$\displaystyle X_{1},...,X_{n} , $ independent exponential distribution random variables, with parameter $\displaystyle \theta. $
show the MLE is biased.
yes. i have done this.
i found the MLE is just $\displaystyle \frac{n}{\sum_{1}^{n}{x_{i}}} $
and how can i calculate its expectation ?
i know sum of Xi is just Erlang Distribution, and i know its p.d.f. $\displaystyle \frac{\theta^{n}x^{n-1}e^{-\theta x}}{(n-1)!} $
so if i integrate it, $\displaystyle \int_{0}^{\infty}{\frac{\theta^{n}{x^{n}e^{-\theta x}}}{(n-1)!}dx} $
then i got some messy terms cancel out, but then i just found its expectation : $\displaystyle \frac{n}{\theta} $. Therefore, this MLE is unbiased.
which is not like the question asks.
If the MLE is biased then you'll get that $\displaystyle E(\hat{\theta}) \neq \theta$. Your answer agrees with my very rough calculation on a scrap bit of napkin but I'm not convinced you got it using a valid approach.
Here is an outline of an approach I considered (there is probably some clever theorem that gives the required result in two lines that will be brought to my attention):
Let $\displaystyle X = X_1 + X_2 + \, .... \, + X_n$. Then the pdf of $\displaystyle X$ is $\displaystyle f(x) = \, ....$
Let $\displaystyle U = \frac{X}{n}$. Then the pdf of $\displaystyle U$ is $\displaystyle g(u) = n f(nu) = \, ....$
Let $\displaystyle W = \frac{1}{U}$. To get the pdf of $\displaystyle W$, first consider the cdf:
$\displaystyle H(w) = \Pr(W < w) = \Pr \left( \frac{1}{U} < w \right) = \Pr \left( U > \frac{1}{w} \right)$ $\displaystyle = \int_{1/w}^{+ \infty} g(u) \, du$ $\displaystyle = 1 - \int^{1/w}_{0} g(u) \, du = \, .... $
therefore the pdf of $\displaystyle W$ is $\displaystyle h(w) = \frac{dH}{dw} = \frac{1}{w^2} \, g \left(\frac{1}{w} \right)$.
Now calculate $\displaystyle E(\hat{\theta}) = E(W) = \int_0^{+\infty} \frac{1}{w} \, g\left(\frac{1}{w} \right) \, dw$.
By exploiting known results from the Erlang distribution the calculation here should not be too arduous.
I got what you got but even when I'm being careful my carelessness is legendary. So be warned.