4
$\begingroup$

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! :)

$\endgroup$
0
$\begingroup$

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 $Unif(0,1)$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.