= Solution
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
$$
\boxed{Y\sim K^t(x,\cdot)\implies F(Y)\sim Q^t(F(x),\cdot).}
$$
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