Does the random variable $f(\tau)M_\tau$, where $M$ is a martingale and $\tau$ is a stopping time, have zero expectation?

Question

Suppose that $M:=\{M_t\}_{t\geq0}$ is a martingale adapted to some filtration $\mathcal{F}:=\{\mathcal{F}_t\}_{t\geq0}$ with $M_0\equiv0$ and that $\tau$ is an $\mathcal{F}_t$-stopping time. Suppose that $f:[0,\infty)\to[0,\infty)$ is a function of time. Is it ever the case that

$$\mathbb{E}\left[f(\tau)M_\tau\right]=0?$$

My thinking is that if we define a process $N_t:=f(t)M_t$, then even though $N$ is no longer a martingale, since

$$\mathbb{E}[N_t|\mathcal{F}_s]=N_s+(f(t)-f(s))M_s,$$

it does have zero mean at every point in time.

I'm particularly concerned with the case $f(t)=1/t$. Any references, or key-words to search for are very welcomed (I'm really not sure where to start here). Many thanks in advanced.

Answer 1

In general, we cannot expect this. Just consider e.g. a Brownian motion $(M_t)_{t \geq 0}$ and the stopping time $$\tau := \inf\{t>0; M_t = 1\}.$$ Then $$\mathbb{E}(f(\tau) M_{\tau}) = \mathbb{E}f(\tau)=0 \iff f=0,$$ i.e. the claim holds only true for the (trivial) function $f:=0$.