Time integral over stochastic process depends on distribution only?

Question

Let $X(t),Y(t)$ be two stochastic processes, integrable on $[0,T]$ with $X(t)\stackrel{d}{=}Y(t),\forall t\in [0,T]$.

Does this imply

$$\int_0^t X(s)ds = \int_0^t Y(s)ds, \qquad \forall t \in [0,T],$$

i.e. the integral depends on the distribution only?

Answer 1

Certainly not. To see why, try $X(t)=U$ for every $t$, $Y(t)=U$ for every $t\leqslant\frac12T$ and $Y(t)=V$ for every $t\gt\frac12T$, for some nondegenerate i.i.d. $U$ and $V$, then $X(T)=T\cdot U$ is not in general distributed like $Y(T)=\frac12T\cdot (U+V)$.