Let an urn contain w white and b black balls. Draw a ball randomly from the urn and return it together with another ball of the same color. Let $b_n$ be the number of black balls and $w_n$ the number of white balls after the n-th draw-and-replacement. Let $X_n$ be the relative proportion of white balls after the n-th draw-and-replacement.

I start with b=w=1, so the total number of balls after the n-th draw-and-replacement is n+2. Now I want to find the limit distribution of $X_n$; I already showed that $X_n$ is a martingale and that it converges a.s.. It is

$X_n = \dfrac{w_n}{n+2}$ for $n \in \mathbb{N}_0$.

I've read that the limit distribution is a beta distribution, but I don't know how to get there.
I could write $w_n$ as the sum of $Y_i$ where $Y_i$ is 0, if the i-th ball is black and 1, if the i-th ball is black. Then I'd have

$w_n = 1+\sum_{i=1}^{n} Y_i$.

Does this help? How can I proceed?

Thanks! :)


Refer to this?

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

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

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

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

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

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 $Unif(0,1)$

  • 2
    $\begingroup$ @Did Edited. I accounted for $t=0$ this time $\endgroup$ – BCLC Feb 12 '16 at 22:59

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