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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 (vuo ) YX 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.
 Spring '08
 Zahrn
 The Land

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