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Unformatted text preview: STA3007 Applied probability (0809) Solution to Assignment 1 Chapter 1 Problem 2.3 Let S r = ξ 1 + ··· + ξ r , where ξ k is the number of additional samples needed to observe the k th distinct element compared to observing the ( k 1)th distinct element. Assume that the k 1 distinct element have already been observed, then, let p k = Pr( ξ k = 1) = N ( k 1) N = 1 k 1 N , we have Pr( ξ k = n ) = p k (1 p k ) n 1 , n = 1 , 2 , ··· , which is a geometric distribution indexed by parameter p k . So, E( ξ k ) = 1 /p k . Finally, the given formula E( S r ) = E( ξ 1 ) + ··· + E( ξ r ) = 1 p 1 + ··· + 1 p r = N ( 1 N + 1 N 1 + ··· + 1 N r + 1 ) is established. Problem 2.7 We are given that Pr { U > u and W > w } = Pr { U > u } Pr { W > w } , which means that Pr { U > u and W > w } = [1 F U ( u )][1 F W ( w )] for all u,w . According to the definition for independence we should show that Pr { U ≤ u and W ≤ w } = F U ( u ) F W ( w ) for all u,w . Taking complements and using the fact that Pr( A ∪ B ) = Pr( A ) + Pr( B ) Pr( A ∩ B ) , 1 we have Pr { U ≤ u and W ≤ w } = 1 Pr { U > u or W > w } = 1 [Pr { U > u } + Pr { W > w }  Pr { U > u and W > w } ] = 1 [(1 F U ( u )) + (1 F W ( w )) (1...
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 Spring '11
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