Neeraj,
Everything you have said is correct. Now, let's go back to the definition of
maximum likelihood estimate. It is the value of the parameter that
maximizes the joint density of the data that were actually obtained. For an
independent sample, the joint density is the product of the individual
densities, f(X_i).
Consider candidates theta: If theta is less than max(X_i), then the
probability of the sample if t = theta is zero, since f(X_i) is zero at
those values of X_i that exceed theta. If theta >= max(X_i), then the
density at each X_i is 1/theta, i = 1,2,..n, so that their product is
(1/theta)^n. This is maximized with respect to theta, subject the constraint
that theta >= max(X_i), by the smallest value of theta that is greater
than or equal to every X_i.
Ellen Hertz
Neeraj Nagarkatti wrote:
> Question:
>
> Let X_1,...,X_n be a random sample from the Uniform U[0,t] distribution,
> i.e. with pdf:
>
> f(x|t) = 1/t (0 < x < t).
>
> Obtain the maximum likelihood estimator of t.
>
> Now the model solution says:
>
> Because it's a non-regular case, you can't differentiate it, so the
> likelihood is:
>
> L(t) = 1/t^n, 0 < x_i < t, (i=1...n)
>
> which is the same as:
>
> L(t) = 1/t^n, 0 < max(X_i) < t.
>
> And clearly:
>
> ^
> t = max(X_i).
>
> Now I don't quite understand why the mle has to be the sample maximum.
> Can any1 shed any light as to why this is the case?
>
> Thank you.
>
> Neeraj.
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