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Functions of Random VariablesPowerPoint Presentation

Functions of Random Variables

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Functions of Random Variables

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Functions of Random Variables

- Distribution function method
- Moment generating function method
- Transformation 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).

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’s t distribution

where

Worked for a distillery

Not allowed to publish

Published under the pseudonym “Student

t distribution

standard normal distribution

Distribution of the Max and Min Statistics

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

Finding the distribution function of M.

Differentiating we find the density function of M.

f(x)

g(t)

Finding the distribution function of m.

Differentiating we find the density function of m.

f(x)

g(t)

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.

Thus if U has a uniform distribution from 0 to 1. Then

has density f(x).

U

Use of moment generating functions

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)

- 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

where

using

or

then

where

thus

We will use

Note:

Also

Note:

Also

Equating coefficients of tk, we get

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

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 .

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

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

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

Thus

has a normal distribution with mean

and variance

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.