Dr. Hackney STA Solutions pg 31

Dr. Hackney STA Solutions pg 31 - 3-4 Solutions Manual for...

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Unformatted text preview: 3-4 Solutions Manual for Statistical Inference Note, the product of these three limits is one. Three other terms are lim M lim M M -x 1/2 = 1 1/2 N -M N -M N - M - (K - x) N -K N x 1/2 = 1 and N lim = 1. The only term left is lim (M - x) (N - M - (K - x)) (N - K) M -x N -K K x K-x M/N p,M ,N = M/N p,M ,N lim N - M - (K - x) N -K K-x = px (1 - p)K-x . b. If in (a) we in addition have K , p 0, M K/N pK , by the Poisson approximation to the binomial, we heuristically get M x N -M K-x N K K x e- x p (1 - p)K-x . x x! c. Using Stirling's formula as in (a), we get lim M x N -M K-x N K -x N,M,K, M 0, KM N N = = e K x ex M x ex (N -M ) N K eK N,M,K, M 0, KM x! N N lim 1 lim x! N,M,K, M 0, KM N N KM N x K-x K-x e N -M N K K-x = = MK 1 x lim 1- N x! N,M,K, M 0, KM K N N e- x . x! 3.12 Consider a sequence of Bernoulli trials with success probability p. Define X = number of successes in first n trials and Y = number of failures before the rth success. Then X and Y have the specified binomial and hypergeometric distributions, respectively. And we have Fx (r - 1) = = = = = = P (X r - 1) P (rth success on (n + 1)st or later trial) P (at least n + 1 - r failures before the rth success) P (Y n - r + 1) 1 - P (Y n - r) 1 - FY (n - r). ...
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