If $(X_i)$ are i.i.d. exponential $\lambda$, then $\hat\lambda=n/\sum{X_i}$ is a biased estimator of $\lambda$ the main problem is that i have no clue on calculating $E(\frac{1}{x})$
let $U = \frac{1}{\frac{1}{n}\sum{X_i}}$ then,
$E(U) = n*E(\frac{1}{\sum{X_i}})$. I think that i'm supposed to calculate:
$\int_0^\infty \frac{1}{x} \lambda e^{-\lambda x}$ but this is $E(\frac{1}{X})$ not $E(\frac{1}{\sum{X_i}})$
Any hints on solving this problem?
Kees
 A: Since $E[X_k]=1/\lambda$ for every $k$, $E[1/U]=1/\lambda$. By convexity, $E[U]\gt1/E[1/U]=\lambda$, thus, $U$ is a biased estimator of $\lambda$.
This applies to every distribution on $(0,+\infty)$, in the sense that $U$ is a biased estimator (biased upwards) of the parameter $1/E[X]$.
A: The distribution $X=\sum X$ is the Erlang distribution with parameters $\lambda$ and $n$. This distribution has the form
$$
f(x;n,\lambda)=\frac{\lambda^n x^{n-1} e^{-\lambda x}}{\Gamma(x)}
$$
(according to the wikipedia page. I don't know how to derive it.)
To find the expected value $E\left(\frac 1X\right)$, you have to evaluate
$$
\int_{x=0}^\infty \frac 1x \frac{\lambda^n x^{n-1} e^{-\lambda x}}{\Gamma(n)}\,dx
=\int_{x=0}^\infty\frac{\lambda^n x^{n-2} e^{-\lambda x}}{\Gamma(n)}\,dx
$$
This is a standard integral, with general solution:
$$
\int_{x=0}^\infty x^{n-1} e^{- x/a}\,dx=(n-1)!a^n
$$
With this, we find
\begin{align}
\int_{x=0}^\infty\frac{\lambda^n x^{n-2} e^{-\lambda x}}{\Gamma(n)}\,dx&=\frac{\lambda^n}{(n-1)!}\int_{x=0}^\infty x^{n-2}e^{-\lambda x}\,dx\\
&=\frac{\lambda^n}{(n-1)!}(n-2)!\left(\frac 1\lambda\right)^{n-1}\\
&=\frac\lambda{n-1}
\end{align}
