= Solution
The acceptance <probability> is the prior average of the Poisson <likelihood>:
$$
\boxed{A_k=E_{\theta,T}\left[e^{-\theta L/2}\frac{(\theta L/2)^k}{k!}\right]=Z_k=P_{\rm prior}(S=k).}
$$
Thus the expected number of proposals per accepted draw is $1/Z_k$. A numerical rate depends on the specified prior, sample size and observed count; it is not determined by $k$ alone.
For a more explicit one-dimensional expression, condition on $\theta$. The <probability generating function>, integrating each independent exponential epoch, is
$$
E[z^S\mid\theta]
=E[e^{-\theta(1-z)L/2}]
=\prod_{j=2}^n\frac{\lambda_j}{\lambda_j+\theta j(1-z)/2}
=\prod_{i=1}^{n-1}\frac{i}{i+\theta(1-z)}.
$$
Let $p_k(\theta)$ be its coefficient of $z^k$. Then $Z_k=\int\pi(\theta)p_k(\theta)\,d\theta$. This identifies the acceptance rate with a fully specified marginal <likelihood> average.
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