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

Question

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 $\lambda=1$.

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(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.

Answer 1

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 $n\to \infty$.

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Your answer for the second part is correct.

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,

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