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I was trying to solve the following exercise:

Suppose that the sequence of independent events $\{A_i\}$ satisfies $\phi(n) = \sum_{i=1}^{n} P(A_i) \rightarrow \infty \text{ as } n \rightarrow \infty$ and let $\tau_k :=\min\{n | \sum_{i=1}^{n} 1(A_i) = k\}$. By applying Doob's stopping time theorem to an appropriate martingale, prove the more quantitative result that $E[\phi(\tau_k)] = k \text{ for all } k \geq 1$

Here, $1(\cdot)$ is the indicator function. By Doob's stopping theorem, I think the book is referring to the following theorem (which is just named "stopping time theorem"):

If $(M_n)$ is a martingale with respect to $(\mathcal{F}_n)$, then the stopped process $(M_{\tau \wedge n})$ is also a martingale with respect to $(\mathcal{F}_n)$.

First of all, let me say that I think my proof is wrong. I don't use the stopping time theorem or the fact that $\phi(n)$ diverges for $n \rightarrow \infty$ and it just looks too easy to be true. In any case, I'd welcome hints and constructive criticism.

My attempt at a proof:

Since $P(A) = E[1(A)]$, we can write $\phi(\tau_k) = \sum_{i=1}^{\tau_k} P(A_i) = \sum_{i=1}^{\tau_k} E[1(A_i)]$ Now, $\tau_k$ is the smallest $n$ such that from the sequence $\{A_1, A_2, \ldots, A_n\}$, exactly $k$ of them occur. So $\sum_{i=1}^{\tau_k} E[1(A_i)] = \sum_{j=1}^{k} E[1(A_{i_j})]$ for indices $i_j$. Since, by definition of $\tau_k$, we know that the events $A_{i_j}$ occur for $j = 1, \ldots, k$, we have that $\phi(\tau_k) = \underbrace{1 + 1 + \ldots + 1}_{k} = k$ and taking the expectation of both sides gives $E[\phi(\tau_k)] = E[k] = k$

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    @Henry: Thanks. It was natural to expect a simple application of the optional stopping theorem...2011-03-16

1 Answers 1

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  1. Doob's stopping theorem is probably intended to mean the optional stopping theorem, named in honour of Joseph Leo Doob.
  2. You implicitly use $\phi(n) \rightarrow \infty$ since if it had some finite upper bound $L$ then for $k>L$ you would not be able to have $\phi(\tau_k) = k$.
  3. When you say $\sum_{i=1}^{\tau_k} E[1(A_i)] = \sum_{j=1}^{k} E[1(A_{i_j})]$, it is unclear whether you are using $A_i$ to mean unknown future events at the start or known past events at the stopping time. It does not matter which because of the optional stopping theorem as applied to martingales (meeting some additional conditions which are satisfied in this case), namely that the expected value of such a martingale at a stopping time is equal to its initial value. I would expect you to find something like $E[X_{\tau \wedge n}]=E[X_0]$ or $E[X_{\tau}]=E[X_0]$ somewhere in your book.
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    I guess I just don't understand stopping times well enough to see how the theory naturally fits into solving this problem. I'll keep trying. Thanks for your time.2011-03-16