Lecture 05-2007 - Distribution Functions for Random...

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Distribution Functions for Random Variables: Lecture VI I. Bivariate Continuous Random Variables A. Definition 3.4.1. If there is a nonnegative function   , f x y defined over the whole plane such that       2 1 2 1 , , 2 1 2 1 y y x x dy dx y x f y Y y x X x P for any 1 x , 2 x , 1 y , and 2 y satisfying 12 xx , yy , then   , XY is a bivariate continuous random variable and   , f x y is called the joint density function. B. Much of the work with distribution functions involves integration. In order to demonstrate a couple of solution techniques, I will work through a couple of examples. 1. Example 3.4.1. If     , exp f x y xy x y   , 0 x , 0 y and 0 otherwise, what is   1, 1 P X Y  :     1 0 1 1 , 1 dy dx e y x Y X P y x First, note that the integral can be separated into two terms:   1 0 1 1 , 1 dy e y dx e x Y X P y x Each of these integrals can be solved using integration by parts:     udv uv vdu udv uv d vdu udv vdu uv d In terms of a proper integral we have b a b a b a udv uv vdu In this case, we have 1 1 1 11 ,1 , 2 0.74 x x x x v x dv xe dx du e dx u e xe e dx e        
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AEB 6933 Mathematical Statistics for Food and Resource Economics Lecture V Professor Charles B. Moss Fall 2007 2     11 1 0 00 1 01 1 2 0.26 y y y yy ye dy ye e dy ye e ee e                    1 10 1, 1 0.735 0.264 0.194 xy P X Y xe dx ye dy 2. Example 3.4.3. This example demonstrates the use of changes in variables. Implicitly the example assumes that the random variables are joint uniform for all 0 x , 1 y . The question is then: What is the probability that 22 1  ? Mathematically, this question is not separable:     1 0 2 1 0 1 0 2 2 1 1 2 dx x dx dy Y X P x As previously stated, we will solve this problem using integration by change in variables. By trigonometric identity 1 sin cos xx  . Therefore, sin 1 cos  . The change in variables is then to let cos . The integration by change in variables is then:       2 1 2 1 ' ) ( t t x x dt t t f dx x f Such that   tx and   . The change in variable then implies that: 4 2 1 cos sin 2 1 sin 1 2 0 2 0 2 1 0 2 d dx x II.
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Lecture 05-2007 - Distribution Functions for Random...

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