hw8_soln

# hw8_soln - Homework 8 Solutions Problem Solutions Yates and...

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Unformatted text preview: Homework 8 Solutions Problem Solutions : Yates and Goodman, 4.8.3 4.8.4 4.8.6 4.9.3 4.9.4 4.10.5 and 4.10.8 Problem 4.8.3 Solution Given the event A = { X + Y ≤ 1 } , we wish to find f X,Y | A ( x, y ). First we find P [ A ] = Z 1 Z 1- x 6 e- (2 x +3 y ) dy dx = 1- 3 e- 2 + 2 e- 3 (1) So then f X,Y | A ( x, y ) = ( 6 e- (2 x +3 y ) 1- 3 e- 2 +2 e- 3 x + y ≤ 1 , x ≥ , y ≥ otherwise (2) Problem 4.8.4 Solution First we observe that for n = 1 , 2 , . . . , the marginal PMF of N satisfies P N ( n ) = n X k =1 P N,K ( n, k ) = (1- p ) n- 1 p n X k =1 1 n = (1- p ) n- 1 p (1) Thus, the event B has probability P [ B ] = ∞ X n =10 P N ( n ) = (1- p ) 9 p [1 + (1- p ) + (1- p ) 2 + ··· ] = (1- p ) 9 (2) From Theorem 4.19, P N,K | B ( n, k ) = ( P N,K ( n,k ) P [ B ] n, k ∈ B otherwise (3) = (1- p ) n- 10 p/n n = 10 , 11 , . . . ; k = 1 , . . . , n otherwise (4) The conditional PMF P N | B ( n | b ) could be found directly from P N ( n ) using Theorem 2.17. However, we can also find it just by summing the conditional joint PMF. P N | B ( n ) = n X k =1 P N,K | B ( n, k ) = (1- p ) n- 10 p n = 10 , 11 , . . . otherwise (5) From the conditional PMF P N | B ( n ), we can calculate directly the conditional moments of N given B . Instead, however, we observe that given B , N = N- 9 has a geometric PMF with mean 1 /p . That is, for n = 1...
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hw8_soln - Homework 8 Solutions Problem Solutions Yates and...

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