Supremum of a supermartingale

Question

For a nonnegative supermartingale $(M_n)_{n\in\mathbb{N}}$ I would like to prove that $$E[\sup_n M_n] \leq E[M_0]$$ I have that $E[M_n] \leq E[M_0]$ for every $n$. This combined with nonnegativity gives me $L^1$-boundedness so then I can apply Doob's convergence theorem, i.e. $$\lim_n M_n = M_{\infty} \quad \text{a.s.}$$ and $E[M_\infty] < \infty$. I thought of using Scheffe's lemma from here on but couldn't arrive at much.

Answer 1

I doubt it. Definitely we have $\sup_n M_n \geq M_0$, which means that your inequality forces the relation $\sup_n M_n = M_0$ a.s., which seems very unlikely.

For example, consider the martingale $(M_n)$ defined by $M_0 = 1$ and $M_{n+1} = X_{n+1} M_n$ for an i.i.d. sequence $(X_n)$ with $\Bbb{P}(X_n = 0) = \Bbb{P}(X_n = 2) = 1/2$. Then we have

$$\textstyle\Bbb{P}(\sup_n M_n = 2^k) = 2^{-(k+1)}$$

and hence the expectation of the supremum is infinite:

$$\textstyle\Bbb{E}(\sup_n M_n) = +\infty.$$