Is the logarithm of a martingale still a martingale?

Question

For example $S_t = S_0\cdot \exp(\sigma W^*_t - 0.5\sigma^2t)$ is martingale (it's a part of Black-Scholes).
Is $\ln S_t$ a martingale?

Answer 1

$$ln(S_t)=ln(S_0)-0.5\sigma^2 t + \sigma W^{*}_t$$

When you look at the above, you see the process $ln(S_t)$ has a drift $-0.5\sigma^2 t$, so it cannot possibly be a martingale. Just taking the expectation we see that:

$$\mathbb{E}[ln(S_t)]=ln(S_0)-0.5\sigma^2 t$$

On the other hand, when we exponentiate, the expectation is centered on the initial value (because of the mean property of lognormally distributed random variable):

$$\mathbb{E}[S_t]=\mathbb{E}[S_0e^{-0.5\sigma^2 t + \sigma W^{*}_t}]=S_0$$

So the above looks like a much more plausible Martingale candidate!

In general, intuitively, if you see a process with a (deterministic) drift, it cannot possibly be a Martingale, because it's future expected values will not be centered on the "current" value.