How to find the Mean Squared Error of this estimator?

Question

Let $\lambda >0$ and $\mu = \lambda /2$. The variable $X$ has a Poisson distribution with parameter $\mu$. Let $X_1, ..., X_n$ be a sample from $X$ and $\theta = e^{- \mu}$. How to find the MSE of the estimator $G_n(X_1, ..., X_n) = e^{- \bar{X}_n}$? I already found that $\mathbb{E}G_n = \text{e}^{- \mu n(1-e^{-1/n})}$ and I know that $MSE(G_n) = \mathbb{E}(G_n - \theta )^2$, but I am not sure what to do next.

Answer 1

I think it is more convinient to find MSE directly without use of variance $$ MSE(G_n) = \mathbb{E}(G_n - \theta )^2 = \mathbb{E}(G_n^2) - 2\theta \mathbb{E}G_n +\theta^2=e^{-\mu n\left(1-e^{-2/n}\right)}-2\theta e^{-\mu n\left(1-e^{-1/n}\right)}+\theta^2. $$ Here $$ \mathbb{E}(G_n^2) = \mathbb{E}(e^{-2\bar X_n})=\mathbb{E}(e^{-\frac{2}{n}Y}) = M_Y\left(-\frac{2}{n}\right)=e^{-\mu n\left(1-e^{-2/n}\right)} $$ since $Y=X_1+\ldots+X_n$ has Poisson distribution with parameter $\mu n$ and $M_Y(t) = e^{\mu n (e^t-1)}$ is moment generating function of $Y$.