Slides05-2010

# Slides05-2010 - Distribution Functions for Random Variables...

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Unformatted text preview: Distribution Functions for Random Variables: Lecture V Charles B. Moss August 31, 2010 Charles B. Moss () Distribution Functions for Random Variables: Lecture V August 31, 2010 1 / 25 Outline 1 Bivariate Continuous Random Variables 2 Marginal Density 3 Independence 4 Distribution Functions 5 Change of Variables Charles B. Moss () Distribution Functions for Random Variables: Lecture V August 31, 2010 2 / 25 Bivariate Continuous Random Variables Bivariate Continuous Random Variables Definition 3.4.1. If there is a nonnegative function f ( x , y ) defined over the whole plane such that P ( x 1 ≤ X ≤ x 2 , y 1 ≤ Y ≤ y 2 ) = Z y 2 y 1 Z x 2 x 1 f ( x , y ) dxdy (1) for x 1 , x 2 , y 1 , and y 2 satisfying x 1 ≤ x 2 , y 1 ≤ y 2 , then ( X , Y ) is a bivariate continuous random variable and f ( X , Y ) is called the joint density function. 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. Example 3.4.1. If f ( x , y ) = xy exp ( − x − y ) , x > 0 , y > 0 and 0 otherwise, what is P ( X > 1 , Y < 1) P ( X > 1 , Y < 1) = Z 1 Z ∞ 1 xye − ( x + y ) dxdy (2) Charles B. Moss () Distribution Functions for Random Variables: Lecture V August 31, 2010 3 / 25 Bivariate Continuous Random Variables First, note that the integral can be separated into two terms P ( X > 1 , Y < 1) = Z ∞ 1 xe − 1 dx Z 1 ye − y dy (3) Each of these integrals can be solved using integration by parts: d ( uv ) = v du + u dv v du = d ( uv ) − u dv R v du = uv − R u dv (4) In terms of a proper integral we have Z b a v du = ( uv | b a − Z b a u dv (5) Charles B. Moss () Distribution Functions for Random Variables: Lecture V August 31, 2010 4 / 25 Bivariate Continuous Random Variables In this case, we have Z ∞ 1 xe − x dx ⇒ v = x , dv = 1 du = e − x , u = − e − x Z ∞ 1 xe − x dx − ( − xe − x ⎧ ⎧ ∞ 1 + Z ∞ 1 e − x dx = 2 e 1 = 0 . 74 (6) Charles B. Moss () Distribution Functions for Random Variables: Lecture V August 31, 2010 5 / 25 Bivariate Continuous Random Variables Working on the second part of the integral Z 1 ye − y dy = ( − ye − 1 ⎧ ⎧ 1 + Z 1 e − y dy = ( − ye − 1 ⎧ ⎧ 1 + ( − e − y ⎧ ⎧ 1 = ( − e − 1 + 0 ) + ( − e − 1 + 1 ) (7) Putting the two parts together P ( X > 1 , Y < 1) = Z ∞ 1 xe − x dx Z 1 ye − y dy = (0 . 735) (0 . 264) = 0 . 194 (8) Charles B. Moss () Distribution Functions for Random Variables: Lecture V August 31, 2010 6 / 25 Bivariate Continuous Random Variables 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 , y < 1 . The question is then: What is the probability that x 2 + y 2 < 1 ? Mathematically, this question is not separable: P...
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## This note was uploaded on 07/15/2011 for the course AEB 6180 taught by Professor Staff during the Spring '10 term at University of Florida.

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Slides05-2010 - Distribution Functions for Random Variables...

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