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Unformatted text preview: MA 316: MATHEMATICAL PROBABILITY ASSIGNMENT #7 SOLUTIONS: SECTIONS 2.1, 2.2, 2.3, and 2.4 FALL 2003 2.1: If f ( x 1 ,x 2 ) = 4 x 1 x 2 for 0 < x 1 < 1 , < x 2 < 1 is the joint density function of X 1 and X 2 then P (0 < X 1 < 1 / 2 , 1 / 4 < X 2 < 1) = Z 1 / 2 Z 1 1 / 4 4 x 1 x 2 dx 2 dx 1 = 15 64 . For the second part and as the hint suggests, notice that P ( X 1 = X 2 ) is the volume under the surface f ( x 1 ,x 2 ) = 4 x 1 x 2 and above the line segment 0 < x 1 = x 2 < 1 in the x 1 x 2 plane, which necessarily must be 0. For the third part, we have P ( X 1 < X 2 ) = Z 1 Z x 2 4 x 1 x 2 dx 1 dx 2 = 1 2 and the integral for the fourth part is identical to the third part, so P ( X 1 X 2 ) = 1 2 . 2.5: If R g ( x ) dx = 1 then the function f ( x 1 ,x 2 ) = 2 g ( p x 2 1 + x 2 2 ) p x 2 1 + x 2 2 for 0 < x 1 < and 0 < x 2 < is a density function since converting this integral to one involving polar coordinates using the standard substitution r 2 = x 2 1 + x 2 2 gives Z / 2 Z 2 g ( r ) r rdrd = 2 Z / 2 Z g ( r ) drd = 2 Z / 2 d = 1 . 2.9: For part a), given the table as in the text we can compute the density function for X 1 as f 1 (0) = 7 / 12 and f 1 (1) = 5 / 12 and for X 2 as f 2 (0) = 4 / 12, f 2 (1) = 5 / 12, and f 2 (2) = 3 / 12. For part b) notice that P ( X 1 + X 2 = 1) = 3 / 12 + 2 / 12 = 5 / 12 since the event X 1 + X 2 = 1 happens when x 1 = 0 ,x 2 = 1 and x 1 = 1 ,x 2 = 0 . ....
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This note was uploaded on 09/10/2009 for the course STATS 517 taught by Professor Song during the Fall '07 term at Purdue UniversityWest Lafayette.
 Fall '07
 Song
 Probability

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