Show that if $(M_t)_{t \in [0,T]}$ is a non-negative and strict local martingale on [0,T] we have $\mathop{\mathbb{E}}[M_T] < M_0$.

Question

I was stuck in proving this statement.

There is a hint: By contradiction, using the tower property and the fact that if a random variable $X \leq 0$ a.s. and $\mathop{\mathbb{E}}[X] = 0$ then $X = 0$ a.s.

As the hint suggests, if we do it by contradiction, then we assuming that $(M_t)_{t \in [0,T]}$ is a non-negative and strict local martingale on $[0,T]$ and that $\mathop{\mathbb{E}}[M_T] \geq M_0$ and we should derive a contradiction from here. But, I don't know how to proceed from here and how the tower property can be applied.

I was wondering if the definition of strict local martingale could be useful? But I just cant put all these things together.

Glad if anyone can help!

score 2 · Accepted Answer · answered Mar 22 '22 at 02:25

Lemma If $M_t$ is a non-negative local martingale, then it is a supermartingale.

Proof. This is a simple application of Fatou's lemma. See here. $\blacksquare$

Now, arguing by contradiction, suppose $\mathbb{E}(M_T) = M_0$. Since $M_t$ is a supermartingale, this implies that $\mathbb{E}(M_t) = M_0$ for all $t\in[0,T]$. But then we have that for any $0 \leq s \leq t \leq T$, $M_s - \mathbb{E}(M_t \, | \, \mathcal{F}_s) \geq 0 $ and $\mathbb{E}(M_s - \mathbb{E}(M_t \, | \, \mathcal{F}_s)) = 0$, implying that $M_s - \mathbb{E}(M_t \, | \, \mathcal{F}_s) = 0$. But then, $M_t$ is a true martingale, so it cannot be a strict local martingale.

Sorry, but how does $M_t$ is a supermartingale implies that E($M_t$)=$M_0$ for all $t∈[0,T]$? — math_killer, Mar 25 '22 at 05:43

Show that if $(M_t)_{t \in [0,T]}$ is a non-negative and strict local martingale on [0,T] we have $\mathop{\mathbb{E}}[M_T] < M_0$.

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