# biasedness/unbiasedness of an MLE.

To show whether an MLE I just found is biased/unbiased, would I need to find the expectation of the answer? Plus would I do this by integrating $\text{MLE} \cdot \text{pdf}$.

My MLE is $\frac{1}{\bar x}$ I've heard the expectation of this is the same as of the expectation of $\frac{1}{x}$

http://www2.imperial.ac.uk/~ayoung/m2s1/Exercises8.PDF question 6 part 2, I differentiated th log likelihood and set to zero to get $\hat\theta = \frac{n}{sum...} = \frac{1}{\bar x}$

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Do you want to state what pdf is in your case and, maybe, show how you derived the MLE? – Sasha Nov 11 '12 at 6:50
see edit, but looks like you beat me to it. – cheeseman123 Nov 11 '12 at 7:11
Could you nevertheless make your post self-contained? One SHOULD NOT have to refer to the content of the link to understand what is going on. – Did Nov 11 '12 at 8:02

I take it that you are dealing with the exponential distribution with $$f_X(x) = \lambda \mathrm{e}^{-\lambda x} [x > 0]$$ Assuming all elements of the sample $\{x_1,x_2,\ldots,x_n\}$ are positive, the log-likelihood reads: $$n \log \lambda - \lambda \sum_{k=1}^n x_k$$ which attains its maximum exactly at $\lambda = \frac{n}{\sum_{k=1}^n x_k}$.
Now to computation of the expectation of the MLE: $$\begin{eqnarray} \mathbb{E}\left(\frac{n}{X_1+X_2+\cdots+X_n}\right) &=& n \mathbb{E}\left(\frac{1}{X_1+X_2+\cdots+X_n}\right) \\ &=& n \mathbb{E}\left( \int_0^\infty \exp\left(-t(X_1+\cdots+X_n)\right) \mathrm{d}t \right) \\ &=& n \int_0^\infty \mathbb{E}\left( \exp\left(-t(X_1+\cdots+X_n)\right) \right) \mathrm{d}t \\ &\stackrel{\text{indep.}}{=}& n \int_0^\infty \left(\mathbb{E}\left( \exp\left(-t X_1\right) \right)\right)^n \mathrm{d}t \\ &=& n \int_0^\infty \left(\frac{\lambda}{t+\lambda}\right)^n \mathrm{d}t = \left. -\frac{n}{n-1} \frac{\lambda^n}{(t+\lambda)^{n-1}} \right|_{0}^\infty \\ &=& \frac{n}{n-1} \lambda \end{eqnarray}$$ Thus the MLE is biased.
Do you mind showing me how you got $$\begin{eqnarray} \\ &\stackrel{\text{indep.}}{=}& n \int_0^\infty \left(\mathbb{E}\left( \exp\left(-t X_1\right) \right)\right)^n \mathrm{d}t \\ \end{eqnarray}$$ – cheeseman123 Nov 11 '12 at 7:21
plus from there to $$\begin{eqnarray} &=& n \int_0^\infty \left(\frac{\lambda}{t+\lambda}\right)^n \mathrm{d}t = \left. -\frac{n}{n-1} \frac{\lambda^n}{(t+\lambda)^{n-1}} \right|_{0}^\infty \\ &=& \frac{n}{n-1} \lambda \end{eqnarray}$$ – cheeseman123 Nov 11 '12 at 7:59
@cheeseman123 Independence of $X_k$ was used here:$$\mathbb{E}\left(\exp(-t(X_1+\cdots+X_n))\right) = \mathbb{E}\left( \mathrm{e}^{-t X_1} \cdots \mathrm{E}^{-t X_n} \right) \stackrel{\text{indep.}}{=} \mathbb{E}\left( \mathrm{e}^{-t X_1} \right) \cdots \mathbb{E}\left( \mathrm{e}^{-t X_n} \right) \stackrel{\text{i.d.}}{=} \left(\mathbb{E}\left( \mathrm{e}^{-t X_1} \right)\right)^n$$ where the last equality follows because $X_k$ are identically distributed, thus means are equal. – Sasha Nov 11 '12 at 15:16