Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 28J (Principles of Statistics)

For $X_1\sim \mathrm{Poisson}(\theta )$, its probability generating function isThus $\mathbb{E}[X_1]=G^{\prime }(1)=\theta$ andThe log-likelihood is, up to an additive constant,so the maximum-likelihood estimator is ${\hat{\theta }}_{\mathrm{MLE}}=\overline{X}$. The central limit theorem gives

The mass function is the mixtureso it is nonnegative and sums to one. Its first two moments areThereforeFor $0<\pi <1$ this is strictly larger than the mean, so the Zero-inflated Poisson distribution models overdispersion. At $\pi =1$ it reduces to an ordinary Poisson distribution, and at $\pi =0$ it is degenerate at zero.

Let $N_{+}=\#\{i:Y_i>0\}$, $N_0=n-N_{+}$, and $a=1-e^{-\lambda }$. Ignoring factors independent of $\pi$, the likelihood isThe unconstrained critical point satisfiesso, after enforcing $0\le \pi \le 1$,

Under the null model, $\pi =1$, which is a boundary point of the parameter space. Standard interior-point asymptotic normality of the maximum-likelihood estimator therefore need not apply.

Indeed, put $a=1-e^{-\lambda }$. Under $\pi =1$, $N_{+}\sim \mathrm{Binomial}(n,a)$. For the unconstrained estimator $\tilde{\pi }=N_{+}/(na)$, the central limit theorem givesSince $\hat{\pi }=\min \{1,\tilde{\pi }\}$, the continuous mapping theorem yieldsThis distribution has probability $1/2$ at zero and a continuous normal half-density on the negative half-line, so it is not normal.