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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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    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.2012-11-11

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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.

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    Oh right, so yo$u$ just used the known result. I should have worked that one out backwards. Thank you greatly for the hel$p$.2012-11-12