Total variation is invariant under a measurable bijection with measurable inverse. Using part (b),
$$
\begin{aligned}
\|Q^t(F(x),\cdot)-\nu\|_{\mathrm{TV}}
&=\sup_A|K^t(x,F^{-1}(A))-\pi(F^{-1}(A))|\\
&=\|K^t(x,\cdot)-\pi\|_{\mathrm{TV}}
\leq2^{-t}.
\end{aligned}
$$
Every $z\in\mathbb R^d$ equals $F(x)$ for a unique $x$, so
$$
\boxed{\|Q^t(z,\cdot)-\nu\|_{\mathrm{TV}}\leq2^{-t}
\quad\text{for every }z.}
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 216 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-216/2/c/solution