Let $X_n\ge0,n\ge0$, be a supermartingale. Show that $CP(\sup X_n>C)\le EX_0$.
I tried to use the inequality supermartingale satisfies, which is $E(X_n|\cal {F_{n-1}})$$\le X_{n-1}$. However, I have no clue to proceed.
Could someone kindly give some hint? Thanks!
Let $T= \inf\{n \geq 0; X_n >C\}$, where $C>0$. Then $T$ is a stopping time. By the optional stopping theorem, $E(X_{T\wedge k}) \leq E(X_0)$, for all $k>0$. However, note that, for $k>0$, \begin{align*} E(X_{T\wedge k}) &\geq E\left(X_{T\wedge k} \mathbb{I}_{\{\sup_{n\leq k} X_n >C\}} \right)\\ &\geq CP(\sup_{n\leq k} X_n >C). \end{align*} That is, for $k>0$, \begin{align*} CP(\sup_{n\leq k} X_n >C) \leq E(X_0). \end{align*} The final result follows immediately.