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Let $$X_1, \ldots, X_n \overset{iid}{\sim} Exp(\lambda), \quad \lambda > 0$$ The Maximum-Likelihood-Estimator is given by $$\widehat{\lambda} = \frac{1}{\frac{1}{n}\sum_{i=1}^{n}{X_i}} = \frac{n}{\sum_{i=1}^{n}{X_i}}$$ and it's not unbiased:

Using $$X_1, \ldots, X_n \overset{iid}{\sim} Exp(\lambda) \quad \Rightarrow \quad \sum_{i=1}^{n}{X_i} \sim \Gamma(\lambda, n)$$ and $$\Gamma(n+1) = n\cdot\Gamma(n) \quad \text{ for } \quad n \in \mathbb{N}$$ we get:

\begin{align*} E\bigg(\frac{n}{\sum_{i=1}^{n}{X_i}}\bigg) &= \int_{0}^{\infty}{\frac{n}{x}\cdot \frac{\lambda^n}{\Gamma(n)}\cdot x^{n-1}\cdot e^{-\lambda x}dx}\\ &= \int_{0}^{\infty}{\frac{n}{x}\cdot \frac{\lambda^{n-1}\cdot\lambda}{(n-1)\cdot\Gamma(n-1)}\cdot x^{n-1}\cdot e^{-\lambda x}dx}\\ &= \lambda\cdot\int_{0}^{\infty}{\frac{n}{n-1}\cdot\frac{\lambda^{n-1}}{\Gamma(n-1)}\cdot x^{(n-1)-1}\cdot e^{-\lambda x}dx}\\ &= \lambda\cdot\frac{n}{n-1}\cdot\int_{0}^{\infty}{\frac{\lambda^{n-1}}{\Gamma(n-1)}\cdot x^{(n-1)-1}\cdot e^{-\lambda x}dx}\\ &= \lambda\cdot\frac{n}{n-1} \end{align*} So $\widehat{\lambda}$ is not unbiased.

\begin{align*} E\bigg(\frac{n-1}{\sum_{i=1}^{n}{X_i}}\bigg) &= \int_{0}^{\infty}{\frac{n-1}{x}\cdot \frac{\lambda^n}{\Gamma(n)}\cdot x^{n-1}\cdot e^{-\lambda x}dx}\\ &= \int_{0}^{\infty}{\frac{n-1}{x}\cdot \frac{\lambda^n}{(n-1)\cdot\Gamma(n-1)}\cdot x^{n-1}\cdot e^{-\lambda x}dx}\\ &= \int_{0}^{\infty}{\frac{n-1}{x}\cdot \frac{\lambda^n}{\Gamma(n-1)}\cdot x^{n-1}\cdot e^{-\lambda x}dx}\\ &= \int_{0}^{\infty}{\frac{\lambda^{n-1}\cdot\lambda}{\Gamma(n-1)}\cdot x^{(n-1)-1}\cdot e^{-\lambda x}dx}\\ &= \lambda\cdot\int_{0}^{\infty}{\frac{\lambda^{n-1}}{\Gamma(n-1)}\cdot x^{(n-1)-1}\cdot e^{-\lambda x}dx}\\ &= \lambda \end{align*} So $\frac{n-1}{\sum_{i=1}^{n}{X_i}}$ is unbiased.

Is that correct?

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  • $\begingroup$ You have two typos. Firstly $Γ(n-1)=n!$ is not correct. But it is a typo since later you expand it correctly. Secondly, (not actually a typo) the third line of the first integral calculation is not necessary (you multiply with (n-1)/(n-1) but why? You already have the (n-1) term from writing (n-1)! as (n-2)!(n-1)). Otherwise it is fine! $\endgroup$ – Jimmy R. Dec 2 '14 at 13:19
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    $\begingroup$ Thank you. I corrected the typos. $\endgroup$ – GeMir Dec 2 '14 at 13:24
  • $\begingroup$ Hm, and now I started wondering why it is so: $\Gamma(\lambda,n)$ is the distribution of $\sum_{i=1}^{n}{X_i}$ (assuming $X_1,\ldots,X_n \overset{\text{iid}}{\sim} Exp(\lambda)$) but we've got $\frac{n}{\sum_{i=1}^{n}{X_i}}$... $\endgroup$ – GeMir Dec 6 '14 at 8:41
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    $\begingroup$ It is like when you want to calculate E[1/X] und you know the distribution of X. Don't worry. $\endgroup$ – Jimmy R. Dec 6 '14 at 8:53
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    $\begingroup$ Ok, I found this: Is $g(a)$ a continuous function and $E(g(X))$ exists, so is $E(g(X)) = \int_{-\infty}^{\infty}{g(x)\cdot f(x)dx}$ and $\frac{n}{x}$ is continuous on $\mathbb{R}^{+}$. $\endgroup$ – GeMir Dec 6 '14 at 18:23

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