Suppose $(x\_t)\_t$ is a bounded $\mathbb F\_t$ martingale and $f(t,x)$ is continuous, bounded, and concave in $x$. So, for any $s \ge t$, $$\mathbb E\_t f(s,x\_s) \le f(s,\mathbb E(x\_s)) = f(s,x).$$


Does an analogous property also hold for any bounded $\mathbb F\_t$ stopping time? More precisely, suppose $T \in (0,\infty)$ is a constant. Then, given any $\mathbb F\_t$ stopping time $\tau$ such that $\tau \le T$ a.s., is it the case that, for any $t$ $$\mathbb E\_t f(\tau, x\_\tau) \le \mathbb E\_t f(\tau, \mathbb E\_t (x\_\tau)) = \mathbb E\_t f(\tau, x)$$



Concavity, martingales and stopping time
I assume $\mathbb{F}\_t$ is the underlying filtration and $\mathbb{E}\_t(\cdot)$ stands for $\mathbb{E}(\cdot | \mathbb{F}\_t)$. Also, $f(s, x)$ in the right-hand side should read $f(s, x\_t)$.


I also leave aside technicalities, such as whether $x\_\tau$ is measurable (which, as far as I remember, may fail to be true unless one chooses a, say, càdlàg version of $x\_t$ and extend the filtration in the usual way), and assume that the filtration and the martingale satisfy the usual regularity conditions.


Even assuming all the necessary regularity, the answer is negative, as shown by the following counter-example. Take $x\_t$ to be the process which starts at $0$ and makes a $\pm 1$ jump at $t = 1$. Set $\tau = 2$ if this jump was positive, and $\tau = 3$ otherwise. Finally, define $f(t, x) = g(t) x$, where $g$ is a continuous function satisfying $g(2) = 1$ and $g(3) = 0$. Then
$$ \mathbb{E}\_0 f(\tau, x\_\tau) = \mathbb{E}(g(\tau) x\_\tau) = \tfrac{1}{2} g(2) \times 1 + \tfrac{1}{2} g(3) \times (-1) = \tfrac{1}{2} , $$
but
$$ \mathbb{E}\_0 f(\tau, x\_0) = \mathbb{E}(g(\tau) \times 0) = 0 . $$


