I have a probability space $(\mathbb{P}, \Omega, \mathcal{F})$ and in this space, I have a submartingale $(X_n)_n$ with the following two properties:
- $\inf_n X_n < 0$
- $\mathbb{E}[X_0] \geq 0$
and I am supposed to show that $\mathbb{E}[\sup_n X_n] = \infty$.
I am not very familiar with the language of probability but the main idea I am using is the fact that, the expectations form an increasing sequence: $\mathbb{E}[X_{n+1}] \geq \mathbb{E}[X_n]$.
Since the infimum is a negative function we know that the set $\{ \omega \in \Omega : X_j(\omega) < 0\}$ has positive measure (probability) for some $j$. Otherwise, the countable union of these would have measure zero and almost surely our random variables would be positive which contradicts the first assumption in the hypothesis. So, let $j_1$ be the minimum natural number with this property. We know that expectations are increasing and before $j_1$, all random variables are almost surely positive. So, $X_{j_1}$ must compensate this "going below zero" by "going up high somewhere else" which means that the supremum of $X_{j_1}$ over all of $\Omega$, is strictly larger than supremum of earlier random variables. My goal is to somehow continue this process and reach that $\sup_n X_n$ is too big to have finite expectation. I can continue one more step: Since $X_{j_1}$ has non-negative expectation, it can not be negative everywhere. Therefore, it must be positive in a set of positive measure:$\mathbb{P}\{ \omega \in \Omega: X_{j_1}(w) > 0 \} > 0$. By the above argument, there is an $X_{j_2}$ such that there is a positive measure subset of this set where $X_{j_2}$ is negative.
Can I continue doing this? Is it possible that first two random variables in my submartingale take care of the infimum condition and then the rest in the sequence are just constant? In particular, what is wrong with the following counterexample: $\Omega = [0,1]$, $X_n = 4$ (constant) for $n \geq 2$, $X_0 = -1$ on first half of the interval, $X_0 =1$ on the second half. Finally, $X_1 = 4$ on the first half and $X_1 = -1$ on the second. I satisfy that $X_0$ has zero (non-negative) expectation. And I satisfy that expectations increase. However, the supremum of these functions is constant 4. Then, the only problem should be that my example is not a submartingale. And this tells me that to solve my original problem, I should somehow use the submartingale property, with filtrations and everything, which I am not very comfortable with. (I also learned things like Doob's Inequality etc., but I can not see how that can be useful.)
Should I change my point of view and try something else? I don't know how to attack this problem, other than the way I am used to from my measure theory class.
Hint: Doob's Optional Stopping Theorem. Consider the stopping time $T:=\min\{n: X_n<0\}$. By Optional Stopping, $0\le\Bbb E[X_0]\le\Bbb E[X_{\min(T,n)}]$ for $n=1,2,\ldots$. The latter expectation is equal to $$ \Bbb E[X_T; T\le n]+\Bbb E[X_n; T>n]. $$ The second of these expectations is at most $\Bbb E[M; T>n]$, where $M:=\sup_nX_n$. What would happen to the upper bound $$ \Bbb E[X_T; T\le n]+\Bbb E[M; T>n], $$ in the limit as $n\to\infty$, were it true that $\Bbb E[M]<\infty$? (Notice that $\{\inf_nX_n<0\}=\{T<\infty\}$ and that $X_T<0$ on $\{T<\infty\}$.)