Point estimation help please

[RyuVI]

Hello,

I would appreciate if someone could explain more thoroughly the reasoning behind the answer to Question 10.10. I've read through the answer and even though the techniques seem to make sense I know I wouldn't have been able to do this myself. Perhaps it's because I am unfamilar with truncated data - I've never worked with truncated probabiilty distributions before.

I don't understand why in part (i) we calculated E(X) using this truncated poisson distribution for the method of moments estimator. (It kinda makes sense to me that we use the truncated poisson for the MLE).

Why can't we just calculate E(X) using (Sigma Xi)/n?

I thought that x bar = E(X) = the estimate for lambda for method of moments. But part (ii) implies this is not the case.

I would really appreciate some clarification if anyone can offer some.

RyuVI

 

[bobbathejobba]

Hello,

I would appreciate if someone could explain more thoroughly the reasoning behind the answer to Question 10.10. I've read through the answer and even though the techniques seem to make sense I know I wouldn't have been able to do this myself. Perhaps it's because I am unfamilar with truncated data - I've never worked with truncated probabiilty distributions before.

I don't understand why in part (i) we calculated E(X) using this truncated poisson distribution for the method of moments estimator. (It kinda makes sense to me that we use the truncated poisson for the MLE).

Why can't we just calculate E(X) using (Sigma Xi)/n?

I thought that x bar = E(X) = the estimate for lambda for method of moments. But part (ii) implies this is not the case.

I would really appreciate some clarification if anyone can offer some.

RyuVI

Hi ther RyuVI

Not quite sure what Q10.10 is - but I'll hazard a guess:

Method of moments equates E(X) - the mean of the distribution
to X bar - the mean of the sample

Since I guessing your sample data comes from the truncated distribution - that means your sample mean is of the truncated distrubiton. So you should stick it equal to the E(X) of the truncated distribution.

If you had data from the whole distribution (and so a sample mean of the whole distribution) then you could stick this equal to X bar.

Let me know if I'm totally up the spout - or someone else out there help out (though I suspect most are not that far through the course yet).

PS Bit worried you said you could work out E(X) using sigmaX/n. As that is the formula for the sample mean. For E(X) you use sigma xP(X=x) or the integral of xf(x).

 

[RyuVI]

Hi ther RyuVI

Not quite sure what Q10.10 is - but I'll hazard a guess:

Method of moments equates E(X) - the mean of the distribution
to X bar - the mean of the sample

Since I guessing your sample data comes from the truncated distribution - that means your sample mean is of the truncated distrubiton. So you should stick it equal to the E(X) of the truncated distribution.

If you had data from the whole distribution (and so a sample mean of the whole distribution) then you could stick this equal to X bar.

Let me know if I'm totally up the spout - or someone else out there help out (though I suspect most are not that far through the course yet).

bobba, you are a legend! Thanks so much for your reply, it makes sense to me now. You were spot on: x bar has to be equated to the expectation of the truncated distribution (poisson in this case) which was confusing me, especially the way this was calculated. (ie sigma [x.k.truncP(X=x)] where k is a constant of proportionality.) I didn't really get how k was then determined but it makes sense now, again I guess I'm just unfamiliar with truncated distributions.

PS Bit worried you said you could work out E(X) using sigmaX/n. As that is the formula for the sample mean. For E(X) you use sigma xP(X=x) or the integral of xf(x).

Yeah sorry I meant to ask why can't we equate E(X) to x bar (sample mean).

Thanks again,
Ryan.

 

[bobbathejobba]

Nice to know the brain cells still work!