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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 !

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  • $\begingroup$ Since $x\mapsto \frac{1}{x}$ is convex, $(\frac{1}{M_t})_t$ is always a sub-martingale. $\endgroup$
    – Surb
    Dec 20, 2020 at 14:14
  • $\begingroup$ I guess there should be something like for every $t$, $f(M_t)$ might be integrable. For $M_t$ I think it should be ok as I showed above by checking these conditions, but not 100% sure. As for Brownian motion, I doubt we can just apply the convexity... $\endgroup$
    – uru
    Dec 20, 2020 at 14:27
  • $\begingroup$ of course you need $\frac{1}{M_t}$ being integrable for all $t$. $\endgroup$
    – Surb
    Dec 20, 2020 at 14:37

2 Answers 2

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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.

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  • $\begingroup$ Thanks for your comment, I agree. I have one more question for this condition $M_t \in (0, \infty)$. Do we need it being true for all $t \geq 0$ or the argument is sufficient in the case of fixing a $t$, and $\mathbb{E}(\frac{1}{M_t})$ is finite for this fixed $t$ ? $\endgroup$
    – uru
    Dec 25, 2020 at 22:52
  • $\begingroup$ @rururururururu As $E(\frac1{M_t} | \mathcal{F}_s) \ge \frac1{M_s}$ for $t \ge s$ and as we must have a condition $E \frac1{M_t} < \infty$ it is sufficient to have condition $E \frac1{M_t} < \infty$ for some $t_k \to \infty$. But if we have this condition only for fixed $t$ then how are you going to get it for all $t$? $\endgroup$
    – Botnakov N.
    Dec 25, 2020 at 23:46
  • $\begingroup$ @rururururururu, consider $\xi$ such that $P(\xi = 4^{-n}) = P(\xi = 1 - 4^{-n}) = 2^{-n}$, $n \ge 2$. Put $M_t = 1$ for $t \le 1$ and $M_t = \xi$ for $t > 1$ and consider natural filtration. Thus $E \frac{1}{M_t}$ is finite for $t = 1$, but not for all $t$. So it's not sufficient to have $E \frac{1}{M_t}$ at fixed $t$. $\endgroup$
    – Botnakov N.
    Dec 25, 2020 at 23:53
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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".

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