I've got this problem related to conditional probability and the law of total probability.
A particle moves randomly around a network consisting of three sites. At each time $0,1,2,\dots$, its position is one of the points $X$, $Y$ or $Z$. If the particle’s current position is X, then at the next step it stays at X with probability 1/2, and moves to Y with probability 1/2. If the particle’s current position is Y , then at the next step it moves to X with probability 3/8, and moves to Z with probability 5/8. Once the particle is at Z, it stays at Z for ever. In all cases, the next step is made independently of the path by which the particle reached its current position.
At time 0 the particle is at position X. Let $M(n)\in \{X, Y, Z\}$ denote the position of the particle at time n. Let $p(n) = P(M(n) = X)$, the probability that the particle is at position X at time n.
I need to find an expression for the following conditional probability: $P(M(n)=X | M(2n)=X)$.
I have so far managed to find a recurrence relation for p(n):
$16p(n)=8p(n-1)+3p(n-2)$, and the initial conditions $p(0)=1$, $p(1)=1/2$, to give the solution: $p(n)=(-1)^n (1/4)^{n+1}+(3/4)^{n+1}$.
However, I am not sure how to find an expression for the required probability. Any ideas?
I too get $p(n)=(3^{n+1}+(-1)^n)/4^{n+1}$
Use Bayes' Rule. Let $p(s\mid t):=\mathsf P(M(s)=X\mid M(t)=X)$ . $$p(n\mid 2n)=\dfrac{p(2n\mid n)~p(n)}{p(2n)}$$
Also consider that $p(n)$ is the probability that the position is a $X$ after $n$ time units passes from initially being at position $X$ (ie at time $0$), while $p(2n\mid n)$ is the probability that the position is at $X$ after $n$ time units passes from being at position $X$ at time $n$. So...