 Solution

Use the same acceptance indicator functions and set $K_n={\sum }_{i=1}^n{A}_i$, $S_n={\sum }_{i=1}^n{A}_i\varphi (Y_i)$. The rejection sampling estimator based on these $n$ proposals is the mean over the accepted observations:For $K_n=0$, any fixed convention, such as zero, is harmless asymptotically. The $A_i$ are independent and identically distributed random variables with Bernoulli distribution and success probability $1/M$, so the strong law of large numbers gives $K_n/n\to 1/M$ almost surely. In particular $K_n>0$ eventually almost surely. The exact identitythen holds whenever $K_n>0$. Wherever $S_n\ne 0$, it follows thatA qualification is missing from the printed claim: the quotient must be defined eventually. A convenient sufficient assumption is ${\mathbb{E}}_p|\varphi |<\infty$ and $\theta \ne 0$. Indeed, another application of the strong law of large numbers gives $S_n/n\to \theta /M$, ensuring that $S_n$ is eventually nonzero. This proves the stated quotient limit under that assumption.

As printed, the assertion is false as a universally defined quotient. For example, $\varphi \equiv 0$ makes both estimates zero for every $n$. Even excluding that example is insufficient: take $p=q$ to be the uniform distribution on $(0,1)$, $M=1$, and $\varphi (y)=1$ for $y<1/2$ and $-1$ otherwise. Then $S_n$ is a simple symmetric random walk, which returns to zero infinitely often almost surely, leaving the quotient undefined infinitely often.

With integrability but no assumption $\theta \ne 0$, the exact multiplicative identity remains valid, and both estimates converge to $\theta$ almost surely. In particular ${\hat{\theta }}_n^{\mathrm{IS}}-{\hat{\theta }}_n^{\mathrm{RS}}\to 0$ almost surely. One may alternatively define the quotient's removable $0/0$ cases as $M{K}_n/n$; that explicitly extended quantity always converges to one. These are valid repairs of the source's unrestricted statement.