2
$\begingroup$

I am trying to understand the assumptions under which I am allowed to apply Wald's equation for a sum of a random number $N$ of random variables $X_n$, $1\leq n\leq N$. There seem to be several versions of Wald's equation, and I am interested in the case where $N$ is a stopping time and the $X_t$ are not IID.

For example, assume that we have a simple random walk $X_n = X_{n-1} + Z_n - 1$, with $X_0 = 1$ and all $Z_n$ are IID binomial. Assume that I define some stopping time $N$ for $(X_n)$ which has finite expectation. One requirement for Wald's equation is that $\mathbb{E}[X_n 1_{\{N \geq n\}}] = \mathbb{E}[S_n] P(N\geq n)$ for all $n$. Is this assumption automatically fulfilled for the above random walk because $N$ is a stopping time (i.e., the indicator random varible $1_{\{N=n\}}$ is a function of $X_1, \ldots, X_n$)? My confusion comes from the fact that sometimes these requirements are stated in terms of filtrations, which I am not familiar with. For example, $N$ has to be a stopping time with respect to a filtration, and $X_n$ and $\mathcal{F}_{n-1}$ are independent for every $n$ (this is from Wikipedia for example). How do I define a filtration here and how do I check if $X_n$ is independent of $\mathcal{F}_{n-1}$? Is it possible to restate this somehow without resorting to a filtration?

$\endgroup$
1
$\begingroup$

xivaxy, I don't think that's correct.

Suppose $X_n = X_{n-1}$ with probability .5, and $X_n = 0$ with probability .5. We set $X_0 \ne 0$.

Let $N$ be the first $n$ such that $X_n = 0$. $N$ is a stopping time because $1_{\{N=n\}}$ is a function of $(X_1,...,X_n)$ only.

Then we have $\mathbb{E}[X_n 1_{\{N \ge n\}}] = \mathbb{E}[X_{n}]$ because $X_n \ne 0$ implies $1_{\{N \ge n\}} = 1$, but clearly $\mathbb{E}[X_{n}] > \mathbb{E}[X_{n}] P(N \ge n)$ for $n \ge 2$.

$\endgroup$
0
$\begingroup$

For the non-IID case the requirement $\mathbb{E}[X_n 1_{\{N \geq n\}}] = \mathbb{E}[X_n] P(N\geq n)$ is also automatically satisfied, since $1_{\{N \geq n\}}=1-1_{\{\cup_{m=1}^{n-1} (N = m) \}}$ is independent of $X_n$ by the definition of stopping time. However, $\mathbb{E}[S_n]=\mathbb{E}[X_1]\mathbb{E}[N]$ is not true anymore. The only thing that can be said is that $\mathbb{E}[S_n]=\mathbb{E}[\sum_{n=1}^N \mathbb{E}[X_n]]$.

$\endgroup$

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.