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Just in case I will remind what is called Polya urn: Suppose you have an urn containing one red and one blue ball. You draw one at random. Then if the ball is red, put it back in the urn with an additional red ball, otherwise put it back and add a blue ball.

Now denote by $N$ number of balls drawn before first appearance of a blue one. What will be $\operatorname{E}(N+2)^{-1}$?

I can find a mean for the $N$ itself just by $$\operatorname{E}(N) = \left(\frac 1 2 \cdot\frac 1 3 \right)\cdot1+ \left( \frac 1 2 \cdot \frac 2 3 \cdot\frac 1 4 \right)\cdot2+\cdots = \sum_{i=1}^\infty \left( \prod_{k=1}^i \frac{k}{k+1}\cdot\frac{1}{i+2}\right)\cdot i$$

but get stuck what to do with $\operatorname{E}(N+2)^{-1}$. Can anyone explain it?

Michael Hardy
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Hasek
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3 Answers3

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Applying the martingale optimal stopping theorem one can show that $\mathbb{E}(T+2)^{-1} = \frac{1}{4}$ (here $T$ stands for the number of balls drawn until the first blue ball appears)

Theorem (Optimal Stopping theorem)

Let $Y_{n}$ be a martingale adapted to filtration $\{ \mathcal{F_{n}} \}$ and let $T$ be the stopping time. Then if

(1) $Y_{n}$ is uniformly integrable.

(2) $\mathbb{P}(T < \infty) = 1$

(3) $\mathbb{E}(|Y_{T}|) < \infty$

then $\mathbb{E}(Y_{n}) = \mathbb{E}(Y_{0})$

So, since $Y_{n}$ is uniformly integrable it suffcies to check the second condition, indeed $$P(T < \infty) = \frac{1}{2} \cdot \frac{2}{3} \ldots \cdot \frac{n-1}{n} = \frac{1}{n} \to 0$$ as $n \to \infty$. By the optimal stopping theorem, which is to say that $$\mathbb{E}(\frac{T}{T+2}) = \frac{1}{2}$$ the result follows.

hyperkahler
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1

Let $X_k$ be the number of red balls in the $k$th trial, so that $X_k = 0 \text{ or } 1.$

Here's a useful lemma:

Let $R$ be uniformly distributed in the interval $(0,1)$ and let the conditional distribution of $Y_k$, for $k=1,2,3,\ldots$ be given by $$ \Pr(Y_k = 1\mid R) = R \text{ and } \Pr(Y_k = 0 \mid R) = 1-R $$ and $Y_1,Y_2,Y_3,\ldots$ are conditionally independent given $R.$ Then the marginal (i.e. "unconditional") distribution of the sequence $Y_1,Y_2,Y_3,\ldots$ is the same as that of $X_1,X_2,X_3,\ldots\ $.

Thus we can regard $N$ as the number of indices $k$ for which $Y_k=1$ before the first index $k$ for which $Y_k=0.$ Thus we have \begin{align} \Pr(N=n) & = \operatorname{E}(\Pr(N=n \mid R)) = \operatorname{E}( R^n(1-R) ) = \int_0^1 r^n (1-r)\,dr = \int_0^1 (r^n - r^{n+1})\,dr. \end{align} Then the expected value: \begin{align} \operatorname{E} \left( \frac 1 {N+2} \right) & = \sum_{n=0}^\infty \frac 1 {n+2} \Pr(N=n) = \sum_{n=0}^\infty \frac 1 {n+2} \left( \frac 1 {n+1} - \frac 1 {n+2} \right) \\[10pt] & = \sum_{n=0}^\infty \left( \frac 1 {(n+1)(n+2)} - \frac 1 {(n+2)^2} \right) \\[10pt] & = \sum_{n=0}^\infty \left( \frac 1 {n+1} - \frac 1 {n+2} \right) - \sum_{n=0}^\infty \frac 1 {(n+2)^2} \text{ (partial fractions)} \\[10pt] & = 1 - \left( \frac {\pi^2} 6 - 1 \right) = \cdots \end{align} The first term, $1,$ comes from telesciping. The fraction $\pi^2/6$ is not easy to derive from scratch but is widely known.

And you still have the problem of proving the lemma.

Michael Hardy
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0

Let $N$ represents the first time that a white ball being drawn, we have $$\{N=n\}=\left\{R_0=1, R_1=2,R_2=3,\dots,R_{n-1}=n,R_{n}=n\right\}$$ which is finite. So, $N$ is a stopping time. Then, \begin{align*} E\left[\frac{1}{N+2}\right]&=\sum_{n=1}^{\infty}\left(\frac{1}{n+2}\right)P(N=n)\\ &=\sum_{n=1}^{\infty}\left(\frac{1}{n+2}\right)\left(\frac{1}{n(n+1)}\right)\\ &=\sum_{n=1}^{\infty}\frac{1}{2}\left(\frac{1}{n+1}\right)\left(\frac{1}{n}-\frac{1}{n+2}\right)\\ &=\frac{1}{4} \end{align*}

Simple
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