Let $(X_n)_{n\geq 1}$ be a Markov process on a finite state space $S$, and let $N$ be a stopping time with respect to the natural filtration $(\mathcal{F}_n)_{n\geq 1}$ defined by \begin{equation} \mathcal{F}_n=\sigma(X_1,\ldots,X_n). \end{equation}
I now want to compute the probability \begin{equation} P(X_{2}=i,\,X_{3}=j|N=4). \end{equation}
I want to write this as \begin{align} & P(X_{2}=i|N=4)\cdot P(X_{3}=j|X_{2}=i,N=4)\\ & =P(X_{2}=i|N=4)\cdot P(X_{3}=j|X_{2}=i). \end{align}
But something is stopping me from writing the last line above. Given $X_{2}$, nothing else should matter in order to say something about $X_{3}$, since $X_{n}$ forms a Markov process. However, the event $\{N=4\}$ gives something more than merely everything up to $X_{2}$.
Is my argument correct, in that I am wrong in writing neglecting the $N=4$ term in the conditioning? Can someone provide a stronger and perhaps more rigorous justification for why I might be wrong?
Thanks in advance.
The event $N=4$ gives you information about, potentially, $X_1,X_2,X_3,X_4$. So it still gives you additional information about $X_3$ even if $X_2$ is known. The Markov property only lets you drop conditioning about variables prior to $X_2$.
If you had something like $\Pr[X_{10} = i \mid X_9 = j, N=7]$, then you could use the Markov property to simplify this to $\Pr[X_{10} = i \mid X_9 = j]$ because knowing $X_9$ lets you forget about $X_1, \dots, X_8$, and whether $N=7$ or not can be determined entirely from $X_1, \dots, X_7$.
As a particular example, suppose you are taking a random walk on $\mathbb Z$ with $N$ being the stopping time $\inf\{n : X_n = 4\}$. Then $$\Pr[X_3 = 3 \mid X_2 = 2] = \frac12$$ (given that $X_2 = 2$, $X_3$ can be either $1$ or $3$ with equal probability) but $$\Pr[X_3 = 3 \mid X_2 = 2, N = 4] = 1$$ (given that $N=4$ and $X_2=2$, the sequence must continue $X_3=3$ and $X_4=4$).