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
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.