Functions and Transformations of Random Variables

Section 09

Transformation of continuous X Suppose X is a continuous random variable

with pdf and cdf Suppose is a one-to-one function with

inverse ; so that

The random variable is a transformation of X with pdf:

If is a strictly increasing function, then and then

Transformation of discrete X

Again, Since X is discrete, Y is also discrete

with pdf

This is the sum of all the probabilities where u(x) is equal to a specified value of y

Transformation of jointly distributed X and Y

X and Y are jointly distributed with pdf and are functions of x and y This makes and also random variables

with a joint distribution

In order to find the joint pdf of U and V, call it g(u,v), we expand the one variable case Find inverse functions and so that and Then the joint pdf is:

Sum of random variables

If then

If Xs are continuous with joint pdf

If Xs are discrete with joint pdf

Convolution method for sums

If X1 and X2 are independent, we use the convolution method for both discrete & cont.

Discrete:

Continuous:

Sums of random variables

If X1, X2, …, Xn are random variables and

If Xs are mutually independent

Central Limit Theorem

X1, X2, …, Xn are independent random variables with the same distribution of mean μ and standard deviation σ

As n increases, Yn approaches the normal distribution

Questions asking about probabilities for large sums of independent random variables are often asking to use the normal approximation (integer correction sometimes necessary).

Sums of certain distribution

This table is on page 280 of the Actex manualDistribution of Xi Distribution of Y

Bernoulli B(1,p) Binomial B(k,p)

Binomial B(n,p) Binomial B( ,p)

Poisson Poisson

Geometric p Negative binomial k,p

Normal N(μ,σ2) Normal N( ,)

There are more than these but these are the most common/easy to remember

Distribution of max or min of random variables

X1 and X2 are independent random variables

Mixtures of Distributions

X1 and X2 are independent random variables We can define a brand new random variable X as

a mixture of these variables! X has the pdf

Expectations, probabilities, and moments follow this “weighted-average” form

Be careful! Variances do not follow weighted-average! Instead, find first and second moments of X and subtract

Sample Exam #101

The profit for a new product is given by Z = 3X – Y – 5 . X and Y are independent random variables with Var(X)=1 and Var(Y)=2.

What is the variance of Z?

A) 1 B) 5 C) 7 D) 11 E) 16

Sample Exam #102

A company has two electric generators. The time until failure for each generator follows an exponential distribution with mean 10. The company will begin using the second generator immediately after the first one fails.

What is the variance of the total time that the generators produce electricity?

Sample Exam #103

In a small metropolitan area, annual losses due to storm, fire, and theft are assumed to be independent, exponentially distributed random variables with respective means 1.0, 1.5, and 2.4

Determine the probability that the maximum of these losses exceeds 3.

Sample Exam #142

An auto insurance company is implementing a new bonus system. In each month, if a policyholder does not have an accident, he or she will receive a $5 cash-back bonus from the insurer.

Among the 1000 policyholders of the auto insurance company, 400 are classified as low-risk drivers and 600 are classified as high-risk drivers.

In each month, the probability of zero accidents for high-risk drivers is .8 and the probability of zero accidents for low-risk drivers is .9

Calculate the expected bonus payment from the insurer to the 1000 policyholders in one year.

Sample Exam #123

You are given the following information about N, the annual number of claims for a randomly selected insured:

Let S denote the total annual claim for an insured. When N=1, S is exponentially distributed with mean 5. When N>1, S is exponentially distributed with mean 8.

Determine P(4<S<8).