2
$\begingroup$

Hi I need some help with these questions:

Let $(\Omega,\mathcal{F},P)$ be a probability space with a filtration $(\mathcal{F_n})_{n\geq0}$. Show the following claims:

a) Let $(X_n)_{n\geq 0}$ be a sequence of integrable random variables. $(X_n)_{n\geq 0}$ is an $(\mathcal{F_n})_{n\geq 0}-martingale$ if and only if for all $n\geq 0$ and all stopping times $\tau$ with $\tau\leq n$ a.s $$E(X_n|\mathcal{F_{\tau}})=X_{\tau}$$

I have the next solution:

Since $(X_n)_{n\geq 0}$ is a martingale with respect with to $\mathcal{F_n}$ i know that $E(X_{n+1}|\mathcal{F_n})=X_n$ for all $n$. Aditionally since $\mathcal{F_\tau}\subset\mathcal{F_n}$ by properties of conditional expectation I know that $E(E(X_{n+1}|\mathcal{F_n})|\mathcal{F_\tau}))=E(X_{n+1}|\mathcal{F_\tau})=X_{\tau}$ (using the property of martingale given before.)

But at this stage I don't know If my justification is correct.

b) If $(X_n)_{n\geq 0}$ is an $\mathcal{F_n}$ martingale and $\tau$ is a bounded stopping time, then: $$E(X_\tau)=E(X_0)$$ Here, I know that $\mathcal{F_0}\subset\mathcal{F_n}$ then using a similar argument as before: $E(E(X_{\tau}|\mathcal{F_0})|\mathcal{F_\tau}))=E(X_\tau|X_0)=X_0= E(E(X_n|\mathcal{F_n})|F_\tau)=X_n$, the result is gotten taking expectations.

c) Let $(X_n)_{n\geq 0}$ is a previsible martingale. Then for all $n\geq 0$ $$X_n=X_0$$. With this, I don´t know how to proceed

$\endgroup$
2
$\begingroup$

For part (a) I don't see what property of a martingale given before allows you to justify $E(X_{n+1}\mid \mathcal F_\tau) = X_\tau$ a priori but not $E(X_n\mid \mathcal F_\tau) = X_\tau.$ I would approach by partitioning the sample space into $\{\tau = 0\},\ldots \{\tau = n\}.$

For part (b), you appear to be saying $X_n=X_0.$ That's not right. There are plenty of martingales where the $X_n$ aren't all equal to each other. If you can use the fact that you used in one of the equalities, that $E(X_\tau\mid \mathcal F_0) =X_0$ then just apply the tower law: $$ E(X_\tau) = E(E(X_\tau \mid \mathcal F_0)) = E(X_0).$$ But I feel like the point of the problem is actually to prove $E(X_\tau\mid \mathcal F_0) = X_0$ using part (a) (since this equation is not generally true without some extra conditions... that $\tau$ is bounded is sufficient).

Unlike in part (b), for part (c) you do have $X_n=X_0.$ You can prove it by first showing $X_1 = X_0$ and proceeding similarly by induction. We have $$ E(X_1\mid \mathcal F_0) = X_1E(1\mid \mathcal F_0) = X_1$$ by previsibility and $E(X_1\mid \mathcal F_0) = X_0$ by the martingale property.

$\endgroup$

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.