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## PowerPoint Slideshow about ' Functions of Random Variables' - melyssa-garrett

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### Using of moment generating functions to find the distribution of functions of Random Variables

Methods for determining the distribution of functions of Random Variables

- Distribution function method
- Moment generating function method
- Transformation method

Distribution function method

Let X, Y, Z …. have joint density f(x,y,z, …)

Let W = h( X, Y, Z, …)

First step

Find the distribution function of W

G(w) = P[W ≤ w] = P[h( X, Y, Z, …)≤ w]

Second step

Find the density function of W

g(w) = G'(w).

Example: Student’s t distribution

Let Z and U be two independent random variables with:

- Z having a Standard Normal distribution

and

- U having a c2 distribution with n degrees of freedom

Find the distribution of

The density of Z is:

The density of U is:

Therefore the joint density of Z and U is:

The distribution function of T is:

Then

where

Student – W.W. Gosset

Worked for a distillery

Not allowed to publish

Published under the pseudonym “Student

t distribution

standard normal distribution

Let x1, x2, … , xndenote a sample of size n from the density f(x).

Let M = max(xi) then determine the distribution of M.

Repeat this computation for m = min(xi)

Assume that the density is the uniform density from 0 to q.

Hence

and the distribution function

The probability integral transformation

This transformation allows one to convert observations that come from a uniform distribution from 0 to 1 to observations that come from an arbitrary distribution.

Let U denote an observation having a uniform distribution from 0 to 1.

Let f(x) denote an arbitrary density function and F(x) its corresponding cumulative distribution function.

Find the distribution of X.

Let

Hence.

Definition

Let X denote a random variable with probability density function f(x) if continuous (probability mass function p(x) if discrete)

Then

mX(t) = the moment generating function of X

The distribution of a random variable X is described by either

- The density function f(x) if X continuous (probability mass function p(x) if X discrete), or
- The cumulative distribution function F(x), or
- The moment generating function mX(t)

Properties

- mX(0) = 1

Let X be a random variable with moment generating function mX(t). Let Y = bX + a

Then mY(t) = mbX + a(t)

= E(e [bX + a]t) = eatmX (bt)

- Let X and Y be two independent random variables with moment generating function mX(t) and mY(t) .

Then mX+Y(t) = mX (t) mY (t)

Let X and Y be two random variables with moment generating function mX(t) and mY(t) and two distribution functions FX(x) and FY(y) respectively.

Let mX (t) = mY (t) then FX(x) = FY(x).

This ensures that the distribution of a random variable can be identified by its moment generating function

using

or

Note:

Also

Note:

Also

Example

Suppose that X has a normal distribution with mean mand standard deviation s.

Find the distribution of Y = aX + b

Solution:

= the moment generating function of the normal distribution with mean am + b and variance a2s2.

Thus Y = aX + b has a normal distribution with mean am + b and variance a2s2.

Special Case: the z transformation

Thus Z has a standard normal distribution .

Example

Suppose that X and Y are independent eachhaving a normal distribution with means mX and mY , standard deviations sX and sY

Find the distribution of S = X + Y

Solution:

Now

or

= the moment generating function of the normal distribution with mean mX + mY and variance

Thus Y = X + Y has a normal distribution with mean mX + mY and variance

Example

Suppose that X and Y are independent eachhaving a normal distribution with means mX and mY , standard deviations sX and sY

Find the distribution of L = aX + bY

Solution:

Now

or

= the moment generating function of the normal distribution with mean amX + bmY and variance

Thus Y = aX + bY has a normal distribution with mean amX + bmY and variance

a = +1 and b = -1.

Special Case:

Thus Y = X - Y has a normal distribution with mean mX - mY and variance

Example (Extension to n independent RV’s)

Suppose that X1, X2, …, Xn are independent eachhaving a normal distribution with means mi, standard deviations si (for i = 1, 2, … , n)

Find the distribution of L = a1X1 + a1X2 + …+ anXn

Solution:

(for i = 1, 2, … , n)

Now

or

= the moment generating function of the normal distribution with mean

and variance

Thus Y = a1X1 + … + anXnhas a normal distribution with mean a1m1+ …+ anmn and variance

Special case:

In this case X1, X2, …, Xn is a sample from a normal distribution with mean m, and standard deviations s, and

Summary

If x1, x2, …, xn is a sample from a normal distribution with mean m, and standard deviations s, then

has a normal distribution with mean

and variance

Sampling distribution of

Population

The Central Limit theorem

If x1, x2, …, xn is a sample from a distribution with mean m, and standard deviations s, then if n is large

has a normal distribution with mean

and variance

Proof: (use moment generating functions)

We will use the following fact:

Let

m1(t), m2(t), …

denote a sequence of moment generating functions corresponding to the sequence of distribution functions:

F1(x) , F2(x), …

Let m(t) be a moment generating function corresponding to the distribution function F(x) then if

then

Let x1, x2, … denote a sequence of independent random variables coming from a distribution with moment generating function m(t) and distribution function F(x).

Let Sn = x1 + x2 + … + xn then

Is the moment generating function of the standard normal distribution

Thus the limiting distribution of z is the standard normal distribution

Q.E.D.

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