# Strict local martingale implies $\mathbb E_t[S_u]<S_t\ \forall t<u$

Is it true that if $S$ is a strict local martingale (i.e. it is a local martingale but not a true martingale) such that $S_t\ge 0\ \forall t$, then we have $$\mathbb E_t[S_u]<S_t\quad \forall t<u$$

It is mentioned in a proof in my the lecture notes, so I have been trying to verify it. We know that every non-negative local martingale is a super-martingale, i.e. $\mathbb E_t[S_u]\le S_t\ \forall t<u$. So assuming the above claim is true, then this would mean that if $\exists t',u'$ such that $\mathbb E_{t'}[S_{u'}]=S_{t'}$ then $S$ would have to be a true martingale. But I am unable to prove this (or find a counter-example).

• What is $\mathbb{E}_t$? And is $S$ an arbitrary local martingale or a non-negative local martingale?
– saz
Jun 5, 2014 at 9:24
• $\mathbb E_t[S_u]$ means $\mathbb E[S_u\mid \mathcal F_t]$, sorry about the confusion. And $S$ is an arbitraty non-negative local martingale. Jun 5, 2014 at 12:23

## 1 Answer

This is not true.

For example, take your favorite strict local martingale $X_t$, under the filtration $\mathcal{G}_t$, and suppose without loss of generality that $X_0 = 0$. Let $S_t = 0$ for $0 \le t \le 1$, and $S_t = X_{t-1}$ for $t \ge 1$. Likewise, let $\mathcal{F}_t = \mathcal{G}_0$ for $0 \le t \le 1$ and $\mathcal{F}_t = \mathcal{G}_{t-1}$ for $t \ge 1$. Then $S_t$ is a strict local martingale under the filtration $\mathcal{F}_t$, but for any $0 \le s < t \le 1$, we have $E[S_t | \mathcal{F}_s] = S_s = 0$.

Morally speaking, the martingale property is something that holds for all pairs of times. So knowing a process is not a martingale only tells you that property fails for some pair of times. You can't expect that knowledge to give you information about the process at all pairs of times.

• Great example, thanks! Jun 7, 2014 at 23:48