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The event (m=N) represents no change within the observed sequence: every observation belongs to the first-rate segment, and the second-rate segment is empty. It is not the event $\lambda =\varphi$; under continuous independent rate priors that exact equality has posterior probability zero. The endpoint is instead a distinct index value with positive prior mass.

For retained Markov chain Monte Carlo draws $({\lambda }^{(b)},{\varphi }^{(b)},m^{(b)})$, the direct estimate isIts consistency follows from the ergodic theorem for a positive Harris recurrent Markov chain. The draws are correlated, so an independent-binomial standard error would generally be incorrect; uncertainty should account for the Markov chain Monte Carlo asymptotic variance.

A conditional Monte Carlo alternative averages the conditional probability from the preceding solution:The tower property of conditional expectation shows that its posterior expectation is the same target. This replaces an indicator by its conditional expectation; for correlated chains one should assess efficiency rather than automatically assume its entire asymptotic variance is smaller.

An exact finite-sum check is available by integrating out both rates. The marginal posterior weights of a Poisson count change point areThese follow from the Gamma function integral ${\int }_0^{\infty }{z}^{a-1}{e}^{-bz}dz=\Gamma (a)/b^a$. All omitted prior-normalization and factorial factors are independent of $j$, so they cancel from this ratio. This exact calculation provides a useful independent check on either simulation estimate.

The printed identity needs an expectation on its right-hand side. For fixed $r\ge 1$, condition on $X_2,\dots ,X_r$ and use independence of $X_1$:The expression inside this expectation is generally random and cannot equal the unconditional probability by itself.

For each independent Monte Carlo method replicate, draw $R_j$ from its given distribution. If $R_j=r\ge 1$, draw $r-1$ independent $F$-variables, form $W_j={\sum }_{i=2}^r{X}_{j,i}$ and set $H_j=F(x-W_j)$. For $r=1$ the empty sum is $0$. If $R$ may equal zero, use $H_j=1_{\{x\ge 0\}}$ when $R_j=0$, since $S_0=0$. The law of total expectation givesSince $0\le H_j\le 1$, the estimator has statistical consistency by the strong law of large numbers. It is the conditional Monte Carlo estimator obtained by Rao-Blackwellization of the direct indicator $1_{\{S_R\le x\}}$, and the law of total variance gives $\mathrm{Var}(H_j)\le l(1-l)$.

This version assumes that the cumulative distribution function $F$ can be evaluated. An easy sampler alone does not automatically supply an easy cumulative distribution function evaluation. If only sampling is available, use the direct indicator estimator, or replace each conditional cumulative distribution function by the average of several independent indicators $1_{\{X_{j,1,k}\le x-W_j\}}$; the nested version remains an unbiased estimator. With $L$ such inner draws its variance per outer replicate is $\mathrm{Var}(H_j)+L^{-1}E[H_j(1-H_j)]$.

Replace an estimator by its conditional expectation given selected information. The mean is unchanged and the variance cannot increase, by the law of total variance. In simulation this produces conditional Monte Carlo estimators. Sufficiency is needed for the classical inferential strengthening, but not for the elementary variance reduction identity.