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Unformatted text preview: Homework Solutions Assignment Number 1 Due January 22, 2007 Section 3.7 Problem 20. (a) • f X ( x ) = R ∞∞ f X,Y ( x,y ) dy = R 2 x 1 2 dy = y 2 2 x = 1 x 2 , ≤ x ≤ 2 • f Y ( y ) = R ∞∞ f X,Y ( x,y ) dx = R y 1 2 dx = x 2 y = y 2 , ≤ y ≤ 2 (b) • f X ( x ) = R ∞∞ f X,Y ( x,y ) dy = R x 1 x dy = y x y = x y =0 = 1 , ≤ x ≤ 1 • f Y ( y ) = R ∞∞ f X,Y ( x,y ) dx = R 1 y 1 x dx = ln  x  y =1 x = y = ln y, ≤ y ≤ 1 (c) • f X ( x ) = R ∞∞ f X,Y ( x,y ) dy = R 1 x 6 xdy = 6 xy  y =1 x y =0 = 6 x (1 x ) , ≤ x ≤ 1 • f Y ( y ) = R ∞∞ f X,Y ( x,y ) dx = R 1 y 6 xdx = 3 x 2 1 y = 3(1 y ) 2 , ≤ y ≤ 1 Problem 23. p X,Y ( x,y ) = 4! x ! y !(4 x y )! 1 2 x 1 3 y 1 6 4 x y , ≤ x + y ≤ 4 The way to approach this problem is to try to just work the problem out directly from the definition: p X ( x ) = X all y p X,Y ( x,y ) = 4 x X y =0 4! x ! y !(4 x y )! 1 2 x 1 3 y 1 6 4 x y The problem here is that the value of x depends on y , since, for example, if y = 3, x can only be 0 or 1. There are two ways to handle this: The obvious way and the clever way. Probably the more obvious way is to just calculate the sum for every value. For example: p X (0) = ∑ all y p X,Y (0 ,y ) = ∑ 4 y =0 4!...
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This note was uploaded on 03/23/2008 for the course MATH 181A taught by Professor Xu,lily during the Spring '07 term at UCSD.
 Spring '07
 Xu,Lily
 Statistics

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