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?
