Luke, I think you probably have it right. Do you mean you get p = 1/n? If so, this is plausible.

I am having trouble with Q3 where we need to express the distribution P(x1 | x1 + ... + xn)
If I let this equal ( P(x1 = a) * P(x2 + ... + xn = N-a) )/ P(x1 + ... + xn = N), I almost get a binomial distribution like:
(N choose a) * p^a * q^N-a
but it seems that I'm not getting a probability for p and q but instead I have n, the number of random variables

Anyone know what I'm doing wrong?

For problem 2.4, you don't have to evaluate var[S^2] explicitly. You can get it by looking up properties of the chi-squared distribution, following part (b) of Silvey 1.1.

From Nuno in /m11: "I have a question about problem 2.1. I have estimated the parameter theta which gives me directly the expectation of e^(-theta) but not necessarily its variance. Can the variance of e^(-theta) also be expressed as a function of the variance of theta or do we have to estimate it directly?"


This question is not phrased right because e^(-theta) is not a random variable, so it does not have an expectation.

An important fact to note: Suppose g(x) is an unbiased estimator of theta. Usually f(g(x)) is *not* an unbiased estimator of f(theta).

Edited by author 01-28-2004 05:23 PM
Andrew asked /m6:

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)?


Technically, we have a different estimator g_n for each n. And yes, the expectation of g_n is defined using the pdf for (x_1, ..., x_n) which is the n-fold product of p(x_i).

Jay's comment /m9 "I think a preciser definition of p* should be the joint probability of each x of (x_i)" is correct.

I started to say I got much more traction for 2.4 by relating variances, but I found an error in my proof. Back to square one...

Edited by author 01-28-2004 05:24 PM
Hint to Jay for 1.3 /m8: Think coin flips.

Edited by author 01-28-2004 05:26 PM
Jay, yes, you can use the results you mention in /m12.

Doug, thanks for pointing out the confusion about the insect question. I did not know that the current printing of Silvey is different!

Here is the question:

An entomologist samples at random from a large pop. of a species. She records the sex of each insect and stops sampling after obtaining M > 1 males, by whichtime the total sample size is x. Find an MVUE of theta, the proportion of males in the pop.

For 2.4, can I use the conclusion from 1.1(b) and "MSE = Variance + Bias" as granted? This makes the life a lot easier.

It is not hard to prove the second one, though.

Edited by author 01-28-2004 01:54 AM
I have a question about problem 2.1. I have estimated the parameter theta which gives me directly the expectation of e^(-theta) but not necessarily its variance. Can the variance of e^(-theta) also be expressed as a function of the variance of theta or do we have to estimate it directly?


Doug & Jay:

I believe that an easier way to prove 2.4 is to express one of the variances as a function of the other - start with the expression of one of the variances and transform it into the other. What remains will tell you which of the two is greater than the other.

Prof Elkan: The question about insect sex is 2.7 in my copy of Silvey. Question 2.8 is about the "Leaves of a plant". You may want to write out the question on this message board so that there is no confusion.

Edited by author 01-28-2004 12:58 AM
Doug:

I believe there's an easier way, although I don't know either. Mathematicians are lazy, they want everyting LOOKS simple and elegant.

When I work on this assignment, most the time I feel that it is almost there, I just need a little tricky manipulation to get it done.

I am looking for that trick to make this question easier.

Andrew:
"define a p* over (X^n) to be the product of each of the p(x_i)?"

This is true only when each x is independent to others.

I think a preciser definition of p* should be the joint probability of each x of (x_i).

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.