At time $0$, an urn contains $1$ black ball and $1$ white ball. At each time $1,2,3,...,$ a ball is chosen at random from the urn and is replaced together with a new ball of the same colour. Just after time $n$, there are therefore $n+2$ balls in the urn, of which $B_n+1$ are black, where $B_n$ is the number of black balls chosen by time $n$. Let $M_n = \frac{B_n + 1}{(n + 2)}$, the proportion of black balls in the urn just after time $n$. Prove that (relative to a natural filtration which you should specify) $M$ is a martingale.

There are several ways of solving the problem. One proof starts off by letting $\mathcal{F}_n=\sigma(B_1,...,B_n)$ and then stating $$E[M_n|\mathcal{F}_{n-1}]=P(B_n=B_{n-1})\frac{B_{n-1}+1}{n+2}+P(B_n=B_{n-1}+1)\frac{B_{n-1}+2}{n+2}$$ Intuitively speaking, I agree completely with the equation. It reminds of the basic definition of conditional expectation for discrete rv $X$ given $Y=y$: $$E(X|Y=y)=\sum_xx\cdot P(x|y)$$ How can one prove the equation rigorously? There are similar problems like De Moivre's martingale (https://en.wikipedia.org/wiki/Martingale_(probability_theory)) which make use of the same property of conditional expectation to state: $$E(Y_{n+1}|X_1,...,X_n)=p\left(\frac{q}{p}\right)^{X_n+1}q\left(\frac{q}{p}\right)^{X_n-1}$$ from which it is easy to then show that $(Y_n)_{n\in\mathbb{N}}$ is a martingale. What is the general formula that I am missing which applies to both of these problems?


1 Answer 1


We set $\mathscr{F}_k:=\sigma(B_u,u\leq k)$. The differences $(B_n-B_{n-1})_{n \in \mathbb{N}}$ can only take the values $\{0,1\}$. We have, for $f$ measurable and under the integrability condition $E[|f(B_n-B_{n-1})|]<\infty$ $$\begin{aligned}E[f(B_n-B_{n-1})|\mathscr{F}_{n-1}]&=\sum_{k \in \{0,1\}}f(k)P(B_n-B_{n-1}=k|\mathscr{F}_{n-1}) \end{aligned}$$ If we define $f(x)=\frac{x+1}{n+2}$ we obtain $$\begin{aligned}E[f(B_n-B_{n-1})|\mathscr{F}_{n-1}]&=\underbrace{P(B_n-B_{n-1}=0|\mathscr{F}_{n-1})}_{:=p_{0,n-1}}\frac{1}{n+2}+\underbrace{P(B_n-B_{n-1}=1|\mathscr{F}_{n-1})}_{:=p_{1,n-1}}\frac{2}{n+2}\end{aligned}$$ By rearranging the terms $$\begin{aligned}E\bigg[\frac{B_n+1}{n+2}\bigg|\mathscr{F}_{n-1}\bigg]&=\frac{B_{n-1}}{n+2}+p_{0,n-1}\frac{1}{n+2}+p_{1,n-1}\frac{2}{n+2}=\\ &=p_{0,n-1}\frac{B_{n-1}+1}{n+2}+p_{1,n-1}\frac{B_{n-1}+2}{n+2}\end{aligned}$$

Similarly, in the other problem we obtain for $\mathscr{G}_k:=\sigma(X_u,u\leq k)$

$$\begin{aligned}E[Y_nY_{n-1}^{-1}|\mathscr{G}_{n-1}]&=E\bigg[\bigg(\frac{q}{p}\bigg)^{X_n-X_{n-1}}\bigg|\mathscr{G}_{n-1}\bigg]=\\ &=\sum_{k \in \{-1,1\}}\bigg(\frac{q}{p}\bigg)^kP(X_n-X_{n-1}=k|\mathscr{G}_{n-1})=\\ &=\bigg(\frac{q}{p}\bigg)^{-1}q+\bigg(\frac{q}{p}\bigg)p\end{aligned}$$ and so $$E[Y_n|\mathscr{G}_{n-1}]=Y_{n-1}\bigg(\frac{q}{p}\bigg)^{-1}q+Y_{n-1}\bigg(\frac{q}{p}\bigg)p$$

  • $\begingroup$ Thanks for the response. How did you come up with the first equation (specifically with respect to the role of the sigma algebra)?. After rearranging the terms, how do you get the final result? $\endgroup$ Oct 2, 2022 at 15:31
  • $\begingroup$ Since probabilities sum to one, $B_{n-1}=(p_{0,n-1}+p_{1,n-1})B_{n-1}$. For the expectation formula, I refer to theorems 8.37 and 8.38 in Klenke's book. @UniformIntegrability $\endgroup$
    – Snoop
    Oct 2, 2022 at 15:45
  • $\begingroup$ Great!. Does Lemma 8.10 in Klenke's book play any role in martingale problems?. I was focusing on that lemma in trying to solve this problem. Thanks again $\endgroup$ Oct 2, 2022 at 15:52
  • $\begingroup$ @UniformIntegrability it depends on the problem itself really, but it is safe to say all properties of conditional expectation are useful in martingale problems $\endgroup$
    – Snoop
    Oct 2, 2022 at 15:57
  • $\begingroup$ Yes, I suppose so. The lemma requires a partition of $\Omega$. I thought it could apply in this problem $\endgroup$ Oct 2, 2022 at 16:01

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