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The maximum likelihood estimator of an exponential distribution $f(x, \lambda) = \lambda e^{-\lambda x}$ is $\lambda_{MLE} = \frac {n} {\sum x_i}$; I know how to derive that by taking the derivative of the log likelihood and setting it equal to zero.

I then read in an article that "Unfortunately this estimator is clearly biased since $<\sum_i x_i>$ is indeed $1/\lambda$ but $<1/\sum_i x_i > \neq \lambda$."

Why does $<\sum_i x_i> = 1/\lambda$? If I am correct in deducing the $< >$ operator means expected value, then I thought $E(x_i) = 1/\lambda$ - that is, the expected value of one such $x_i$, is $1/\lambda$, not the sum of all $x_i$'s. And can someone explain the second of the statement and how these two statements demonstrate the MLE is biased?

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This should read: "This estimator is biased since hence $E(1/\lambda_{MLE})=1/\lambda$ but $E(\lambda_{MLE})\ne\lambda$".

The fact that $E(1/\lambda_{MLE})=1/\lambda$ is direct since $E(x_i)=1/\lambda$ for every $i$ and $1/\lambda_{MLE}=(x_1+\cdots+x_n)/n$.

And, by convexity, $E(\lambda_{MLE})\gt1/E(1/\lambda_{MLE})=\lambda$.

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How do you tell E(MLE) is not lambda? – fool May 30 '14 at 18:21
Read on my lips: by convexity. – Did May 30 '14 at 18:22
Oh, whoops, didn't detect the flow of your answer. – fool May 30 '14 at 18:23

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