MLE (maximum likelihood) question

I have questions about two situations:

I call the mle of a parameter as hat( parameter )


1 - Let X1, ..., xn iid uniform (a , b)

Then I can argue that hat( a) = X(1) and hat( b ) = X(n), where X(i)
is the ith order statistic.

Now put lambda = int{ x dF(x) }, ie the mean.

Does the equivariance of MLEs allow me to simply say that the mean of
the above distribution is (a + b) / 2, so the MLE of the mean is
( hat(a) + hat(b)) / 2? I think so.



2 - X1, ..., Xn iid N(mu, sigma^2) with mu, sigma unknown.

It is straightforward to prove that hat( mu ) = sample average and hat
( sigma^2 ) = n^-1 sum_{i=1}^{n} (xi - hat(mu))^2

Now I want to find the mle of tau = 95th percentile of X. Thus P(X <
tau) = 0.95, so tau = F^{-1}( 0.95 ), where we define F^{-1}( 0.95 ) =
inf{r real | F(r) = 0.95 }

I put \hat( F )(y) = int_{ - infty }^{ y } ( \hat( sigma^2)^-0.5
phi( u - \hat( mu)) du, with phi the standard normal pdf.

\hat( tau ) = \hat( F )^{-1}( .95 )... yes? Is this correct, and is
there a more elegant way to express this?

Thanks all.

earl

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