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\m25 answer: Your approach to part b is sound. For part a, can you start with the definition of conditional probability as a fraction, and see what cancels?

Hint: look for an "n choose k" coefficient; what is the other name for "n choose k"?

For problem 4, can you give some more intuition on how to approach this?

For part a, I can see that the distibution associated with Z = \sum x_i is also a Poisson Distribition, but I can't figure out how to use that to get the Conditional distribution.

For part b, I'm assuming that \sum x_i is the sufficient statistic, since we were supposed to prove that in part a. I'm guessing that we should use an initial estimator of
P(x_i = 0 event) = 1 for x1 = 0, 0 for x1 != 0.

This would be an unbiased estimator, and then we could find what P(x_i = 0 event| \sum x_i = A) which should give the MVUE.

I think I have outlined the idea of how to do it, but I can't seem to do the specific computations to get the results. Could you please offer some ideas on how to actually compute these values. Thanks.

/m21 answer: Constant variance does not imply zero covariance. However, you may assume that all estimators are independent here. In other words, each estimator has a different argument x where all the x are independent of each other.

/m19 Here is a useful technique for solving an equation of type:
f'(x)=0 (minimize function by setting derivative equal to zero)
subject to
g(x) - 1 = 0

http://mathworld.wolfram.com/LagrangeMultiplier.html

Induction should also work.

I was trying by induction and it was not working, but I now think I was just trying to prove the wrong thing. I think I have it now.
Thanks,

Samory


< replied-to message removed by QT >

I think induction holds when the covariance of those estimators are zero. Do constant variances of estimators imply zero covariance?

Thanks,

/m19 answer: Can you do the proof by induction on n? Part b gives you the base case.

Hi,

About problem 1, c), I'm having a hard time finding the weights that optimize the variance. I can infer from part b) what the form might be, but I cannot so far prove it, and I'd like some direction if possible.

Thanks,

Samory

/13 answer: The definition of completeness talks about E[f(x)] for all functions f where x follows the distribution P(theta) for any theta. For MVUEs, instead of all functions f(x), we need all functions f(t(x)) so we use the distribution of t(x) induced by the distribution of x.

So yes, completeness is a property of a family of distributions, but for MVUEs we need it to be true of the family of induced distributions, not the original family of distributions of x.

/m9 answer: The answer to your question is no, I think. We discussed this in person. If anyone would like a more detailed answer here, please ask.

/m10 answer: I don't have Silvey in front of me, but if A is a matrix, then yes, A' is its transpose. The prime ' can also mean "derivative."

/m8 answer: You can cite it from a text, but if you do so, make sure that the result is stated carefully, and give a complete source citation.

If you derive it yourself, you will get some extra points.

/m11 Nevermind. I retract my question. I have found a solution that shows that the estimator is unbiased. Thanks.

What exactly does it mean for the family of distributions of t to be complete? I thought completeness was a property of the distribution family, i.e. \theta. What exactly are the family of distributions of t?

Sorry, I don't have time to answer these right now--I will early afternoon tody (Sat.)

For Problem 2a, I can show that we have g(x) can be written as g(t) so it is a function of t only. We've also show in class that Gaussians are complete. However, when I try to show that g(t) is unbiased, I keep getting that E[g(t)] = n/n-1 * \sigma^2. This doesn't seem correct since we are trying to show it is an MVUE. Can you comment on this? Are we supposed to assume that we are asymptotically unbiased. It seems that the MVUE should be the one suggested in the second case where we divide by n.

In the Silvey text, chapter three, what is A'? I forgot what this notation means. Is it the transpose?

Edited by author 01-28-2005 12:19 AM
For {\theta = \theta1, \theta2}, if we know that the family of distributions of t=(t1, t2) is complete, can we always say that when \theta1 is known, the family of distr. of t2 is complete?

For problem #2, do we have to derive the variance of the sample variance or can we cite it from a text?

/m6 answer: Constant variance means the variance is the same regardless of what the true theta is.

What does it mean for an estimator to have constant variance?

/m4 answer: This is actually about Problem 3. The biologist does just one experiment, with M fixed in advance, and obtains one integer, namely x.

Question about Problem 2:

Can she repeat the experiment, or is the experiment done once and only one value of X is obtained?

Yes, "constant" means the variance does not depend on theta.

In Problem 1,
What do we mean by constant variance?
Do we mean that the variance does not depend on theta?

Thanks

Please ask questions here about the second assignment for CSE 291, Winter 2005.