The problem 20.a asks to prove that the modified Polya urn process described in Example 5 leads to Bose Einstein Statistics.
Then problem 20.b asks to prove the modified polya url process yields MB and FD statistics if the cards are replaced and not replaced respectively.
I am sure of the derivation of 20.a below (i.e. $P_{Bose-Einstein}$). Can someone please help checking the remaining 2 (i.e. $P_{Maxwell-Boltzmann}$ and $P_{Fermi-Dirac}$).
$N = \sum\limits_{i =1}^nN_i$
$P_{Bose-Einstein} = \bigg[\frac{1}{n}\frac{2}{n+1}...\frac{N_1}{n+N_1-1}\bigg].\bigg[\frac{1}{n+N_1}\frac{2}{n+N_1+1}...\frac{N_2}{n+N_1+N_2-1}\bigg]...\bigg[\frac{1}{n+N_1+N_2..N_{n-1}}\frac{2}{n+N_1+N_2..N_{n-1}+1}...\frac{N_n}{n+N_1+N_2..N_{n-1}+(N_{n}-1)}\bigg].\frac{N!}{N_1!.N_2!...N_n!} = \frac{N!}{P(n+N-1,n-1)} = \frac{1}{C(n+N-1, N)}$
$P_{Maxwell-Boltzmann} = \big[\frac{1}{n}\frac{1}{n}...\frac{1}{n}\big].\frac{N!}{N_1!.N_2!...N_n!} = \frac{N!}{n^N.N_1!N_2!...N_n!}$
$P_{Fermi-Dirac} = \big[\frac{1}{n}\frac{1}{(n-1)}...\frac{1}{(n-(N_1+N_2..N_{n-1}+N_{n}-1))}\big].\frac{N!}{N_1!.N_2!...N_n!} = \frac{N!}{\frac{n!}{(n-N)!}} = \frac{1}{C(n, N)}$ (since $N_i \in \{0,1\}$)
=== Here is the Section 4, Problem 20 ===
- (a) In Example 5, a mathematical model is discussed which claims to give a dis- tribution of identical balls into boxes in such a way that all distinguishable arrangements are equally probable (Bose-Einstein statistics). Prove this by showing that the probability of a distribution of N balls into n boxes (accord- ing to this model) with N1 balls in the first box, N2 in the secoPnd, · · · , Nn in the nth, is 1/C(n−1+N, N) for any set of numbers Ni such that ni=1 Ni = N.
(b) Show that the model in (a) leads to Maxwell-Boltzmann statistics if the drawn card is replaced (but no extra card added) and to Fermi-Dirac statistics if the drawn card is not replaced. Hint: Calculate in each case the number of possible arrangements of the balls in the boxes. First do the problem of 4 particles in 6 boxes as in the example, and then do N particles in n boxes (n > N) to get the results in Problem 19.
==== Here is the referred Example 5 =====
Example 5. In Example 4 we found the same answer for the probability that two partic- ular boxes were empty whether or not we considered the balls distinguishable. This was true because the allowed distinguishable arrangements were equally probable. Without the restriction of one ball or none per box, all distinguishable arrangements are not equally probable according to the methods of Examples 3 and 4. For exam- ple, the probability of all balls in box 1 is 1/64; compare this with the probability of no balls in the first 2 boxes and one ball in each of the other 4 boxes, which is 4! ÷ 64 = 1 . We see that the concentrated arrangements (all or several balls in one 54 box) are less probable than the more uniform arrangements. Now we are going to try to imagine a situation in which all distinguishable ar- rangements are equally probable. Suppose the 6 boxes are benches in a waiting room and the 4 balls are people who are going to come in and sit on the benches. Then if the people are friends, there will be a certain tendency for them to sit together and the probabilities we have been calculating will not apply—the probabilities of the concentrated arrangements will increase. Consider the following mathematical model. (This is a modification of P ́olya’s urn model.) We have 6 boxes labeled 1 to 6, and 4 balls. From 6 cards labeled 1 to 6 we draw one at random and place a ball in the box numbered the same as the card drawn. We then replace the card and also add another card of the same number so that there are now 7 cards, two with the number first drawn. We now select a card at random from these 7, put a ball in the corresponding box and again replace the card adding a duplicate to make 8 cards. We repeat this process two more times (until all balls are distributed). Then the probability that all balls are in box 1 is 1 · 2 · 3 · 4. The probability 6789 ofoneballineachofthefirst4boxesis 1 ·1 ·1 ·1 ·4!(here 1 ·1 ·1 ·1 isthe 6789 6789 probability that the first ball is in box 1, the second in box 2, etc.; we must add to this the probability that the first ball is in box 3, the second in box 1, etc.; there are 4! such possibilities all giving one ball in each of the first 4 boxes). We see that the distributions “all balls in box 1” and “one ball in each of the first 4 boxes” are equally probable. Further calculation (Problem 20) shows that all distinguishable arrangements are equally probable.