Showing that supermartingale is a u.i. martingale

Question

Let $M_t$ be a non-negative supermartingale, $EM_0 < \infty$ and $M_t \to M_\infty$ a.s. I want to show that if $E M_\infty = EM_0$, then $M$ is a uniformly integrable martingale. I thought about doing so by showing that $M_t = E ( M_\infty | \mathcal{F}_t )$ but couldn't figure out the necessary steps. Any hints? Just somewhere to start would be great.

Edit: An additional question: Can I follow from the above that $E (sup_{t\geq0} M_t) < \infty$? That would also give uniform integrability, right?

Answer 1

In a supermartingale $EM_t$ is decreasing. If $EM_{\infty} =EM_0$ then $EM_t$ is independent of $t$. For $s >t$ the non-negative random variable $M_t-E(M_s|\mathcal F_t)$ has mean $0$ and hence it vanishes a.s. This proves that $(M_t)$ is a martingale. Also, $M_t \to M_{\infty}$ a.s and $EM_t \to EM_{\infty}$ together imply convergence in $L^{1}$ because of non-negativity. [See https://en.wikipedia.org/wiki/Scheff%C3%A9%27s_lemma ] This also implies uniform integrability.

Answer 2

By the optional stopping theorem, we know that for any stopping time $T$, $$ M_T \ge E[M_\infty \mid \mathcal F_T]. $$ Moreover, since $M$ is a supermartingale, again via the optional stopping theorem, $$ E[M_0] \ge E[M_T]. $$ Thus, $$E[M_0] \ge E[M_T] \ge E[M_\infty], $$ which means that $$ E[M_0] = E[M_T] $$ for all stopping times $T$. The fact that $M$ is a UI martingale now follows from the following result.

Theorem. Suppose $\{M_t\}_{t\in[0,\infty]}$ is a right-continuous process adapted to the right-continuous filtration $\{\mathcal F_t\}_{t\in[0,\infty]}$. Then, $M$ is a uniformly integrable martingale if and only if for every stopping time $T$ we know that $E[\vert M_T\vert]<\infty$ and $E[M_T]=E[M_0]$.

Clearly, you only need the "if" part of this result. You don't explicitly assume right-continuity, but this is a standard assumption, so I hope it works for you.

Reference. Cohen, S. N., & Elliott, R. J. (2015). Stochastic calculus and applications (Vol. 2). New York: Birkhäuser.