$EM_T=1 \implies M_t$ is a martingale

Question

Fix $T>0$ and let $M_t$ be a non negative supermartingale for all $t\leq T$. I read somewhere that $EM_0=EM_T=1$ implies that $M_t$ is a martingale. Now I know that $EM_t=1$ for all $t$ implies that $M_t$ is a martingale. I saw this proved somewhere on MSE too. However, I cannot prove $EM_T=1 \implies M_t$ is a martingale. Any help is appreciated.

Answer 1

You have to assume that $EM_0=1$. Otherwise the result is false: just take any supermartingale which is not a martingale and divide it by the constant $EM_T$ to get a counter-example. Now assume that $EM_0=EM_T=1$. From the definition of a supermartingale it follows immediately (by taking expectation)that $EM_t$ is decreasing in $t$: $EM_{t+1} \leq EM_t$. Hence $EM_0=EM_T=1$ implies that $EM_t=1$ for all $t \in [0,T]$ ($1=EM_T\leq EM_t \leq EM_0=1$ so equality must hold throughout) and you already know why this implies that $(M_t)$ is a martingale.