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I am having a bit of an argument with my study group about a Rao-Blackwell problem that we have for our statistical theory class. The problem goes like this:

Let X~U(0,$\theta$), and suppose we have a sample of size n, {$X_1,..X_n$} We know that both $T_1=Xbar$ and $T_2= \frac {n+1}{n} Max(X_1,...X_n)$ are unbiased estimators for $\theta$. Which of the two is a better estimator in terms of variance?

Now, I know that Rao-Blackwell states that E[$T_1|T_2]=T*$.

I have found other examples out there that show that:

$\frac{2}{n}E[\Sigma X_n|Max(X_1,...X_n)]=\frac{n+1}{n}Max(X_1,...,X_n)$

What I am dealing with is:

$\frac{2}{n}E[\Sigma X_n|\frac{n+1}{n}Max(X_1,...X_n)]=???$

What my other classmates in my study group came up with is that the answer is:

2 $E[Xbar|\frac{n+1}{n}Max(X_1,...X_n)]$

= 2[$\frac{{\frac{n+1}{n}Max(X_1,...X_n)}}{2}$]

=${\frac{n+1}{n}Max(X_1,...X_n)}$

Which I don't understand their reasoning of why to just divide the whole thing by two and everything just cancels out.

From what I have been reading, I know that the P($X_i=Max)=\frac{1}{n}$ and that P($X_i \ne Max)=Max/2$. Also, that one of the $X_i$ must be equal to the maximum and the other $n-1$ are half of the maximum.

So then what I get from trial and error is:

$\frac{2}{n}E[\Sigma X_n|Max(X_1,...X_n)]$

=$\frac{2}{n}(\frac{n+1}{n}X_{max}+\frac{n-1}{2}\frac{n+1}{n}X_{max})$

=$\frac{2}{n}(\frac{n+1}{n}X_{max}+\frac{n^2-1}{2n}X_{max})$

=$\frac{2}{n}(\frac{2n+2}{2n}X_{max}+\frac{n^2-1}{2n}X_{max})$

=$\frac{2}{n}(\frac{2n+2+n^2-1}{2n}X_{max})$

=$\frac{2}{n}(\frac{n^2+2n+1}{2n}X_{max})$

=$\frac{1}{n}(\frac{(n+1)(n+1)}{n}X_{max})$

=$\frac{{n+1}^2}{n^2}X_{max}$

=$({\frac{{n+1}}{n}})^2X_{max}$

Now, I am not sure if this is right, but it would seem that by this theorem, this variance is lower than $T_1$ by the Rao-Blackwell theorem, but I am not sure if this means that $T_2$ is a better estimator of the variance.

Could someone please check on this for me and let me know their thoughts?

Thank you!

Additional Question:

We were looking at the variance of $T_2$ vs. T*, and we are thinking that since $Var[T_2]<Var[T*]$ that this proves that $T_2$ is the better statistic in terms of variance.

Any thoughts on this one?

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1 Answer 1

An estimator is said to perform better compared to another if its MSE is lower. Since MSE=Bias+Vriance, it explicitely implies that the estimator with lower variance should be preferred.

Among all unbiased estimators, the estimator which achieves the Cramer-Rao lower bound has the lowest variance and is the best estimator. However, one cannot claim that such an estimator always exists.

In your problem $Max\{X_1,X_2,...X_N\}$ is the maximum likelihood estimator of $\theta$. However it is biased. $\frac{n+1}{n}$ is the bias correction term. With only this informations and without any further analysis, one can deduce that this estimator is at least asymptotically the best estimator. However in your problem you already have higher variance with $T_1$, therefore you are done.

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So we should have just looked at each of the variances individually and compared them instead of going through the whole Rao-Blackwell theorem? –  Perdue Mar 10 '13 at 0:40
    
yes. You dont need Rau-Blackwell theorem to show that one is better than the other. You will check MSE of both as I told you. –  Seyhmus Güngören Mar 10 '13 at 0:52
    
Well, I just feel silly about it now because that was what I originally thought and I just thought it was too easy that way. –  Perdue Mar 10 '13 at 14:13
    
@Perdue: You have accidentally created two accounts, which is why you were not able to comment on this post directly. Here is the process to merge your accounts: From any page footer -> 'contact us' >> 'Merge user profiles'. –  Zev Chonoles Mar 10 '13 at 15:44

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