Truncated Poisson Asymptotics

Question

This is a homework problem. I have figured out part (a) but I need help with part (b). I include part (a) for completion.

Suppose $X_1,\ldots,X_n$ are iid Poisson random variables. Furthermore, let $Z_n$ be the proportion of zeroes observed i.e. $Z_n = n^{-1}\sum_{i=1}^n 1\{X_j=0\}$.

$(a)$ Find the joint asymptotic distribution of $\left(\bar{X}_n,Z_n\right)$

Since $\text{E}[X_1]=\theta$ and $\text{Var}[X_1]=\theta$, by the central limit theorem we have $$\sqrt{n}(\bar{X}_n-\theta) \overset{D}{\longrightarrow} Z_1,\quad Z_1\sim N(0,\theta)$$ and since $\text{E}[1\{X_1=0\}] = P(X_1=0) = e^{-\theta}$ and $$\text{Var}[1\{X_1=0\}] = \text{E}[1\{X_1=0\}^2] - \text{E}[1\{X_1=0\}]^2=e^{-\theta}-e^{-2\theta}=e^{-\theta}(1-e^{-\theta})$$ by the central limit theorem we have $$\sqrt{n}(Z_n-e^{-\theta}) \overset{D}{\longrightarrow} Z_2,\quad Z_2\sim N(0,e^{-\theta}(1-e^{-\theta}))$$ Furthermore we have $$\text{Cov}[X_1,1\{X_1=0\}] = 0 - \theta e^{-\theta}$$ Therefore, by the multivariate central limit theorem $$\sqrt{n}\begin{pmatrix} \bar{X}_n-\theta \\ Z_n - e^{-\theta}\end{pmatrix} \overset{D}{\longrightarrow} \mathbf{Y}, \quad \mathbf{Y} \sim \text{MVN}\left( \begin{pmatrix} 0 \\ 0 \end{pmatrix}, \begin{pmatrix} \theta & -\theta e^{-\theta}\\-\theta e^{-\theta} & e^{-\theta}(1-e^{-\theta})\end{pmatrix}\right)$$

$(b)$ Based on your answer in (a), find the asymptotic distribution of $\sum_{i=1}^n X_i \big/ \sum_{i=1}^n 1\{X_i>0\}$. This is an estimate of the mean $\text{E}[X|X\geq 1]$ from a truncated Poisson.

We have $$\dfrac{\sum_{i=1}^n X_i}{\sum_{i=1}^n 1\{X_i>0\}}=\dfrac{n\bar{X}_n}{n-nZ_n} = \dfrac{\bar{X}_n}{1-Z_n}$$ I do not know how to proceed from here! I have a ratio of two normal distributions (marginal normal, and jointly normal).

$(c)$ Compute the exact mean and variance from a truncated Poisson$(\theta)$ with zero values truncated; i.e. $X\sim \text{Poisson}(\theta)$, compute $\text{E}[X|X\geq 1]$ and $\text{Var}[X|X\geq 1]$. Compare this to the asymptotic result in (b).

Answer 1

I figured out that the teacher wanted us to use the multivariate delta method:

Part (b)

We have $$T=\dfrac{\sum_{i=1}^n X_i}{\sum_{i=1}^n 1\{X_i>0\}}=\dfrac{n\bar{X}_n}{n-nZ_n} = \dfrac{\bar{X}_n}{1-Z_n}$$Let $h(x,y)=x/(1-y)$ with gradient evaluated at $(\theta, e^{-\theta})'$ $ \mathbf{D} = \nabla h(\theta, e^{-\theta}) = \left(\dfrac{1}{1-e^{-\theta}}, \dfrac{\theta}{(1-e^{-\theta})^2}\right)' $. Then by the multivariate delta method $$\sqrt{n}\left( \dfrac{\bar{X}}{1-Z_n} - \dfrac{\theta}{1-e^{-\theta}}\right) \overset{D}{\longrightarrow} W, \quad W\sim \text{N}\left( 0, \mathbf{D}\Sigma \mathbf{D}'\right)$$ where $$ \mathbf{D}\Sigma \mathbf{D}' = \begin{pmatrix}\dfrac{1}{1-e^{-\theta}} & \dfrac{\theta}{(1-e^{-\theta})^2}\end{pmatrix}\begin{pmatrix} \theta & -\theta e^{-\theta}\\-\theta e^{-\theta} & e^{-\theta}(1-e^{-\theta})\end{pmatrix}\begin{pmatrix}\dfrac{1}{1-e^{-\theta}} \\ \dfrac{\theta}{(1-e^{-\theta})^2}\end{pmatrix} = \dfrac{\theta-\theta e^{-\theta}-\theta^2e^{-\theta}}{(1-e^{-\theta})^3}$$

Part (c)

$\begin{align*} \text{E}[X|X\geq 1] &= \dfrac{\sum_{k=1}^\infty kP(X=k)}{P(X\geq 1)}\\ &= \dfrac{\sum_{k=0}^\infty kP(X=k) - 0*P(X=0)}{1-P(X=0)}\\ &= \dfrac{\theta}{1-e^{-\theta}}\\ \text{Var}[X|X\geq 1] &= \text{E}[X^2|X\geq 1] - (\text{E}[X|X\geq 1])^2\\ &= \dfrac{\theta+\theta^2}{1-e^{-\theta}} - \left(\dfrac{\theta}{1-e^{-\theta}}\right)^2\\ &= \dfrac{\theta+\theta^2}{1-e^{-\theta}} - \dfrac{\theta^2}{(1-e^{-\theta})^2}\\ &= \dfrac{\theta-\theta e^{-\theta}-\theta^2e^{-\theta}}{(1-e^{-\theta})^2}\\ \dfrac{\text{aVar}[T]}{\text{Var}[X|X\geq 1]} &= \dfrac{\theta-\theta e^{-\theta}-\theta^2e^{-\theta}}{(1-e^{-\theta})^3} \Bigg/ \dfrac{\theta-\theta e^{-\theta}-\theta^2e^{-\theta}}{(1-e^{-\theta})^2}\\ &= \dfrac{1}{1-e^{-\theta}} < 1 \end{align*}$ The asymptotic variance of the estimator in part (b) is smaller than the exact variance