I am trying to understand the implications of the optimal stopping theorem, which is why I tought of the following problem.

Consider the continuous-time Martingale $X = (X_t)_{t \geq 0}$ and the stopping time $T$. Assume that all conditions of the optimal stopping theorem are satisfied, i.e., we know $\mathbb{E}[X_T] = X_0$.

What can we say about the conditional expectation $\mathbb{E}[X_T | T \leq t]$?

I think $\lim_{t\rightarrow \infty} \mathbb{E}[X_T | T \leq t] = X_0$ holds.

But what about finite $t$? I think that $\mathbb{E}[X_T | T \leq t] \neq X_0$, but I can't come up with an example.

A related question, what can we say about $\mathbb{E}[X_t | T \geq t]$?


1 Answer 1


A case when $X_T$ and $T$ are not independent is when $T$ is the hitting time of $\{a,b\}$ by $(X_t)$ a Brownian motion or a symmetric random walk, and $X_0\ne\frac12(a+b)$.

The simplest example might be $(X_n)$ the simple symmetric random walk starting from $X_0=1$ and $T=\inf\{n\mid X_n\in\{0,3\}\}$.

Then, the event $[T\gt n]$ corresponds to the unique path $(1,2,1,2,\ldots)$ hence $E[X_T\mid T\gt n]=E[X_T\mid X_0=x_n]=x_n$ where $x_{2n}=1$ and $x_{2n-1}=2$. Thus, $$ E[X_T;T\leqslant n]=E[X_T]-E[X_T\mid T\gt n]P[T\gt n]=1-2^{-n}x_n, $$ and $$ E[X_T\mid T\leqslant n]=\frac{2^n-x_n}{2^n-1}, $$ Thus, $E[X_T\mid T\leqslant n]\ne X_0$ for every odd integer $n$.

Likewise, $E[X_n\mid T\geqslant n]=E[X_{n-1}+X_n-X_{n-1}\mid T\gt n-1]=x_{n-1}$.

  • $\begingroup$ Thank you very much. Could you please explain the notation $E[A ; B]$, I have never seen that before. Does it mean $E[A 1_{B}]$, with $1$ the indicator-function? $\endgroup$
    – Hank
    Commented Nov 22, 2013 at 9:48
  • $\begingroup$ Exactly. Sorry for the inconvenience. $\endgroup$
    – Did
    Commented Nov 22, 2013 at 9:49

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