Checking consistency of an estimator.

Question

The problem

Let $X_1,X_2,...$ be a i.i.d with Poisson distribution with parameter $\lambda$.

Check if estimator $$Y_n=(1-\frac{a}{n})^{X_1+X_2+...+X_n}$$

is consistent and unbiased with parameter $y=e^{-a\lambda},\;\;a>0$.

My work so far

Consistent

We want to show that $Y_n\rightarrow y \;\;\;\;a.s.$ Let's note that we can equivalently show that $(Y_n)^\frac{1}{n} \rightarrow y^\frac{1}{n}. \;\;\;\;a.s.$

So

$(Y_n)^\frac{1}{n}=(1-\frac{a}{n})^\frac{{X_1+X_2+...+X_n}}{n} \rightarrow (1-\frac{a}{n})^{\lambda}\rightarrow e^{-a \lambda} =y \neq y^\frac{1}{n}$.

I used the law of great number's above.

So ($Y_n)$ is not a consistent estimator.

Am I thinking correctly ?

Answer 1

For consistency note that by taking logs we get $$ a\times\frac{X_1+\dotsb+X_n}{n}\times\frac{\log(1-\frac{a}{n})}{a/n}\stackrel{\text{a.s.}}{\to}a\times\lambda\times-1=-a\lambda $$ as $n\to \infty$ by the strong law of large numbers and the fact that $$ \lim_{x\to0}\frac{\log(1-x)}{x}=-1. $$ Hence $$ Y_n=(1-\frac{a}{n})^{X_1+X_2+...+X_n} \stackrel{\text{a.s.}}{\to}e^{-a\lambda} $$ (convergence in probability also follows).

For unbiasedness we can argue as follows. By independence and the identically distributed condition $$ EY_n=\left(E\left(1-\frac{a}{n}\right)^{X_1}\right)^n $$ but $$ E\left(1-\frac{a}{n}\right)^{X_1}=\sum_{k=0}^\infty \left(1-\frac{a}{n}\right)^{k}\frac{\lambda^k}{k!}e^{-\lambda}=e^{-\lambda} \sum_{k=0}^\infty \left(\lambda-\frac{\lambda a}{n}\right)^{k}\biggr/k!=e^{-\lambda}e^{\lambda(1-a/n)}=e^{-\lambda a/n } $$ whence the result follows.