1
$\begingroup$

If $X_n$ is a martingale and $T$ is a stopping time with $P(T \leq n) = 1$ for some $n$, then the optional stopping theorem applies: $\mathbb{E}(X_T) = \mathbb{E}(X_0)$. Sometimes it's hard to show that the stopping time $T$ is bounded a.s, so instead I believe you can show that $|X_{\min(T,n)}| \leq k$ for some $k$. Is this enough to get the conclusion of the OST (that the expectations at time $T$ and $0$ are equal)? I know that in this case the DCT implies that $EX_{\min(T,n)}$ exists, but so what?

(I have looked everywhere for such a definition but I can't find it)

$\endgroup$
5
  • $\begingroup$ You should be able to apply dominated convergence to $\mathbf{E} [X_{\min(T,n)}-X_0]$. $\endgroup$
    – ShawnD
    Commented Feb 16, 2012 at 18:07
  • $\begingroup$ So DCT says that $\mathbb{E}[X_{\min(T,n)} - X_0] \to \mathbb{E}[X_T - X_0].$ Still don't see where to go.. $\endgroup$
    – Court
    Commented Feb 16, 2012 at 18:25
  • $\begingroup$ $X_{\min(T,n)}$ is a martingale, so $\mathbf{E}[X_{\min(T,n)}-X_0]=0$. $\endgroup$
    – ShawnD
    Commented Feb 16, 2012 at 18:26
  • 1
    $\begingroup$ Could you be more specific about everywhere? $\endgroup$
    – Did
    Commented Feb 16, 2012 at 18:39
  • 1
    $\begingroup$ Durrett, lots of Googling online, etc etc etc. $\endgroup$
    – Court
    Commented Feb 16, 2012 at 19:39

1 Answer 1

3
$\begingroup$

Since $X_{\min(T,n)}$ is a martingale (see for instance the book of David Williams, "Probability with Martingales"), we have $\mathbf{E}[X_{\min(T,n)}-X_0]=0$ for all $n$. If your condition holds, we can use dominated convergence to take the limit as $n \rightarrow \infty$ and conclude $\mathbf{E}[X_T-X_0]=0$.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .