Consider the Ehrenfest urn model in which M molecules are...

Consider the Ehrenfest urn model in which $M$ molecules are distributed among two urns, and at each time point one of the molecules is chosen at random and is then removed from its urn and placed in the other one. Let $X_{n}$ denote the number of molecules in urn 1 after the $n$ switch and let $\mu_{n}=E\left[X_{n}\right] .$ Show that (i) $\mu_{n+1}=1+(1-2 / M) \mu_{n}$ (ii) Use (i) to prove that $$ \mu_{n}=\frac{M}{2}+\left(\frac{M-2}{M}\right)^{n}\left(E\left[X_{0}\right]-\frac{M}{2}\right) $$

Transcript

00:01 Once again, welcome to new problem.

00:06 Consider a probability model, and this is the rnfest and model.

00:18 And remember, this model has m molecules, and these molecules are distributed between.

00:28 So these molecules are distributed between two arms.

00:34 This is the first one.

00:35 This is the second one.

00:40 And at some specific point in time, one of the molecules, one of the molecules is chosen.

01:05 It's chosen at random.

01:08 It's chosen at random.

01:10 And is then removed from its urn, it's then removed from its urn and placed in another one.

01:35 Xn represents number of molecules in the first on after the first on after the the nth switch.

01:56 And remember, mu sub n is the same as the expected value of xn.

02:03 So in this case, in part a, prove that mu n plus 1 equals to 1 plus 1 minus 2 all over m, mu sub n, and then in part b, you also want to show that mu sub n is the same is m over 2.

02:37 Remember m stands for the number of molecules m minus 2 all over n raised to n times the expected value of x knot minus n all over 2.

02:54 So we want to show that that's the problem, that's the case.

03:00 So we're going to go right ahead and start on the first part where we know that x to the n is the number of molecules in 1 after nth switch.

03:22 So you're switching the uns and the molecules and so that's what it is.

03:26 And this is the total number of molecules in on 1 and 1 and 2.

03:37 So we have 1 and 2.

03:43 So we're going to start by saying piii equals to probability of xn1 equals to i, given that the probability of xn equals to i, and that probability is 0.

03:58 And then p .i.

04:00 1 plus i.

04:03 Remember, you have two uns.

04:06 And then we're doing the probability for this one equals to i plus 1.

04:12 So you can see when these two are is, we just have i here...

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