Let X be a square integrable martingale with square variation process $\langle X\rangle$. Let $\tau$ be a finite stopping time. Show the following:
If $E(\langle X\rangle)<\infty$, then
$E((X_\tau-X_0)^2)=E(\langle X\rangle_\tau)$ and $E(X_\tau)=E(X_0)$
In my attempted solution I used the following definitions: Let's consider the probability space $\Omega,\mathbb{F},P$, where $\mathscr{F}_k$ for $k=0,1,2,3...$ are the filtrations of $\mathbb{F}$.
$\langle X\rangle_n=\sum\limits_{k=1}^{n}E((X_k-X_{k-1})^2|\mathscr{F}_{k-1})$
Attempted proof:
So $E(\langle X\rangle_\tau)=E(\sum\limits_{k=1}^{\tau}E((X_k-X_{k-1})^2|\mathscr{F}_{k-1}))=E(\mathbb{1}_\Omega\sum\limits_{k=1}^{\tau}E((X_k-X_{k-1})^2|\mathscr{F}_{k-1}))=\sum\limits_{k=1}^{\tau}E((X_k-X_{k-1})^2)=E(\sum\limits_{k=1}^{\tau}(X_k-X_{k-1})^2)=E((X_\tau-X_0)^2)$
Regarding $E(X_\tau)=E(X_0)$, I have read in a similar question that I should use the martingale definition since $X_{\tau\wedge n}$ is a martingale by the optional sampling theorem. However I am not seeing how to solve it.
Questions:
Is my solution insofar right? How should I prove the second point $E(X_\tau)=E(X_0)$?
Thanks in advance!