Past exam of the simulation and statistical programming course of the University of Oxford / 2021 / 2 / b

Because $U\in [0,1]$, the conditional acceptance probability is $a_k=\min (1,{\lambda }^k)$, even when the printed threshold exceeds one. Repeating rejected trials gives the clipped acceptance reweighting identity:For $0<\lambda <1$, $a_k={\lambda }^k$, including $a_0=1$. The denominator is $e^{-\lambda }{\sum }_k{\lambda }^{2k}/k!=e^{{\lambda }^2-\lambda }$, soFor $\lambda \ge 1$ every proposal is accepted, so reweighting by ${\lambda }^k$ would be incorrect. At $\lambda =1$ both formulas agree. Termination has probability one since the acceptance probability is $e^{{\lambda }^2-\lambda }>0$ in the first case and one in the second.

Define $a_y=\min \{1,{\lambda }^y/7\}$. Given $Y=y$, the direct-return branch has probability $a_y$, while the fallback branch distributes its remaining mass uniformly over $0,\dots ,y$. The law of total probability therefore gives, for integer $x\ge 0$,The summation includes $y=x$: a failed direct-return test can still return $x$ through the discrete uniform distribution draw. There is no rejection loop and no additional normalization. Summing the displayed probability mass function over $x$ gives ${\sum }_y{p}_y[a_y+(1-a_y)(y+1)/(y+1)]=1$. At $\lambda =1$, for example, $a_y=1/7$ at every $y$; it does not become one. This is an instance of the accept-or-fallback mixture identity.