Intro Part A -C Functions of Random Variables and Order Statistics

# Intro Part A -C Functions of Random Variables and Order Statistics

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Topic A: Functions of Random Variables In many applications, interest lies in obtaining information about a function of a random variable, say, g ( Y ), instead of the random variable Y itself. Example 1. In a medical experiment, Y denotes the systolic blood pressure for a group of patients. However, oftentimes measurements are observed on the log scale. How is g ( Y ) = log Y distributed? Example 2. In a manufacturing setting, Y denotes machine down-time. How is the associated cost function g ( Y ) = 2 Y 2 + Y + 50 distributed? An agricultural experiment is undertaken to study Y , the diameter of a certain species of eggs. How is g ( Y ) = Y distributed? This topic deals with finding distributions of functions of random variables. We will investigate three main techniques for doing this: (1) Method of distribution functions (2) Method of transformations (3) Method of moment generating functions. The method of distribution functions SETTING : Suppose that Y is a continuous random variable with cdf F Y ( y ). The general approach is to compute F U ( u ) = P ( U u ), the cdf of U = g ( Y ), and then to differentiate it (with respect to u ) to find the pdf of U . Example 4. Suppose that Y ~ U (0 ; 1). Find the distribution of U = g ( Y ) = - log Y . PAGE 1

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Solution. Using the method of distribution functions, we have F U ( u ) = P ( U · u ) = P (- log Y · u ) = P (log Y -u ) = P ( Y e -u ) = 1 - P ( Y e -u ) = 1 - F Y ( e -u ) We know that F Y ( y ) = y for 0 < y < 1. Thus, for 0 < exp(-u) < 1, we have F U ( u ) = 1 - F Y ( exp(-u) = 1 – exp(-u) . Hence, taking derivatives, we get f U ( u ) = d du F U ( u ) = d du (1 - e -u ) = e -u Now, when 0 < y < 1, it must be that u = - log y > 0. Thus, f U ( u ) = e -u ; u > 0 0 ; otherwise. That is, U ~ exponential(1). Example 5. Suppose that Y ~ exponential(1). Find the distribution of U = g ( Y ) = Y + , for > 0. Solution. Using the method of distribution functions, we have F U ( u ) = P ( U u ) = P ( Y + · u ) = P ( Y · u- ) = F Y ( u - ) We know that F Y ( y ) = 1 –exp(-y) for y > 0. Thus, for u - > 0; i.e., u > , we have F U ( u ) = F Y ( u - ) = 1 – exp(-(u- )) . Hence, taking derivatives, we get f U ( u ) = d du F U ( u ) = d du [ 1 – exp(-(u- ))=exp(-(u- )) Thus, PAGE 2
f U ( u ) = ( ) , u e u The random variable U is said to have a shifted exponential distribution , where denotes the shift parameter. Example 6. In Example 5, what is E ( U )? V ( U )?

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( )) ( )) d g u g u du 1 42 43 0. If g is decreasing, g -1 is as well and d du 1 g ( u ) < 0. To handle both cases (as to avoid negative density functions), we use an absolute value
• Spring '11
• lisa
• Statistics, Probability distribution, Probability theory, probability density function

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