# Biased Estimator of Exponential Distribution with Poisson Rate

I have this problem that asks to prove that an estimator is biased:

Here's the problem:

Let $X_1, X_2, ... X_n$ be a random sample from an exponential distribution with Poisson rate $\lambda > 0$ (and mean $1/\lambda$). Consider the estimator $\hat \Lambda = \frac{1}{\bar X} = \frac{n}{\sum X_i}.$

I'm supposed to show that it's biased, find MSE and find a $C$ that will make it unbiased but this is what I got:

$E(\hat\Lambda) = E(\frac{n}{\sum X_i})$

• It seems that you think that $\mathsf E\frac1{Y}$ equals $\frac1{\mathsf EY}$. That however is not true in general. – drhab Nov 22 '17 at 16:15
• Can you please explain why? Doesn't it fall under the property $E(aX) = aE(X)$? – iamnobody Nov 22 '17 at 16:25
• No, it does not fall under linearity of expectation. Note that $x\mapsto\frac1{x}$ is not a linear map. It will explain itself if you try it out on some random variable $Y$ and in a certain sense you are asked to do that ("prove that the estimator is biased"). – drhab Nov 22 '17 at 16:31
• Thank you for pointing that out! That will be a really big help for me. However, I'm still stuck solving this problem. I'm guessing I have to integrate it instead to get the expected value? But I've never encountered integrating a summation before so I'm unsure how to proceed. – iamnobody Nov 22 '17 at 17:11

## 1 Answer

Here is one way to do it. First, start by noting that a sum of iid exponential random variables is a gamma random variable. More precisely, if $X_1, \ldots, X_n \sim \exp(\lambda)$, then $$\sum X_i \sim \textrm{Gamma}(n, 1/\lambda)$$ Setting $Y := \sum_{i=1}^n X_i$ for a moment, this means that the PDF of $Y$ is: $$f_Y(y) = \frac{1}{\Gamma(n)(1/\lambda)^n} y^{n-1} e^{-y/(1/\lambda)}$$ Then the expected value of $1/Y$ is the following integral: $$E(1/Y) = \int_0^{\infty} \frac{1}{\Gamma(n)(1/\lambda)^n} y^{n-1} e^{-y/(1/\lambda)} \cdot \frac{1}{y} \; dy$$ Doing a little algebra here and using the fact that $\Gamma(n) = (n-1)!$, you can rewrite this as $$E(1/Y) = \frac{\lambda}{n-1} \cdot \int_0^{\infty} \frac{1}{\Gamma(n-1)(1/\lambda)^{n-1}} y^{(n-1)-1} e^{-y/(1/\lambda)} dy$$ Now the trick is to recognize that the integrand is just the PDF of a Gamma($n-1$, $1/\lambda$) random variable, so it integrates to one and therefore you're just left with $E(1/Y) = \lambda/(n-1)$. Now you can do the rest.