I am stuck with the following problem: We know that $x_1 \cdots x_n$ ~ Ex$(1/\lambda)$ and we are given the sum $\sum_{i=1}^n x_i$. We want to find the maximum likelihood estimate of $\lambda$, as well as its variance. Also we want to find a $(1-\alpha)$ - confidence interval.
I know that E $ [x_i] = 1/\lambda $ and that the likelihood is $L(\frac{1}{\lambda} | x_1\cdots x_n) = \lambda^n e^{-\sum_i x_i\lambda}$. Therefore the ML estimate should be \begin{align} \hat{\lambda}= \frac{n}{\sum_i x_i}. \end{align} However, I don't know how to calculate the expectation value or the variance of $\hat{\lambda}$ as this involes $E[\frac{1}{\sum_i x_i}]$ and I know that normally \begin{align} E [\frac{1}{X}] > \frac{1}{E[X]}. \end{align} Therefore I guess that E$[\hat{\lambda}] > \lambda$, but I don't know how I would calculate it. Thank you already.
$S=\sum X_i \sim Gamma(n, 1/\lambda).$ If $S\sim Gamma(n,a)$ then $E(S^t)=\int_0^\infty x^t \frac{1}{\Gamma(n)a^n}e^{-x/a}x^{n-1}dx=\frac{\Gamma(n+t)a^{n+t}}{\Gamma(n)a^n}\int_0^\infty \frac{1}{\Gamma(n+t)a^{n+t}}e^{-x/a}x^{n+t-1}dx=\frac{\Gamma(n+t)a^{n+t}}{\Gamma(n)a^n}$ for all $t>-n$. $Var(n/S)=E(n^2/S^2)-E^2(n/S)=n^2\left(\frac{\Gamma(n-2)a^{n-2}}{\Gamma(n)a^n}-\left(\frac{\Gamma(n-1)a^{n-1}}{\Gamma(n)a^n}\right)^2\right)=n^2\left(\frac{1}{(n-1)(n-2)a^2}-\left(\frac{1}{(n-1)a}\right)^2\right)=\frac{n^2}{a^2(n-1)^2(n-2)}.$