hw6 - MATH 471 HW 6 Solutions Pengsheng Ji TA Office Hours...

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Unformatted text preview: MATH 471 HW 6 Solutions Pengsheng Ji TA Office Hours: 4:05-5:05 PM Tuesday 5:20-6:20 PM Thursday 218 Mallot Hall 3.3.5 X has density function f (x) = λe−λx and r(x) = x1/α has the inverse s(y ) = y α . By equation (3.1) the density function of r(X ) is given by f (s(y )) · s (y ) = λe−λy · αy α−1 , for y ≥ 0. 3.3.11 (a)For y > 0, P (Y ≤ y ) = P (−y ≤ X ≤ y ) = F (y ) − F (−y ). Differentiating it on y , we get the density function of Y is f (y ) + f (−y ) for 0 < y < 1 and 0 elsewhere. √ √ √ √ (b) For y > 0, P (Y ≤ y ) = P (− y ≤ X ≤ y ) = F ( y ) − F (− y ). By differentiating it on y , we get the density of Y is 1 √ √ [f ( y ) + f (− y )] · · y −1/2 , 2 for 0 < y < 1 and 0 elsewhere. 3.4.10 P (X + Y < 1) = 3.4.12 The event that no piece is longer than 1/2 is {0 < X < 1/2, Y − X < 1/2, 1 − Y < 1/2}, which is a triangle with vertices (1/2, 0), (1/2, 1/2) and (0, 1/2). probability is 2 times the area of the triangle or 1/4. 4.1.10 So the 1 0 0 1−x α 6xy 2 dydx = 1 0 2x(1 − x)3 dx = 1/20 For every Sunday the probability that at lease one kite flies is 1 − (1 − 1/5) · (1 − 1/3) = 7/15. Let X = the number of days we have to wait until at least on kite flies. Then X has a geometric distribution with success probability 7/15. Hence by Example 1.11 we have EX = 15/7. ...
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This note was uploaded on 09/22/2010 for the course MATH 413 at Cornell.

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hw6 - MATH 471 HW 6 Solutions Pengsheng Ji TA Office Hours...

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