# 412 since the cdf completely determines probabilities

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: the base. The base in this case is the support of fX,Y , which is a triangular region with area 1/2. The height is c. So the volume is c/6, which should be one, so c = 6. The marginal pdf of X is given by ∞ fX (uo ) = fX,Y (u0 , v )dv −∞ = 1−uo 0 c(1 − uo − v )dv = 0 c(1−uo )2 2 0 ≤ uo ≤ 1 otherwise. Note that if we hadn’t already found that c = 1/6, we could ﬁnd c using the expression for fX just found, because fX must integrate to one. That yields: ∞ 1= 1 fX (u)du = −∞ 0 c(1 − u)2 c(1 − u)3 du = − 2 6 )2 1 0 c =, 6 or, again, c = 6. Thus, fX has support [0, 1) with fX (uo ) = 3(1 − uo for 0 ≤ uo ≤ 1. The conditional density fY |X (v |uo ) is deﬁned only if uo is in the support of fX : 0 ≤ uo &lt; 1. For such uo , 2(1−uo −v ) 0 ≤ v ≤ 1 − uo (1−uo )2 fY |X (v |uo ) = 0 else. 126 CHAPTER 4. JOINTLY DISTRIBUTED RANDOM VARIABLES That is, the conditional pdf of Y given X = uo has a triangular shape, over the interval [0, 1 − u0 ], as shown in Figure 4.5. This makes sense geometrically–a slice through the three dimensional region 2 1!u o f (v|uo ) Y|X v 0 1!u 1 o Figure 4.5: The conditional pdf of Y given...
View Full Document

## This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Institute of Technology.

Ask a homework question - tutors are online