Pitman MTH 135/ STA 104 Probability Week 4
Read:  Pitman sections 2.4-2.5

Poisson & Hypergeometric Distributions

* Poisson Approximation to the Binomial
-----------

Shortcomings of CLT:  if p is close to 0 or 1, then \sqrt{n p q} is small.
Normal is symmetric, for example, while binomial for small p is packed up
against 0.

Text looks at binomial with n large and mu=1, i.e., p=1/n; I can do fishing
example, and sneak in exponential distribution.

Recall exponential sum, \sum_{k=0}^oo   a^k/k!  =  exp(a)

---------------

What's the mode?

Is there ever a double-mode?

                            k-1              k
                          mu              mu
  P(k-1) < P(k)  <==>  -----------  <  ----------
                         (k-1) !          k !


                                           mu
                 <==>       1       <  ----------
                                            k

                 <==>       k < mu

SO, mode is largest integer <= mu; if mu is an integer, then (mu-1, mu) is a
double mode.

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What happens when mu gets big?

Looks like the normal, with mean mu and variance sigma^2 = mu.

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* Random Sampling:  WITH and WITHOUT Replacement


With replacement:  Binomial,  mu = G/(G+B)   ("good" and "bad" in
                                                       population), so

      P[g good b bad in n=b+g tries] =  C(n,g) G^g  B^b  / N^n

Without replacement:
      Two approaches:
       one-at-a-time:

                                     G   G-1     G+1-g    B   B-1       B+1-b
                             C(n,g) ---- --- ... ----- x ---- ----- ...------ 
                                     N   N-1     N+1-g    N-g N-g-1     N+1-n


                                n!      G!       B!    (N-n)!
                           =  ------ -------- -------- ------
                              g! b!   (G-g)!   (B-b)!    N!

and "numbers of subsets" approach:

                             C(G,g)    C(B,b)
                           --------------------
                                 C(N,n)

                                G!         B!      (N-n)! n!  
                           =  -------- ---------- -----------
                              g!(G-g)!  b! (B-b)!      N!