I have a stationary process $(X_t)_{t \geq 0}$ with distribution $$\mathbb{P}[X_t \in A ] = \int_A f(x) \, dx$$ for any measurable set $A$ and any $t \geq 0$.
I want to compute $$ \mathbb{P}[X_\tau \in B \mid X_0 \in A ].$$
Now I have $$ \mathbb{P}[X_t \in B \mid X_0 \in A ]= \int \mathbb{E}_x [1_{B}(X_\tau)] \, g(x) \, dx$$ with $$g(x) = \frac{1}{\int_A f(x) \, dx} 1_A(x) f(x). $$
If I approximate it with standard Monte Carlo Method, then I would need some random variables $X_1,\dots,X_n$ distributed according to $g$, and could compute $$ \frac{1}{N} \sum_{i=1}^N \mathbb{E}_{X_i}[1_B(X_\tau)] $$ which converges towards $\mathbb{P}[X_t \in B \mid X_0 \in A ]$ in the $L^2$ norm. However, I would need to estimate $\mathbb{E}_{X_i}[1_B(X_\tau)] $, thus I have to compute multiple trajectories for each $X_i$.
My question is - if I only compute for every point $X_i$ a single trajectory with end point $X_i'$, would then $$ \frac{1}{N} \sum_{i=1}^n 1_B(X_i')$$ also converge to $\mathbb{P}[X_t \in B \mid X_0 \in A ]$?