Because $f$ is a continuous bijection $\mathbb R\to\mathbb R$, it is a Borel isomorphism. Conditional on $X_{-i}=x_{-i}$, the $i$th coordinate of $F(X)$ has the pushforward of $\pi(dx_i\mid x_{-i})$ under $f$. Therefore applying one $K$-update and then $F$ has exactly the same law as applying one $Q$-update to $F(x)$, using the same random coordinate.

Formally, for every Borel set $A$,
$$
Q(F(x),A)=K(x,F^{-1}(A)).
$$
Composition preserves this conjugacy, so induction on $t$ gives
$$
Q^t(F(x),A)=K^t(x,F^{-1}(A)).
$$
Hence
$$
\boxed{Y\sim K^t(x,\cdot)\implies F(Y)\sim Q^t(F(x),\cdot).}
$$


Solution

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

 Solution

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