If $X_n$ is a supermartingale. Prove that for $\lambda>0$ and $n\geq 0$,
$$\lambda P(\inf_{k\leq n} X_k < -\lambda) \leq \int_{\{\inf_{k\leq n} X_k < -\lambda\}}(-X_n)dP$$
This is the same as:
$$E(-X_n).1_{\{\sup_{k\leq n} (-X_k) > \lambda\}} \geq \lambda P(\sup_{k\leq n} (-X_k) > \lambda)$$
and since $-X_n$ is a submartingale, its expectation is non-decreasing and so if the $\sup_{k\leq n} (-X_k)>\lambda$, then $EX_n\geq \lambda$. But I am not sure if that is a sufficient explanation or if that can be stated formally particularly as n could be $\infty$?
Best Answer
Your explanation is not a proof and can't be made into one. The problem isn't that $n$ may be $\infty$ since this is excluded (or requires control by a separate limiting argument).
The first problem is that you're meant to consider $\mathbb{E}[-X_n 1_{\{\sup_{k\leq n} (-X_k) > \lambda\}}]$ but you instead consider $\mathbb{E}[-X_n]$ under the assumption that $\sup_{k\leq n} (-X_k) > \lambda$. These are different since you may have something like e.g. $X_k = - \frac{\lambda}{2}$ for all $k$. Then $\mathbb{E}[X_n] = - \frac{\lambda}{2}$ but $-X_n 1_{\{\sup_{k\leq n} (-X_k) > \lambda\}} = 0$.
Even if this were correct, the next problem is that you try to use the fact that the expectation of $-X_n$ is decreasing in $n$ to conclude that $\mathbb{E}[\max_{k \leq n} (- X_k)] \leq \mathbb{E}[-X_n]$. To do this, you'd need to somehow get the $\max$ out of the expectation which you cannot do (even with the assumption that $X_n$ is a supermartingale).
This result is called Doob's martingale inequality and the usual proof relies on using a suitably chosen stopping time. I give this proof below.
It's slightly more convenient to prove this in terms of the submartingale $Y_n = -X_n$. The trick is to consider the stopping time $T = \min\{k \geq 0: Y_k > \lambda\} \wedge n$. Then $$\{\max_{k \leq n} Y_k > \lambda\} = \{Y_T > \lambda\}$$ This means we can write \begin{align*} \mathbb{P}(\min_{k \leq n} X_k < -\lambda) = \mathbb{P}(\max_{k \leq n} Y_k > \lambda) =& \mathbb{P}(Y_T > \lambda) \\\leq& \lambda^{-1} \mathbb{E}[Y_T 1_{\{Y_T > \lambda\}}] \\\leq& \lambda^{-1} \mathbb{E}[Y_n 1_{\{Y_T > \lambda\}}] \\=& \lambda^{-1}\mathbb{E}[-X_n 1_{\{\min_{k \leq n} X_k < - \lambda\}}] \end{align*}
where the last inequality is by Optional Stopping Theorem.