and for x ∈ (0, 1)

fX(x) =Γ(α + β)

Γ(α)Γ(β)xα−1(1− x)β−1

We write this as X ∼ Beta(α, β)

The first thing that we need to check is that∫ 1

0fX(x)dx = 1, i.e. we need

to check the following identity for α, β > 0.∫ 1

0

xα−1(1− x)β−1dx =Γ(α)Γ(β)

Γ(α + β). . . (∗)

but we are omitting this proof for now. Instead let’s do the mean and vari-ance,

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Page 3: Example - Department of Statisticsabhisheksaha/sta4321/lect24.pdfLecture 24 Agenda 1.Examples from Normal Distribution 2.Beta distribution 3.Moment Generating Function Example Mainly

Mean and Variance

E(X) =

∫ 1

0

xfX(x)dx

=

∫ 1

0

x× Γ(α + β)

Γ(α)Γ(β)xα−1(1− x)β−1dx

=Γ(α + β)

Γ(α)Γ(β)

∫ 1

0

xα(1− x)β−1dx

=Γ(α + β)

Γ(α)Γ(β)× Γ(α + 1)Γ(β)

Γ((α + 1) + β)[from (*)]

=Γ(α + β)

Γ((α + 1) + β)× Γ(α + 1)

Γ(α)

=Γ(α + β)

(α + β)Γ(α + β)× αΓ(α)

Γ(α)

α + β

Similarly,

V (X) =αβ

(α + β)2(α + β + 1).

How does the density look like ?

We have plotted the desity for four combinations of α and β. More will bediscussed in lecture. Please note that α = β = 1 means the uniform density.

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Page 4: Example - Department of Statisticsabhisheksaha/sta4321/lect24.pdfLecture 24 Agenda 1.Examples from Normal Distribution 2.Beta distribution 3.Moment Generating Function Example Mainly

Moment generating function

For a random variable X (discrete or continuous), the moment generatingfunction is a special function associated with it. The moment generatingfunction although intuitively looks strange at first is a powerful theoriticaltool. The name “moment generating function” comes from the fact that itcan be used to generate the “moments” of X. We clarify this shortly.

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Page 5: Example - Department of Statisticsabhisheksaha/sta4321/lect24.pdfLecture 24 Agenda 1.Examples from Normal Distribution 2.Beta distribution 3.Moment Generating Function Example Mainly

For any t ∈ R,

E(etX) =

∫ ∞−∞

etxfX(x)dx [If X is continuous]

=∑

x∈Range(X)

etxP (X = x) [If X is discrete]

Now if t 6= 0, there is no guarantee that the above sum or integral will befinite.

Definition 2. Let X be a random variable and A = {t : E(etX) <∞}. Thenwe define the moment generating function of X, as the function MX : A →(0,∞) where for t ∈ A,

MX(t) = E(etX)

Now for any random variable X,

E(X), E(X2), E(X3), . . .

are known as the moments of the random variable. They provide useful in-formation about the random variable. The following result tells us why thisfunction is called the moment generating function.

Lemma 1. If for a random variable X, MX can be defined for all values inany interval (−ε, ε) around 0 then for k ≥ 1,

dk

dtkMX(t)|t=0 = E(Xk)

Thus the moment generating function can be used to generate moments.

Homework::

1. If X ∼ Beta(α, β), then Y = 1−X ∼ Beta(β, α).

2. Prove the formula for variance of beta distribution.

3. 4.123 a, 4.124, 4.125

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