If $M_n$ is a Martingale, How to Show that $\log M_n$ is a Supermartingale?

Question

I think that the problem should use Jensen's inequality.

This is because my textbook states that with Jensen's inequality, if $\phi$ is a convex function and $M_n$ a martingale, then $\phi(M_n)$ is a submartingale.

The problem is that $\log$ is not a convex function and I am trying to show the opposite direction.

Answer 1

You have that $-\log x$ is a convex function. Then

$\mathbb{E}[-\log M_{n+1} | \mathcal{F_n}] \geq -\log(\mathbb{E}(M_{n+1} | \mathcal{F}_n) = -\log(M_n)$

(I used Jensen in the first inequality)

Then, $-\log(M_n)$ is a submartingale and then $\log(M_n)$ is a supermartingale!

Answer 2

A supermartingale can be thought of as a gambling game which is "super" for the casino, but not for those playing; i.e.

$$\mathbb{E}[X_{n+1}|\mathscr{F}_n]\le X_n.$$

Note that for concave functions (like the logarithm), Jensen's inequality reverses, hence we have:

$$\mathbb{E}[\log M_{n+1} | \mathscr{F}_n] \le \log(\mathbb{E}[M_{n+1}|\mathscr{F}_n]) $$

Since $M_n$ is a martingale, $\mathbb{E}[M_{n+1}|\mathscr{F}_n]=M_n$, thus:

$$\mathbb{E}[\log M_{n+1} | \mathscr{F}_n] \le \log(\mathbb{E}[M_{n+1}|\mathscr{F}_n]) = \log M_n$$

$$\implies \mathbb{E}[\log M_{n+1} | \mathscr{F}_n] \le \log M_n$$

This is exactly what we needed to show, as one sees when comparing with the definition of supermartingale given above.