= Solution
The <Markov property> and time homogeneity give
$$
\mathbb P(M_n=X,M_{2n}=X)=p_n\mathbb P(M_{2n}=X\mid M_n=X)=p_n^2.
$$
Since $p_{2n}>0$, <conditional probability> yields the <Markov chain return bridge> expression
$$
\boxed{\mathbb P(M_n=X\mid M_{2n}=X)=\frac{p_n^2}{p_{2n}}
=\frac{(3^{n+1}+(-1)^n)^2}{4(3^{2n+1}+1)}.}
$$
The explicit formula gives $p_n\sim\frac34(3/4)^n$, so
$$
\boxed{\lim_{n\to\infty}\mathbb P(M_n=X\mid M_{2n}=X)=\frac34.}
$$
The conditioning event has positive probability at each finite time even though those return probabilities tend to zero. Taking an undefined conditional on an infinite-time zero-probability event is not needed.
\Image[/past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2017/probability-return-bridge.png]
{title=Unconditional returns decay while midpoint return conditioning tends to 3/4}
{description=Exact values from the two-root recurrence. The conditional midpoint probabilities oscillate toward 3/4, even as the unconditioned probability of being at X decreases to zero.}
Back to article page
