$X_t$ and $Y_t$ are id for each $t$. $T_1$, $T_2$ are iid. Are $X_{T_1}$, $Y_{T_2}$ id?

Question

I am working on a problem and as an intermediary step I think I need to use the following:

$X_t$ and $Y_t$ are identically distributed for each $t\in[0, \infty)$. $T_1$, $T_2$ are iid, continuous RVs. Then, $X_{T_1}$, $Y_{T_2}$ are identically distributed.

Intuitively I think that should be true but I don't know how to prove it.

If I am not mistaken, to show that property for $T_1$, $T_2$ discrete I could go through the characteristic function + the law of total expectation. What would be the analogue of this for $T_1, T_2$ continuous?

Answer 1

I don't think this needs to be true. If we take $X_t$ and $Y_t$ to be independent Brownian motions, $T_1 := \inf \{t : X_t = 1 \}$, and $T_2 := \inf\{t : Y_t = -1 \}$ then $T_1$ and $T_2$ are i.i.d. by the symmetry properties of Brownian motion, but $X_{T_1} = 1$ and $Y_{T_2} = -1$ so $X_{T_1}$ and $Y_{T_2}$ are not identically distributed.