If $M_t$ is a martingale, when $\frac{1}{M_t}$ be a sub-martingale?

Question

I am wondering what are constrains for using the argument that "martingale + convex function -> sub-martingale".

The problem I have is: if $M_t$ is within $]0, +\infty[$, then whether $(\frac{1}{M_t}, t\geq 0)$ a sub-martingale ?

Generally, let $f$ a convex function, then by Jensen's inequality, we have: $$\forall t\geq 0, \mathbb{E}[f(M_t)] \geq f(\mathbb{E}[M_t])$$

Here for this problem, $f$ will be $f(x) = 1/x$ and since $M_t$ is positive, we have the convexity condition satisfied. And since $M_t$ cannot be $0$ or $+\infty$, we have $f(\mathbb{E}[M_t])$ well defined.

So am I right to write: $$\mathbb{E}[\frac{1}{M_t}|\mathcal{F_s}] \geq \frac{1}{\mathbb{E}[M_t|\mathcal{F}_s]} = \frac{1}{M_s}, \quad 0\leq s \leq t$$
so that $(\frac{1}{M_t}, t\geq 0)$ is a sub-martingale ? Or I have overmitted some important points ?

Besides, If we discard the constrain that $M_t$ is within $]0, +\infty[$, and take brownian motion $(B_t, t\geq 0)$ as an example, can I say that $(\frac{1}{B_t}, t\geq 0)$ a sub-martingale ? If not, what's the nature of $(\frac{1}{B_t}, t\geq 0)$ ?

Thanks !

Answer 1

If you can live with the fact that $M^{-1}_t$ may not be integrable, so that $\Bbb E[M^{-1}_t\mid\mathcal F_s]$ has to be understood in a generalized sense, then Jensen can be applied and $M^{-1}_t$ is a "submartingale".

Answer 2

The only condition you omitted is that $E \frac{1}{M_t}$ must be finite.

If we don't have a condition $M_t \in (0, \infty)$ we can't say anything.

For example, $\frac{1}{B_t}$ is not a martingale or submartingale or supermartigale, because $E |\frac{1}{B_t}|$ is not finite.

If $M_t \in (-\infty, 0)$ and $E |\frac{1}{M_t}| < \infty$ we can prove that $\frac{1}{M_t}$ is supermartingale.