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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

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

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! :)

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2 Answers 2

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Suppose that we have started with $w$ white balls and $b$ black balls. Then

\begin{align*} \mathbb{P}(w_n = w+k) &= \binom{n}{k} \frac{w(w+1)\dots(w+k-1)b(b+1)\dots(b+n-k-1)}{(w+b)(w+b+1)\dots(w+b+n-1)} \\ &= \frac{1}{B(w, b)} \binom{n}{k} \frac{\Gamma(w+k)\Gamma(b+n-k)}{\Gamma(w+b+n)} \\ &= \frac{1}{B(w, b)} \frac{k^{w-1} (n-k)^{b-1}}{n^{w+b-1}} \frac{E_k(w)E_{n-k}(b)}{E_n(b+w)} , \end{align*}

where $\Gamma(\cdot)$ is the gamma function, $B(\alpha, \beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$ is the beta function, and

$$ E_n(z) := \frac{\Gamma(n+z)}{n!n^{z-1}}. $$

Note that $E_n(z) \to 1$ as $n\to\infty$. So, if we write $p_k = k/n$, then the m.g.f. of $X_n$ is explicitly given by

\begin{align*} \mathbb{E}[e^{\lambda X_n}] = \frac{1}{B(w, b)} \sum_{k=0}^{n} \exp\biggl( \lambda \frac{p_k + w/n}{1 + (w+b)/n} \biggr) p_k^{w-1}(1 - p_k)^{b-1} \frac{1}{n} \cdot \frac{E_k(w)E_{n-k}(b)}{E_n(b+w)}. \end{align*}

Letting $n \to \infty$, this converges to

\begin{align*} \mathbb{E}[e^{\lambda X_{\infty}}] = \frac{1}{B(w, b)} \int_{0}^{1} e^{\lambda p} p^{w-1}(1 - p)^{b-1} \, \mathrm{d}p. \end{align*}

From this, we read out that the distribution of $X_{\infty}$ has the density

$$ f(p) = \frac{1}{B(w, b)} p^{w-1}(1 - p)^{b-1} \mathbf{1}_{(0,1)}(p), $$

proving that the limit distribution is $\operatorname{Beta}(w, b)$.

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  • $\begingroup$ could you justify the step where you let $n \to \infty$? Since $k$ is fixed, then $p_k \to 0$ as $n\to\infty$. $\endgroup$
    – s l
    Feb 28, 2023 at 4:15
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    $\begingroup$ @sl, Think of the expansion of $\mathbb{E}[e^{\lambda X_n}]$ as a Riemann sum of the form $$\sum_{k=0}^{n} f\left(\frac{k}{n}\right)\frac{1}{n} \cdot a_{n,k},$$ where $f$ is a continuous function on $[0, 1]$ and $(a_{n,k}:0\leq k\leq n<\infty)$ is a bounded double-sequence such that $a_{n,k} \to 1$ whenever $n,k\to\infty$ (hence can be considered as a 'perturbation'). Then we can prove that this indeed converges to $$\int_{0}^{1} f(x) \, \mathrm{d}x. $$ $\endgroup$ Feb 28, 2023 at 4:49
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Refer to this?


Assuming $B_n$ is uniform on $\{0,1,...,n\}$ (proven by induction):

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

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

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

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

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

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 $Unif(0,1)$

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    $\begingroup$ @Did Edited. I accounted for $t=0$ this time $\endgroup$
    – BCLC
    Feb 12, 2016 at 22:59
  • $\begingroup$ This answer takes as granted the fact that $B_n$ has a uniform distribution on the set ${0, \ldots, n}$ from which the fact that the limit in law (in fact, an a.s. limit by martingale cvg) $\Theta$ of $B_n/n+1$ is uniform on [0,1] follows immediately. $\endgroup$
    – Olivier
    May 27, 2020 at 15:15
  • $\begingroup$ In other words this is not an answer. See Durrett "Probability, theory and examples", section 4.3.2 in the last ediition (available online) for the classical method relying on the combinatorics of the problem+ exchangeability. $\endgroup$
    – Olivier
    May 27, 2020 at 15:17
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    $\begingroup$ Please read carefully my comment : the answer assumes $B_n$ is uniform on $\{0...n\}$ to conclude $(B_n+1)/(n+2)$ converges to the uniform distribution (this time on the interval) : this is the trivial part in the analysis of the Polya urn; the "difficulty" here is to prove that $B_n$ is uniform on $\{0...n\}$ $\endgroup$
    – Olivier
    Dec 15, 2020 at 18:16
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    $\begingroup$ That $B_n$ is uniform on $\{0...n\}$ is typically done by induction on n, see any textbook with Polya urn (Durrett book for instance). $\endgroup$
    – Olivier
    Dec 15, 2020 at 18:17

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