412 since the cdf completely determines probabilities

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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 < 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...
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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.

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