Let $(X_t)_t$ be a continuous time Markov process and $ T $ be an almost surely finite stopping time relative to the natural filtration of $(X_t)_t$.
Is the following statement true?
$$ E (X_T | T) = E (X_t) \vert_{t=T}. $$
If so, how does one prove it?
Many thanks for your help.
No, in general this is not true. Consider for example a Brownian motion $(X_t)_{t \geq 0}$ and the stopping time
$$T := \inf\{t \geq 0; X_t=1\}.$$
Since $(X_t)_{t \geq 0}$ has continuous sample paths, we have $X_T=1$ and therefore
$$\mathbb{E}(X_T \mid T) = 1.$$
On the other hand, $f(t)= \mathbb{E}X_t=0$. Consequently, the equality
$$\mathbb{E}(X_T \mid T) = f(T)$$
does not hold.