Lecture 05-2007

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

This preview shows pages 1–9. Sign up to view the full content.

Distribution Functions for Distribution Functions for Random Variables Random Variables Lecture V

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Bivariate Continuous Random Bivariate Continuous Random Variables Variables Definition 3.4.1. If there is a nonnegative function f ( x , y ) defined over the whole plane such that for any x 1 , x 2 , y 1 , 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 ( 29 ( 29 ∫ ∫ = 2 1 2 1 , , 2 1 2 1 y y x x dy dx y x f y Y y x X x P
Example 3.4.1. If f ( x , y )= x y exp(- x - y ), x >0, y >0 and 0 otherwise, what is P ( X >1, Y <1): First, note that the integral can be separated into two terms: ( 29 ( 29 ∫ ∫ + - = < 1 0 1 1 , 1 dy dx e y x Y X P y x ( 29 - - = < 1 0 1 1 , 1 dy e y dx e x Y X P y x

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Each of these integrals can be solved using integration by parts: ( 29 ( 29 - = - = + = udv uv vdu udv uv d vdu udv vdu uv d ( - = b a b a b a udv uv vdu
In this case, we have ( 1 1 1 1 1 , 1 , 2 0.74 x x x x x x v x dv xe dx du e dx u e xe dx xe e dx e x - - - - - - - = = = = - = - + = =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
( ( ( ( 29 ( 29 1 1 1 0 0 0 1 1 0 0 1 1 1 0 1 1 2 0.26 y y y y y ye dy ye e dy ye e e e e - - - - - - - - = - + = - + - = - + + - + = - = ( 29 ( 29 ( 29 1 1 0 1, 1 0.735 0.264 0.194 x y P X Y xe dx y e dy x - - < = = =
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? ( 29 ∫ ∫ - = = < + - 1 0 2 1 0 1 0 2 2 1 1 2 dx x dx dy Y X P x

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
As previously stated, we will solve this problem using integration by change in variables . By trigonometric identity 1=Sin 2 x +Cos 2 x . Therefore, Sin 2 x =1- Cos 2 x . The change in variables is then to let x =Cos x .
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 11/08/2011 for the course AEB 6933 taught by Professor Carriker during the Fall '09 term at University of Florida.

### Page1 / 38

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

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online