Solution. Expected value. By the Law of Total Expectation (conditioning on \(Z\), within the subpopulation \(\{\tilde{X}=\tilde{x}, T=t\}\)):
\[
\begin{aligned}
\operatorname{E}\mathopen{}\left[Y \mid \tilde{X}=\tilde{x}, T=t\right]\mathclose{}
&= \pi \, \operatorname{E}\mathopen{}\left[Y \mid Z=1, \tilde{X}=\tilde{x}, T=t\right]\mathclose{} \\
&\phantom{={}} + (1-\pi) \, \operatorname{E}\mathopen{}\left[Y \mid Z=0, \tilde{X}=\tilde{x}, T=t\right]\mathclose{} \\
&\quad \text{(by Law of Total Expectation; } Z \perp\!\!\!\perp T \mid \tilde{X}\text{)} \\
&= 0 \cdot \pi + \mu_0 (1-\pi) \\
&\quad \text{(substituting the conditional means)} \\
&= (1-\pi) \mu_0 \\
&\quad \text{(simplifying arithmetic)}
\end{aligned}
\]
The substitution \(\operatorname{E}\mathopen{}\left[Y \mid Z=0, \tilde{X}=\tilde{x}, T=t\right]\mathclose{} = \mu_0\) follows immediately from the definition of \(\mu_0\).
Variance. Within this derivation, write \(\operatorname{E}\mathopen{}\left[\,\cdot \mid Z\right]\mathclose{}\) and \(\operatorname{Var}\mathopen{}\left(\cdot \mid Z\right)\mathclose{}\) for the moments conditional on \(Z\) and on \(\tilde{X}=\tilde{x}, T=t\); the outer operators keep their conditioning explicit. By the Law of Total Variance:
\[
\begin{aligned}
\operatorname{Var}\mathopen{}\left(Y \mid \tilde{X}=\tilde{x}, T=t\right)\mathclose{}
&= \operatorname{E}\mathopen{}\left[\operatorname{Var}\mathopen{}\left(Y \mid Z\right)\mathclose{} \mid \tilde{X}=\tilde{x}, T=t\right]\mathclose{} \\
&\phantom{={}} + \operatorname{Var}\mathopen{}\left(\operatorname{E}\mathopen{}\left[Y \mid Z\right]\mathclose{} \mid \tilde{X}=\tilde{x}, T=t\right)\mathclose{} \\
&\quad \text{(by Law of Total Variance)}
\end{aligned}
\]
For the expected conditional variance term, note that
\[\operatorname{Var}\mathopen{}\left(Y \mid Z=1\right)\mathclose{} = 0 \quad\text{and}\quad \operatorname{Var}\mathopen{}\left(Y \mid Z=0\right)\mathclose{} = \mu_0\]
since the \(Z=0\) arm is Poisson, so:
\[
\begin{aligned}
\operatorname{E}\mathopen{}\left[\operatorname{Var}\mathopen{}\left(Y \mid Z\right)\mathclose{} \mid \tilde{X}=\tilde{x}, T=t\right]\mathclose{}
&= \pi \, \operatorname{Var}\mathopen{}\left(Y \mid Z=1\right)\mathclose{} + (1-\pi) \, \operatorname{Var}\mathopen{}\left(Y \mid Z=0\right)\mathclose{} \\
&\quad \text{(expectation over } Z\text{; } Z \perp\!\!\!\perp T \mid \tilde{X}\text{)} \\
&= 0 \cdot \pi + \mu_0 (1-\pi) \\
&\quad \text{(substituting the conditional variances)} \\
&= (1-\pi)\mu_0 \\
&\quad \text{(simplifying arithmetic)}
\end{aligned}
\]
For the variance of conditional expectation term, \(\operatorname{E}\mathopen{}\left[Y \mid Z\right]\mathclose{}\) takes value 0 (with probability \(\pi\)) or \(\mu_0\) (with probability \(1-\pi\)), so:
\[
\begin{aligned}
\operatorname{Var}\mathopen{}\left(\operatorname{E}\mathopen{}\left[Y \mid Z\right]\mathclose{} \mid \tilde{X}=\tilde{x}, T=t\right)\mathclose{}
&= \pi \mathopen{}\left(0 - (1-\pi)\mu_0\right)\mathclose{}^2 + (1-\pi) \mathopen{}\left(\mu_0 - (1-\pi)\mu_0\right)\mathclose{}^2 \\
&\quad \text{(variance over } Z\text{; } Z \perp\!\!\!\perp T \mid \tilde{X}\text{)} \\
&= \pi(1-\pi)^2 \mu_0^2 + (1-\pi)\pi^2 \mu_0^2 \\
&\quad \text{(expanding squared terms)} \\
&= \pi(1-\pi)\mu_0^2 \mathopen{}\left[(1-\pi) + \pi\right]\mathclose{} \\
&\quad \text{(factoring)} \\
&= \pi(1-\pi)\mu_0^2 \\
&\quad \text{(since } (1-\pi) + \pi = 1\text{)}
\end{aligned}
\]
Combining both terms gives:
\[
\begin{aligned}
\operatorname{Var}\mathopen{}\left(Y \mid \tilde{X}=\tilde{x}, T=t\right)\mathclose{}
&= (1-\pi)\mu_0 + \pi(1-\pi)\mu_0^2 \\
&\quad \text{(summing expected variance and variance of expectation)} \\
&= (1-\pi)\mu_0 \mathopen{}\left(1 + \pi\mu_0\right)\mathclose{} \\
&\quad \text{(factoring out } (1-\pi)\mu_0\text{)}
\end{aligned}
\]
The logistic model puts \(\pi\) strictly between 0 and 1, since \(\operatorname{expit}\) never attains its limits, and \(\mu_0 = t \operatorname{exp}\mathopen{}\left\{\eta(\tilde{x})\right\}\mathclose{} > 0\) whenever \(t > 0\). Then \(1 + \pi\mu_0 > 1\), so
\[(1-\pi)\mu_0 (1+\pi\mu_0) > (1-\pi)\mu_0 = \operatorname{E}\mathopen{}\left[Y \mid \tilde{X}=\tilde{x}, T=t\right]\mathclose{}\]
and a zero-inflated count model is overdispersed relative to a Poisson model with the same mean, at every covariate pattern with positive exposure.