independent exponential distribution random variables, with parameter
show the MLE is biased.
yes. i have done this.
i found the MLE is just
and how can i calculate its expectation ?
i know sum of Xi is just Erlang Distribution, and i know its p.d.f.
so if i integrate it,
then i got some messy terms cancel out, but then i just found its expectation : . Therefore, this MLE is unbiased.
which is not like the question asks.
If the MLE is biased then you'll get that . 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 . Then the pdf of is
Let . Then the pdf of is
Let . To get the pdf of , first consider the cdf:
therefore the pdf of is .
Now calculate .
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.