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Unformatted text preview: 34 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 Kx M/N p,M ,N = M/N p,M ,N lim N  M  (K  x) N K Kx = px (1  p)Kx . 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 Kx N K K x e x p (1  p)Kx . x x! c. Using Stirling's formula as in (a), we get lim
M x N M Kx 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 Kx Kx e N M N
K Kx = = 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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 Spring '12
 Dr.Hackney
 Statistics

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