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First note thatis exponentially distributed with mean
Then we immediately haveand
(or you can prove it easily by integration)
a) For the M.L.E. part, I guess it is a standard work.
First write down the joint log-likelihood function:
Differentiate it with respect to the parameterand set it equal to 0:
b) To check the estimator is unbiased or not, you need to check the equality
which is obviously true in this case, because the sample mean is always an
unbiased estimator of the mean.
c) Again, by using the i.i.d. property of the random sample,
d) You should be able to prove that
either by moment generating function, or convolution and induction
Then you can do a simple transformation to obtain the p.d.f. of
Sorry if I give too detailed hints.
![Var[X] = \mu^2](/~talkmath/tex/img/46e45a64811f1a3bfd22026cfe50d107-1.gif "Var[X] = \mu^2")
![E[\hat{\mu}] = \mu](/~talkmath/tex/img/0916fda5ed7e6d65af764d820ed8e496-1.gif "E[\hat{\mu}] = \mu")
![E[\hat{\mu}] = E\left[\frac {\sum_{i=1}^n X_i} {n}\right]
= \frac {\sum_{i=1}^n E[X_i]} {n} = \frac {n\mu} {n} = \mu](/~talkmath/tex/img/381faaa5432c1b4117427c37a603ce20-1.gif "E[\hat{\mu}] = E\left[\frac {\sum_{i=1}^n X_i} {n}\right]
= \frac {\sum_{i=1}^n E[X_i]} {n} = \frac {n\mu} {n} = \mu")
![Var[\hat{\mu}] = Var\left[\frac {\sum_{i=1}^n X_i} {n}\right]
= \frac {\sum_{i=1}^n Var[X_i]} {n^2}
= \frac {n\mu^2} {n^2} = \frac {\mu^2} {n}](/~talkmath/tex/img/1fcb5006f3969d6fc08e2a221e3ab433-1.gif "Var[\hat{\mu}] = Var\left[\frac {\sum_{i=1}^n X_i} {n}\right]
= \frac {\sum_{i=1}^n Var[X_i]} {n^2}
= \frac {n\mu^2} {n^2} = \frac {\mu^2} {n}")
