Suppose X ~ N(µ, σ^2), then we may have found that the sample mean is unbiased and also the MVUE for population mean µ and hence COV(sample mean, U_0)=0, where U_0 means E(U_0)=0. Now If I consider the sample median which is unbiased for population mean then we may consider as E(sample mean - sample median) = 0. Now if we consider COV(sample mean, {sample mean - sample median}, then what will be the result of this? Is it zero? I am facing a problem in this regard. Please, if possible explain this to me.
Here we can consider (sample mean - sample median)=U_0. Then COV(sample mean, U_0} is what you want. Your former result shows that it is 0 (zero).
Dear @Roshini Sooriyarachchi ,
If I directly calculate the value of COV(sample mean, U_0}, Is it zero?
That means e.g. COV(sample mean, U_0}= COV(sample mean, sample mean)-COV(sample mean, sample median)=VAR(sample mean)-COV(sample mean, sample median) =?...
Dear Dhruba,
Then you have the added problem of calculating the covariance between the sample mean and the sample median and more importantly you are not using the given results!
Yes, Respected @Roshini Sooriyarachchi , why I am unable to get the result COV(sample mean, U_0) directly. In our case it will be zero if COV(sample mean, sample median) = VAR(sample mean). but how it is possible please explain.
Dhruba. As an scheptic -from intuition- about variances and covariances I have no good arguments to contribute to the question -I should have them-. In this moment I am reading an article that is related to the topic: http://www.phil.vt.edu/dmayo/personal_website/Error_Statistics_2011.pdf by Barbarah Mayo and Aris Spanos. It questions why some important statistical theories are centered in "error" but not in "data" and it deals with same terms of your question description. I hope it helps.Thanks, emilio
Dear Druba,
This is a long explanation but I am going to do it in detail. Let me use the notation mu for population mean , xbar for sample mean and xm for sample median.
then Covariance(xbar, xm)=E[(xbar-mu)(xm-mu)=E(xbar.xm)-mu^2.........................(1)
and variance(xbar)=cov(xbar,xbar)=E(xbar^2)-mu^2 --------------(2)
to prove (1)=(2) need to show that E(xbar.xm)=E(xbar^2)
that is to show that E[xbar(xm-xbar)]=0
As (xm-xbar)=0 then from theorem In parametric inference, the necessary and sufficient condition that T is MVUE of Ѱ(Ө), is given by COV(T, U_0)= 0, where E(U_0)=0
and Cov[xbar,(xm-xbar)] =E[xbar(xm-xbar)-E(xbar)E(xm-xbar)==E[xbar(xm-xbar)=0
Hence the result.
Please explain me a little bit about the last term of your calculation E(xbar(xm-xbar))=0....why and how?
You have to use the theorem given. That is Cov[xbar,(xm-xbar)]=0
If you expand Cov[xbar,(xm-xbar)] according to definition then it is equal to
E[xbar(xm-xbar)-E(xbar)E(xm-xbar)] where capital E means expectation. As the latter part of this term contains E(xm-xbar) which can be expanded as E(xm)-E(xbar) and as
E(xm) = E(xbar) = mu then the latter term disappears. Then
Cov[xbar,(xm-xbar)] =E[xbar(xm-xbar) which is the result that leads you to
Covariance(xbar, xm) = variance(xbar) and your proof.
The proof is so confusing for me.....
Actually I am going to prove that
Cov[xbar,(xbar-xm)]=0, Where xbar and xm both are unbiased for mu.
Now, I want to show that xbar is MVUE.
Consider
Cov[xbar,(xbar-xm)] = Var(xbar) - Cov(xbar, xm)
Now if we have to show that Cov[xbar,(xbar-xm)] = 0, then we have to assume that
Var(xbar) = Cov(xbar, xm).
But why this result do not come automatically. Actually I am going to apply the theorem of MVUE by considering xbar, xm where both are unbiased for mu.
Yes, xbar is a UMVUE of mu. This can be proved as xbar is a complete sufficient statistics (see http://en.m.wikipedia.org/wiki/Sufficient_statistic) in a Guess distribution. Then, according to http://en.m.wikipedia.org/wiki/Minimum-variance_unbiased_estimator and considering that xbar is a unbiased estimate of mu, the conclusion is proved.
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