= Solution
Let $Z=1$ mean membership in the structural-zero component. Under the fitted <zero-inflated Poisson regression>, $P(Y=0\mid Z=1)=1$, whereas $P(Y=0\mid Z=0)=e^{-\mu}$. By <Bayes theorem>, the <posterior structural-zero probability> is
$$
P(Z=1\mid Y=0)=\frac{\pi}{\pi+(1-\pi)e^{-\mu}}.
$$
Using the printed predictions for this patient gives
$$
\boxed{\frac{0.4448758}{0.4448758+(1-0.4448758)\,0.8007976}\approx0.50019.}
$$
\b[The fitted conditional probability is about $50.0\%$.] It is larger than the prior fitted structural-zero probability $0.4448758$, because observing no episodes increases the probability of latent membership in that component. The interpretation “never at risk” is the model's structural class; an observed six-month zero alone does not identify the class with certainty.
Back to article page
