Functions of Random Variables Methods for determining the

Functions of Random Variables

Methods for determining the distribution of functions of Random Variables 1. Distribution function method 2. Moment generating function method 3. 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: 1. Z having a Standard Normal distribution and 2. U having a c 2 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

Student – W. W. Gosset 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 x 1, x 2, … , xn denote 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)

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. Let Find the distribution of X. Hence.

Thus if U has a uniform distribution from 0 to 1. Then has density f(x). U

Use of moment generating functions

Definition Let X denote a random variable with probability density function f(x) if continuous (probability mass function p(x) if discrete) Then m. X(t) = the moment generating function of X

The distribution of a random variable X is described by either 1. The density function f(x) if X continuous (probability mass function p(x) if X discrete), or 2. The cumulative distribution function F(x), or 3. The moment generating function m. X(t)

Properties 1. m. X(0) = 1 2. 3.

4. Let X be a random variable with moment generating function m. X(t). Let Y = b. X + a Then m. Y(t) = mb. X + a(t) = E(e [b. X + a]t) = eatm. X (bt) 5. Let X and Y be two independent random variables with moment generating function m. X(t) and m. Y(t). Then m. X+Y(t) = m. X (t) m. Y (t)

6. Let X and Y be two random variables with moment generating function m. X(t) and m. Y(t) and two distribution functions FX(x) and FY(y) respectively. Let m. X (t) = m. Y (t) then FX(x) = FY(x). This ensures that the distribution of a random variable can be identified by its moment generating function

M. G. F. ’s - Continuous distributions

M. G. F. ’s - Discrete distributions

Moment generating function of the gamma distribution where

using or

then

Moment generating function of the Standard Normal distribution where thus

We will use

Note: Also

Equating coefficients of tk, we get

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

Example Suppose that X has a normal distribution with mean m and standard deviation s. Find the distribution of Y = a. X + b Solution: = the moment generating function of the normal distribution with mean am + b and variance a 2 s 2.

Thus Y = a. X + b has a normal distribution with mean am + b and variance a 2 s 2. Special Case: the z transformation Thus Z has a standard normal distribution.

Example Suppose that X and Y are independent each having a normal distribution with means m. X and m. Y , standard deviations s. X and s. Y Find the distribution of S = X + Y Solution: Now

or = the moment generating function of the normal distribution with mean m. X + m. Y and variance Thus Y = X + Y has a normal distribution with mean m. X + m. Y and variance

Example Suppose that X and Y are independent each having a normal distribution with means m. X and m. Y , standard deviations s. X and s. Y Find the distribution of L = a. X + b. Y Solution: Now

or = the moment generating function of the normal distribution with mean am. X + bm. Y and variance Thus Y = a. X + b. Y has a normal distribution with mean am. X + bm. Y and variance

Special Case: a = +1 and b = -1. Thus Y = X - Y has a normal distribution with mean m. X - m. Y and variance

Example (Extension to n independent RV’s) Suppose that X 1, X 2, …, Xn are independent each having a normal distribution with means mi, standard deviations si (for i = 1, 2, … , n) Find the distribution of L = a 1 X 1 + a 1 X 2 + …+ an. Xn Solution: (for i = 1, 2, … , n) Now

or = the moment generating function of the normal distribution with mean and variance Thus Y = a 1 X 1 + … + an. Xn has a normal distribution with mean a 1 m 1 + …+ anmn and variance

Special case: In this case X 1, X 2, …, 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 x 1, x 2, …, 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 x 1, x 2, …, 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 m 1(t), m 2(t), … denote a sequence of moment generating functions corresponding to the sequence of distribution functions: F 1(x) , F 2(x), … Let m(t) be a moment generating function corresponding to the distribution function F(x) then if then

Let x 1, x 2, … denote a sequence of independent random variables coming from a distribution with moment generating function m(t) and distribution function F(x). Let Sn = x 1 + x 2 + … + 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.