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

/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.

/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."

/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.

/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.

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

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

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

Thanks,

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 >

/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.

/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.

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.

\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"?

Can anyone tell me what is the probability of two people writing a 50 word paragraph to be exactly the same?

Gold