# Truncated Poisson Asymptotics

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).

• The ratio of two zero-mean normals is a Cauchy. Maybe you want to pivot from here. Dec 3 '13 at 17:27
• I do know about the ratio of normals is cauchy...but it is not something that we covered in class and so I hesitate going down that route. To be honest I'm not quite sure what my teacher expects. Based off the second part of the question it sounds like they want an 'estimate' (e.g. what does it converge to in probability, not distribution) Dec 3 '13 at 17:31
• A convergence in probability result is rather straightforward (as I guess you know, plim goes places that the expected value can't). Dec 3 '13 at 17:51
• Yes, I will ask the teacher what exactly he is looking for. Dec 3 '13 at 18:32

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 $$\mathbf{D}\Sigma \mathbf{D}' = \begin{pmatrix}\dfrac{1}{1-e^{-\theta}} & \dfrac{\theta}{(1-e^{-\theta})^2}\end{pmatrix}\begin{pmatrix} \theta & -\theta e^{-\theta}\\-\theta e^{-\theta} & e^{-\theta}(1-e^{-\theta})\end{pmatrix}\begin{pmatrix}\dfrac{1}{1-e^{-\theta}} \\ \dfrac{\theta}{(1-e^{-\theta})^2}\end{pmatrix} = \dfrac{\theta-\theta e^{-\theta}-\theta^2e^{-\theta}}{(1-e^{-\theta})^3}$$
\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