Explain in words why $2^{W_t}$ is not a martingale

Question

Question: Given a Brownian motion $(W_t)_t,$ explain in words why $2^{W_t}$ is not a martingale.

If we are allowed to use Ito's lemma, then it is easy to see that its drift term $dt$ is not zero.

However, if I am asked to explain in words, I can only think of the following.

Since $2^x$ is a convex function and $W_t$ is a martingale (which has no tendency to rise or fall in future given current information), so $2^{W_t}$ is going to rise due to its convexity.

Is my explaination above correct?

Answer 1

$\mathbb{E}[2^{W_t}|W_t \ge 0] > 1$ for $t>0$, therefore it grows on average, hence not a martingale. We used the MGF of the $N(0,t)$ evaluated at $s = \ln(2)$ to compute the moment.