# Determine the limiting distribution of $n[g(\bar{X}_n)-1/e]$ of iid Poisson samples with two estimators

Let $$X_1,\dots, X_n\sim$$iid Poisson with mean $$\lambda$$, and consider estimating $$g(\lambda)=P_{\lambda}(X_i=1)=\lambda e^{-\lambda}$$ One natural estimator might be the proportion of ones in the sample: $$\hat{p}_n=\frac{1}{n}\#\{i\le n: X_i=1\}=\frac{1}{n}\sum_{i=1}^nI[X_i=1]$$ Another choice would be the maximum likelihood estimator $$g(\bar{X}_n)$$ with $$\bar{X}_n$$ the sample average.

(1) Find the asymptotic relative efficiency of $$\hat{p}_n$$ w.r.t. $$g(\bar{X}_n)$$.

(2) Determine the limiting distribution of $$n[g(\bar{X}_n)-1/e]$$ under $$\lambda=1$$.

(1) The log-likelihood function is $$l_n(\lambda)=-n\lambda+n\bar{X}_n\log\lambda-\log \prod x_i!$$ with $$l''_n=-n\bar{X}_n/\lambda^2$$.

So $$I(\lambda)=-E[-n\bar{X}_n/\lambda^2]=n/\lambda$$

But how about the $$\hat{p}_n$$?

Note that from CLT $$\sqrt{n}(\hat{p}_n-E[I(X_i)=1])=\sqrt{n}(\hat{p}_n-\lambda e^{-\lambda})\to^d N(0,v)$$ where $$v=Var(I(X_i)=1)=\lambda e^{-\lambda}-(\lambda e^{-\lambda})^2$$.

(2) I have no idea about $$1/e$$...

$$n[g(\bar{X}_n)-1/e]=n[\bar{X}_ne^{-\bar{X}_n}-1/e]$$

Take $$f(x)=xe^{-1}-1/e$$ with $$f'(x)=e^{-x}-xe^{-x}$$ and $$f''(x)=(x-2)e^{-x}$$. Note that $$f(1)=f'(1)=0$$ and $$f''(1)=-1/e$$. Thus, $$f(x)\approx \frac{1}{2}f''(1)(x-1)^2$$ This means $$n[g(\bar{X}_n)-1/e]=nf(\bar{X}_n)\approx n\frac{-1}{2e}(\bar{X}_n-1)^2\to \frac{-1}{2e}\chi_1^2?$$ I am confused here because $$E X_1=\lambda$$ and I cannot use CLT.

Suppose $$\hat g_1(\lambda)=g(\overline X_n)=\overline X_ne^{-\overline X_n}$$ and $$\hat g_2(\lambda)=\frac1n\sum\limits_{i=1}^n I(X_i=1)$$.

Provided $$\lambda\ne 1$$ (so that $$g'(\lambda)\ne 0$$ ), by delta method,

$$\operatorname{Var}(\hat g_1) \approx \frac{\lambda (g'(\lambda))^2}{n}=\frac{\lambda e^{-2\lambda}(1-\lambda)^2}{n} \quad , \text{ for large }n$$

And the exact variance of $$\hat g_2$$ is

$$\operatorname{Var}(\hat g_2)=\frac{\lambda e^{-\lambda}(1-\lambda e^{-\lambda})}{n}$$

Note that $$\hat g_1$$ is asymptotically unbiased for $$g(\lambda)$$ (by delta method) and $$\hat g_2$$ is exactly unbiased.

Asymptotic relative efficiency of $$\hat g_2$$ with respect to $$\hat g_1$$ is the limit of the ratio of the variances of $$\hat g_1$$ and $$\hat g_2$$ as $$n\to \infty$$.

When $$\lambda=1$$, by CLT,

$$\sqrt n(\overline X_n-1) \stackrel{d}\longrightarrow Z \,,\quad\text{ where }Z\sim N(0,1)$$

Delta method in this case says that (provided $$g''(1)\ne 0$$, which holds here)

$$n\left(g(\overline X_n)-g(1)\right)\stackrel{d}\longrightarrow \frac{Z^2}{2} g''(1)$$

[ The proof is similar to the proof for the usual delta method, except here you need a second order approximation. Note that $$n\left(g(\overline X_n)-g(1)\right)=\frac{n(\overline X_n-1)^2}{2!}g''(\overline X_n^*)=\frac{(\sqrt n(\overline X_n-1))^2}{2}g''(\overline X_n^*)\,,$$ where $$\overline X_n^*$$ lies between $$\overline X_n$$ and $$1$$. ]

Therefore, $$n\left(g(\overline X_n)-\frac1e\right)\stackrel{d}\longrightarrow -\frac{1}{2e}\chi^2_1$$

In other words,

$$2ne\left(\frac1e-g(\overline X_n)\right) \stackrel{d}\longrightarrow \chi^2_1$$