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Section 09. Functions and Transformations of Random Variables. 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: - PowerPoint PPT Presentation
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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).