Let $(X_t)_{t \geq 0}$ and $(Y_t)_{t\geq 0}$ be stochastic processes where $X_t$ and $Y_t$ are continuous random variables for all $t \geq 0$ (with continuous densities). Let $T$ be also a continuous random variable. Assume that

$P(X_t \in B | Y_t) = \int_B f(x,Y_t,t)\, dx $

for all measurable sets $B$ and with a continuous density $f$.

Is it true that $P(X_T\in B | Y_T, T=t) = P(X_t \in B| Y_t)$ for $t \geq 0$? Intuitively, it should be true and I would like to prove it using a similar approach like here, i.e. with conditioning on $\{ t \leq T \leq t+\epsilon \}$ instead of $T=t$ and then letting $\epsilon \rightarrow 0$, but I'm having trouble with the dependencies of $X_T$ and $Y_T$ on $T$ and also with how to handle conditioning on $Y_T$ and $T$ at the same time and if a joint density of $X_T, Y_T, T$ or something similar is also needed.

Any help is appreciated! :)


1 Answer 1


First, we should define $P(X_T\in B | Y_T, T=t)$. I think, we should assume that $T$ is independent of the processes $(X_t)_{t \geq 0}$ and $(Y_t)_{t \geq 0}$. Let $g_t$ be the joint density of $X_t, Y_t$, $$P(X_t\in B, Y_t\in C) = \int_B \int_C g_t(x,y) \mathrm{d}y \mathrm{d}x,$$ let $g'_t$ be the marginal density of $Y_t$: $$P(Y_t\in C) = \int_C \int_\Omega g_t(x,y) \mathrm{d}x \mathrm{d}y = \int_C g_t'(y) \mathrm{d}y,$$ and let $h$ be the density of $T$: $$P(T \in D) = \int_D h(t) \mathrm{d}t $$ The probability distribution of $(X_T, Y_T, T)$ is then $$P(X_T\in B, Y_T\in C, T \in D) = \int_D \int_B \int_C h(t) \cdot g_t(x,y) \mathrm{d}y \mathrm{d}x \mathrm{d}t$$ and $i(x,y,t) := h(t)g_t(x,y)$ is the joint density. Now we can construct the conditional density of $X_T$ given $T=t, Y_T = y$: $$ j(x|y,t) = \frac{i(x,y,t)}{h(t)g_t'(y)} = \frac{g_t(x,y)}{g'_t(y)}, $$ which is by definition a.s. equal to the conditional density of $X_t$ given $Y_t=y$. We obtain the desired result, when replacing $j(x|y,t)$ with $j(x|Y_t,t)$, which is the conditional density of $X_T$ given $T=t, Y_T$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .