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).

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 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.

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.

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.

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

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

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 05:23 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.

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).

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.

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?

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