Computing the bias of the estimator $e^{-t/\theta}$ I have a random variable with exponential distribution, X~ Exp($\frac{1}{\theta}$) and I need to estimate $\zeta=\mathbb{P}(x>3)$. I know that $e^{-3/\hat{\theta}}$ is the maximum likehood estimator of $\zeta$, with $\hat{\theta}=\bar{x}=\sum_{i=1}^{n}x_{i}/n$. Now I have to compute the MSE of $e^{-3/\hat{\theta}}$ but I don't know  how to start.
I know that I need the bias and variance and but how can I compute the mean of $e^{-3/\hat{\theta}}$? How can I find its density function?
My theacher said that we could use a symbolic calculator to compute an integral in this exercise so I think the answer will be related to compute $\int\theta\cdot g(\theta)d\theta$ with $g$ the density function of $\theta$.
 A: Consider $X\sim$ Exp($\frac{1}{\theta}$) with $\theta$ being a shape parameter, with $E(X)=\theta$, and $X_1,X_2,\ldots,X_n$ an i.i.d. $n$ sample from $X$, used to compute $\hat \theta=\bar X_n$, the maximum likelihood (ML) estimator for $\theta$.
We are interested in $\text{Pr}(X>3)=\exp(-\frac{3}{\theta})$. By the ML invariance theorem, $\exp(-\frac{3}{\hat \theta})$, with $\hat \theta=\bar X_n$, will be the ML estimator. As for the MSE of this estimator with respect to $\exp(-\frac{3}{\theta})$:
$$\text{MSE}=E\left[\left(\exp(-\frac{3}{\bar X_n})-\exp(-\frac{3}{\theta})\right)^2\right]=V\left[\exp(-\frac{3}{\bar X_n})\right]+\left[E\left(\exp(-\frac{3}{\bar X_n})\right)-\exp(-\frac{3}{\theta})\right]^2$$
If $g(\bar X_n)=\exp(-\frac{3}{\bar X_n})$, it is true that
$$E[(g(\bar X_n))^i]=\int_{\Omega_{\bar X_n}} (g(u))^i f_{\bar X_n}(u) du,\ \ i\in \{1,2\},\ \ \ \text{and}\ \ \ V[g(\bar X_n)]=E[(g(\bar X_n))^2]-E[g(\bar X_n)]^2.$$
The evaluation of these integrals will require the definition of the PDF for $\bar X_n$. Using basic properties of the MGF (moment generating function), is can be shown that $m_{\bar X_n}(t)=[m_X(t/n)]^n$, in which $m_X(t)=\frac{1/\theta}{1/\theta -t}$, so that $m_{\bar X_n}=\left(\frac{n/\theta}{n/\theta -t}\right)^n$, the MGF from a Gamma(scale$=n$,shape=$\theta/n$). Therefore, $\bar X_n\sim$ Gamma(scale$=n$,shape=$\theta/n$) and
$$E[(g(\bar X_n))^i]=\int_0^\infty (e^{\frac{-3}{u}})^i \dfrac{1}{(n-1)!}\left(\dfrac{n}{\theta}\right)^n u^{n-1}e^{-u\frac{n}{\theta}} du,\ \ i\in \{1,2\}.$$
The following table shows the magnitude of the MSE and its components (variance and bias$^2$) for $\theta=1$ and $n\in\{5,20,50,100\}$ considering a numerical evaluation of the integrals needed to define the terms. As for reference: $\text{Pr}(X>3)=\exp(-3/\theta)=0.0497870683679$ when $\theta=1$.

To facilitate replication and comparison, I present below the wxMaxima code used to compute the last line of the table ($n=100$, $\theta=1$).
(%i94)
    n:100$ 
    p:float(exp(-3))$
    e:(n^n)/(n-1)!*quad_qagi(exp(-3/x)*x^(n-1)*exp(-(n*x)), x, 0, inf)$
 e2:(n^n)/(n-1)!*quad_qagi(exp(-3/x)^2*x^(n-1)*exp(-(n*x)), x, 0, inf)$
    var: e2[1]-e[1]^2;
    bias2: (e[1]-p)^2;
    mse:var+bias2;
(%o92)  
    2.220131530374526*10^-4
    5.438613087810388*10^-7
    2.225570143462336*10^-4

