If $(X_i)$ are i.i.d. exponential $\lambda$, then $\hat\lambda=n/\sum{X_i}$ is a biased estimator of $\lambda$

Question

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

Answer 1

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}

Answer 2

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 $1/E[X]$.