I got a close form solution for 1.3. But I couldn't match it with any known famous distribution. Any hint?
Maybe it isn't famouse just for me.:)

Silvey 2.4 - In order to solve this question we need to calculate Var[S^2]. The way we tried solving this problem involves finding a value for the fourth moment of x_i as well as a good amount of additional algebra. Is there another approach that doesn't require so much calculation?

Also, is this calculation necessary for the problem set? This value if given in the recommended text for the course. Is it alright if we proceed knowing the value of Var[S^2].

I have a conceptual question about expectations and estimators (or maybe it's just a notation question...):

Estimators are functions that map a set of independent samples to a real number; they map n-tuples to R, where n is allowed to vary and each value of the tuple is in the sample space X.

The expectation of Y is defined as a sum over the entire sample space, of the density at each point, p(x), times y at that point, y(x).

So, precisely, how is the expectation of an estimator defined, given that p is defined over X and the estimator g is defined over X^n? Do we define a p* over (X^n) to be the product of each of the p(x_i)?

No, I mean Silvey 2.8, about insect sex :-) It's a sort-of fun sort-of real world application--imagine a biologist comes to you with this data and asks you to rescue his/her paper. Researchers often collect data and then only afterwards consult an expert about how to analyze it.

If problem 4 is supposed to be 2.7 and not 2.8 the comments on the assignment would make a lot more sense. Or are we supposed to show that the estimator given is the MVUE, and then find its efficiency?

I assume that problem (4), Silvey 2.8, is supposed to be Silvey 2.7.

This is correct. Fortunately for those who like them :-) MVUE examples, and probability calculations in general, often involve non-trivial indefinite integrals.

Some thought on Example 2.6.1 (Silvey's book, page 35)

I finally figured out how the auther easily spot the unbiased estimator 1/x1. Still not sure if my proof is correct. So I posted it here. Any comment is welcomed.

Here we assume x=(x1). To prove 1/x is a unbiased estimator, we need to prove E[1/x]=theta. By integrating (theta^2n)*exp(-theta*x)*x*(1/x) from 0 to +infinite, we found E[1/x]=theta.