I'm trying to find the asymptotic relative efficiency of a Poisson process:
$$\frac{\lambda^t \exp(-\lambda)}{t!} = P(X=t).$$
When $X = t = 0$, the best unbiased estimator of $e^{-\lambda}$ is ((n-1)/n)$^y$, where y = $\Sigma{X_i}$, the complete sufficient statistic for $\lambda$. When $X = t = 1$, the best unbiased estimator of $\lambda e^{-\lambda}$ is $\frac{y}{n}\left(\frac{n-1}{n}\right)^{y-1}$.
It is stated that the asymptotic relative efficiency for $t=0$ is:
ARE $$\left(\left( \frac{n-1}{n}\right)^{n \hat{\lambda}}, \quad\frac{y}{n}\left(\frac{n-1}{n}\right)^{y-1}\right) = \left[\frac{e^{-\lambda}}{\left(\frac{n-1}{n}\right)^{n\lambda}\log\left( \frac{n-1}{n}\right)^n}\right]^2$$
I'm not able to calculate this.
The asymptotic relative efficiency compares the variance of $T_1=\displaystyle\left(\frac{n-1}{n}\right)^{n\hat{\lambda}}$ to the variance of the second $T_2=\frac{y}{n}\left(\frac{n-1}{n}\right)^{y-1}$. I will find the variance of the first UMVUE $T_1$ using the Delta Method:
Since $T(X)=\sum X_i$ is the sufficient statistic for $\lambda$, and $\sum X_i\sim \mathrm{Poisson}(n\lambda)$, then $\operatorname{Var}\left(\sum X_i\right)=n\lambda (1-\lambda)$ assuming $X_i$'s are iid.
To find the variance of $T_1$, let $g(x)=\left(\frac{n-1}{n}\right)^x$ so using Delta Method, $g'(x)=\displaystyle\left(\frac{n-1}{n}\right)^{x}\log\left(\frac{n-1}{n}\right)$
$(g'(\lambda))^2=\displaystyle\left(\frac{n-1}{n}\right)^{\lambda}\log\left(\frac{n-1}{n}\right)$
$\operatorname{Var}(T_1)=(g'(\lambda))^2\operatorname{Var}(\sum X_i)=\left(\frac{n-1}{n}\right)^{2\lambda}\left(\log\frac{n-1}{n}\right)^2n\lambda (1-\lambda)$.
The variance of $T_2$ can be found similarly.