Put $p_k=P_X(k)$. The assumption $p_0=0$ ensures that $X-1$ still takes values in $\{0,1,\ldots\}$, and its <probability mass function> at $k$ is $p_{k+1}$. Using the paper's unhalved $\ell^1$ convention for the <total variation distance>,
$$
\begin{aligned}
\lVert P_X-P_{X-1}\rVert_{\mathrm{TV}}
&=\sum_{k\geq0}|p_k-p_{k+1}|\\
&=\sum_{k\geq0}\{p_k+p_{k+1}-2\min(p_k,p_{k+1})\}\\
&=1+(1-p_0)-2q=2(1-q).
\end{aligned}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 224 / 1 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-224/1/c/solution