What is the distribution of the limit of random variables in a problem involving Polya's Urn?

Question

I know this has been asked elsewhere, but I think the values or random variables are different or something.

From Williams' Probability with Martingales:

I proved that $M_n$ is a $\sigma(B_1, ..., B_n)-$martingale and $P(B_n = k) = \frac{1}{n+1}$

I guess $\Theta := \lim M_n$ exists $\because M_n$ is a nonnegative martingale.

So to give the distribution of $\Theta$, I tried to find the moment-generating function, though I'm not sure if we know that $M_{\Theta}(t)$ exists (also tried something else in answer posted):

$$M_{\Theta}(t) = E[\exp(t\Theta)]$$

$$= E[\exp(t\lim \frac{B_n + 1}{n+2})]$$

$$\stackrel{(*)}{=} E[\exp(t\lim \frac{B_{n+1} - B_n}{1})]$$

$$= E[\exp(t\lim (B_{n+1} - B_n))]$$

$$= E[\lim \exp(t (B_{n+1} - B_n))]$$

$$\stackrel{(**)}{=} \lim E[\exp(t (B_{n+1} - B_n))]$$

where

$$E[\exp(t (B_{n+1} - B_n))]$$

$$= E[E[\exp(t (B_{n+1} - B_n)) | B_n]]$$

$$= E[e^{t (0)}(1 - M_n) + e^{t (1)} M_n]$$

$$= E[(1 - M_n) + e^{t} M_n]$$

$$= E[(1 - M_n)] + E[e^{t} M_n]$$

$$= (1 - E[M_n]) + e^{t}E[ M_n]$$

$$= (1 - (\frac 1 2)) + e^{t}(\frac 1 2)$$

$$= (\frac 1 2) + e^{t}(\frac 1 2) = (1 + e^{t})(\frac 1 2)$$

Hence, $M_{\Theta}(t) = \lim (1 + e^{t})(\frac 1 2) = (1 + e^{t})(\frac 1 2)$.

Any mistakes?
Any suggestions for alternative solutions? Might I be able to evaluate $\lim (B_{n+1} - B_n)$?
This seems to say that $\Theta \sim Bernoulli (\frac 1 2)$. Is it? I tried $\Theta \sim Beta (\alpha, \beta)$ but it seems like $\alpha = \beta = 0$, which violates the assumptions of $\alpha, \beta > 0$, and I guess $Bernoulli(\frac 1 2)$ is kind of like $Beta(0,0)$ (p. 5-6)?
$(*)$ I used Stolz–Cesàro theorem. Is that right? If not, how else can I approach this problem? If so, is there a way to approach this problem without Stolz–Cesàro (it was never discussed in any class I attended in undergrad and grad)?
$(**)$ I think we can use something like a general bounded convergence theorem as $e^{t (B_{n+1} - B_n)}$ is bounded but not uniformly bounded right? If not, how else can I approach this?

To potential readers: this is 1. flawed, 2. in an obvious way, starting at the $(\ast)$ step. — Did, Jan 23 '16 at 10:48
Is there any reason to believe that $\lim\limits_{n\to\infty} (B_{n+1} - B_n)$ exists? — Daniel Fischer, Jan 23 '16 at 12:08
@DanielFischer Prove it does not then...? If it doesn't exist, how else can I approach this problem? — BCLC, Jan 23 '16 at 13:13
You know that $B_{n+1} - B_n \in {0,1}$. So if that were a convergent sequence, from some point onward, we'd always have $B_{n+1} = B_n$, or always $B_{n+1} = B_n + 1$. What do you know about $P(B_{n+1} = B_n)$ and $P(B_{n+1} = B_n + 1)$? — Daniel Fischer, Jan 23 '16 at 15:54
$(\ast)$ cannot be true since almost surely the sequence $(B_{n+1} - B_n)_{n\in \mathbb{N}}$ contains infinitely many ones and infinitely many zeros, so the right hand side simply doesn't make sense. — Daniel Fischer, Jan 24 '16 at 09:48

BCLC · Accepted Answer · 2016-02-12T22:58:26.703

$$M_{\Theta}(t) = E[\exp(t\Theta)]$$

$$= E[\exp(t\lim \frac{B_n + 1}{n+2})]$$

$$= E[\lim\exp(t \frac{B_n + 1}{n+2})]$$

$$= \lim E[\exp(t \frac{B_n + 1}{n+2})]$$

$$= \lim \frac{1}{n+1}[\exp(t \frac{1}{n+2}) + \exp(t \frac{2}{n+2}) + ... + \exp(t \frac{n+1}{n+2})]$$

Case 1: $t \ne 0$

$$= \lim \frac{a(n)}{(n+1)(1-a(n))} (1-a(n)^{n+1}), \ \text{where} \ a(n) := e^{\frac{t}{n+2}}$$

$$= \lim \frac{a(n)}{(n+1)(1-a(n))} \lim (1-a(n)^{n+1})$$

$$= \lim \frac{a(n)}{(n+1)(1-a(n))} (1-e^t)$$

$$= \frac{1-e^t}{-t}$$

$$= \frac{e^t-1}{t}$$

Case 2: $t = 0$

$$= \lim \frac{1}{n+1}[\exp((0) \frac{1}{n+2}) + \exp((0) \frac{2}{n+2}) + ... + \exp((0) \frac{n+1}{n+2})]$$

$$= \lim \frac{1}{n+1} (1)(n+1) = 1$$

This is the mgf of $Unif(0,1)$

What is the distribution of the limit of random variables in a problem involving Polya's Urn?

1 Answers1