Take $\tau_0=0$ and apply part c recursively. A lazy walk makes a genuine jump with probability $1/2$, so the expected time required for $2j$ jumps is $4j$. Using the stated independence,
$$
\mathbb E_0\tau_k
=4\mathbb E_0\tau_{k-1}+1.
$$
The initial value zero solves this recurrence as
$$
\mathbb E_0\tau_k=\frac{4^k-1}{3}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 215 / 1 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-215/1/d/solution