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Replacing a simulated quantity $H$ by $E[H\mid \mathcal{G}]$ preserves its expectation and reduces variance by the law of total variance. This is Rao-Blackwellization. For a random sum, conditioning on the count and all but one summand replaces an event indicator by a cumulative distribution function evaluation. The resulting estimator needs that conditional expectation to be computable.

Random multiplication increases the noise relative to deterministic gain. If the output from one input electron has mean $g$ and variance $g^2$, a Poisson distribution of input electrons with mean $\mu$ produces output variance $2g^2\mu$, by the law of total variance. The high-gain analog-mode excess-noise factor is $\sqrt{2}$.

A Poisson model assumes independent counts between rats conditional on time, a common mean at a given time, and conditional variance equal to that mean. Comparable observation units and consistent counting are needed; unexplained rat-to-rat heterogeneity is excluded by this simple specification. The quadratic Poisson regression isThe printed nested analysis of deviance tests $H_0:{\beta }_2=0$ against $H_1:{\beta }_2\ne 0$. Its likelihood-ratio statistic is $G^2=3.8548$, approximately ${\chi }_1^2$ under the null hypothesis and regular Poisson asymptotics. The reported $p=0.0496$ gives borderline rejection of the log-linear time effect at 5%, conditional on the Poisson assumptions. This is curvature of the log mean, rather than an ordinary quadratic model for the count itself.

There are only three distinct time values, so the intercept, linear term, and quadratic term span every possible set of three log means. This is Poisson mean saturation at three distinct times: the quadratic fit is saturated for the time-group means, which are $14/7=2$, $17/8=2.125$, and $47/7\simeq 6.7143$. A grouped deviance comparison against unrestricted time-group means has zero statistical degrees of freedom: the quadratic functional form cannot be tested at these three times. The individual-observation deviance 24.515 still describes within-group departure from the Poisson distribution, but small means and zeros make the naive ${\chi }_{19}^2$ goodness-of-fit calibration doubtful. Poisson variation itself can be checked using the replicates, preferably by a parametric bootstrap; it is not rendered untestable by the saturated mean specification.

Overdispersion means conditional variance exceeding the variance assumed by the Poisson mean model. For example, rats may have different latent susceptibility, so if $Y\mid \Lambda \sim \mathrm{Pois}(\Lambda )$, the law of total variance gives $\mathrm{Var}(Y)=\mathbb{E}\Lambda +\mathrm{Var}(\Lambda )>\mathbb{E}Y$. A quasi-Poisson model instead uses $\mathrm{Var}(Y_i)=\varphi {\mu }_i$. The Pearson-residual dispersion estimate is $\hat{\varphi }=1.254068$, using $n-p=22-3=19$.

For the conventional quasi-likelihood comparison with estimated dispersion, useIts upper-tail probability is about 0.096, so the evidence for curvature no longer reaches 5%. The F-distribution is an approximate calibration, not an exact small-sample result for arbitrary overdispersed counts; quasi-likelihood does not supply a full Poisson likelihood. With known dispersion one would instead compare $G^2/\varphi$ approximately with ${\chi }_1^2$. The displayed dispersion code uses the undefined name acf.glm2; its fitted values must be those of aom.glm2 for the stated calculation.

Keep baseline covariates fixed and put $q=1-\pi$, ${\mu }_j=e^{{\beta }_0+{\beta }^T{x}_i+{\alpha }_j}$, and $r=1/\tau$ for $\tau >0$. Integration of a Poisson count over the Gamma component givesThis negative binomial distribution has mean ${\mu }_j$ and variance ${\mu }_j+\tau {\mu }_j^2$. Therefore the unconditional one-term distribution is the zero-inflated negative binomial distributionIn particular its zero probability is $\pi +q(1+\tau {\mu }_j{)}^{-1/\tau }$, not simply $\pi$. As $\tau \downarrow 0$ this becomes a Zero-inflated Poisson distribution.

The mixing effect has $\mathbb{E}{b}_i=q$, $\mathbb{E}{b}_i^2=q(1+\tau )$ and $\mathrm{Var}(b_i)=q(\tau +\pi )$. The law of total variance and law of total covariance now giveEquivalently, for $\mu =({\mu }_1,{\mu }_2,{\mu }_3{)}^T$, $\mathbb{E}{Y}_i=q\mu$ and $\mathrm{Cov}(Y_i)=\mathrm{diag}(q\mu )+q(\tau +\pi )\mu {\mu }^T$. This also supplies raw second moments by adding $(\mathbb{E}{Y}_i)(\mathbb{E}{Y}_i{)}^T$. Dependence arises from both shared heterogeneity and shared structural zeros, even when the positive random effect is constant.

The printed moment variance matches the hierarchical variance at the same parameter values, but its printed off-diagonal covariance is $\tau {\mu }_j{\mu }_k$. Equality would require $\tau =q(\tau +\pi )$, or $\pi (\tau -q)=0$. It is therefore false to claim that the two approaches consistently estimate all the same parameters in general.

The precise distinction is between consistency of mean ratios and identification of structural parameters. In the hierarchical model, with $m_j=q{\mu }_j$, both the quadratic variance coefficient and the covariance coefficient in terms of the marginal mean areFor the moment parameterization, writing its parameters as ${\pi }_Q,{\tau }_Q,q_Q=1-{\pi }_Q$, these coefficients are $A_Q=({\pi }_Q+{\tau }_Q)/q_Q$ and $B_Q={\tau }_Q/q_Q^2$. Its intercept is identifiable from the mean only as ${\gamma }_{0Q}={\beta }_{0Q}+\log q_Q$. Matching the true first two moments demands $A_Q=B_Q=\kappa$. One solution always isFor $\kappa >1$ there is also a solution $q_Q=1/\kappa$, ${\pi }_Q=1-1/\kappa$, ${\tau }_Q=1/\kappa$, with the corresponding shifted intercept. Indeed eliminating ${\tau }_Q$ gives ${\pi }_Q({\tau }_Q-q_Q)=0$. This is moment aliasing in a shared zero-inflated count model: matching the moments does not identify the actual structural-zero probability.

For a concrete counterexample take $\pi =0.2,\tau =0.3$. Then $q=0.8$ and $\kappa =0.625$. The only admissible moment-matching choice is ${\pi }_Q=0,{\tau }_Q=0.625$, with an intercept shifted by $\log 0.8$; it cannot recover the true structural-zero proportion 0.2. Correct marginal means can give consistent term and baseline-covariate slopes with sandwich standard errors; they do not justify consistent estimation of the intended $\pi ,\tau$, or latent intercept. Nor does a mean-only estimating equation identify $\pi$ at all. Full-distribution inference provides information beyond these first two moments.

For a single claim $x$, quota share reinsurance with retained fraction $\alpha \in [0,1]$ makes the direct insurer pay $\alpha x$ and the reinsurer pay $(1-\alpha )x$. Under excess of loss reinsurance with retention $M\ge 0$, the direct insurer pays $\min (x,M)$ and the reinsurer pays the positive part $(x-M{)}_{+}$. Thus the concise payout pairs areThe cap in excess of loss reinsurance applies separately to every claim; it is not a cap on the entire annual aggregate.

For the following variance calculations take $\lambda >0$ and $\mathbb{E}[X^2]<\infty$, so the displayed variances are finite. For any per-claim payout $Y$, the law of total variance in a compound Poisson distribution givesThe final term is the raw second moment, not the single-claim variance. Both parties' totals are retained compound Poisson aggregates, with different payout functions of the same claims; they are generally dependent.

For the probability generating function $G(t)=\mathbb{E}[t^X]$, differentiation givesThe second derivative is a factorial moment, not the raw second moment.

In the Galton-Watson process, condition on $X_n=m$. The next generation is a sum of $m$ independent offspring counts, and so its conditional generating function is $G(t{)}^m$. Averaging over $X_n$ givesThe conditional mean is $m\mu$ and the conditional variance is $m{\sigma }^2$. Thus iterated expectation gives $\mathbb{E}[X_n]={\mu }^n$ from one ancestor, and the law of total variance gives $v_{n+1}={\mu }^2{v}_n+{\sigma }^2{\mu }^n$, $v_0=0$. Iterating yields the branching-process generation momentsFor the critical case $\mu =1$, the correct limiting value is $v_n=n{\sigma }^2$. If $\mu =0$, nonnegative offspring are zero almost surely, so every later generation has zero mean and variance; handle this directly rather than use an ambiguous zero power. At generation zero the mean is one and variance zero.

The branching-process extinction criterion states that extinction probability is the smallest fixed point of $G$ in $[0,1]$: probabilities of extinction by successive generations are the iterates of $G$ from zero, increasing to that fixed point. Here the equation is $4q^3-7q+3=0$, factoring as $(q-1)(2q-1)(2q+3)=0$. The two admissible roots are $1/2$ and one, so