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Let $f(x)=\theta x^{-(\theta+1)}$ for $x>1$, where $\theta>1$ is an unknown parameter. Can we find an unbiased estimator for $\theta$ in this case which reaches the Cramer Rao Lower Bound?

My attempt:

To find such an estimator, I think I first need to find a sufficient statistic. After that, I think the Rao-Blackwell theorem should be applied. Therefore, I tried to use the following theorem: (Screenshot from the Wikipedia website). Fisher-Neyman Factorization Theorem (Wikipedia)

In my case this would imply $\prod\limits_{i=1}^n \theta x_i^{-\theta-1}=\prod\limits_{i=1}^n x_i^{-1}\theta x_i^{-\theta}$.

I am in doubt whether this is in the form required by theorem or not (I would say no). Is it in the required form? If not, would there be an other way to factor the function properly?

Michael Hardy
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dreamer
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    Was the support of the distribution specified? This works as a density on the interval from $1$ to $\infty$, but not on the interval from anything else to $\infty$. It also works on the interval from $a<1$ to some finite number bigger than that. Probably what was intended was $(1,\infty)$. I'd state that in posing the problem. – Michael Hardy May 29 '13 at 14:52
  • My bad, I forgot to include that. I will add the information to the O.P. – dreamer May 29 '13 at 14:54
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    Your joint density for an i.i.d. sample of size $n$ is $\prod_{i=1}^n\left(\theta x_i^{-(\theta+1)}\right)=\theta^n \left(\prod_{i=1}^n x_i\right)^{-(\theta+1).}$ By Fisher's factorization, the sufficient statistic is just the product of the $n$ observations. – Michael Hardy May 30 '13 at 03:26
  • Thank you @MichaelHardy , that was exaclty what I wanted to be confirmed. – dreamer May 30 '13 at 09:09

1 Answers1

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You can always find an estimator which attains the Cramer Rao (Fischer Information) lower bound by scaling your MLE estimator to attain the right variance. However, this risks making the estimator biased. Sometimes, the unbiased estimator has minimum variance but other times an unbiased one does. Try calculating the CR bound for your function and the variance of your estimator.

To be clear, sufficiency is a wholly different concept. Sufficiency says that you can't squeeze out extra information about your statistic. For example if you have 10 observations but only use 6 of them to estimate the mean that would be not sufficient since there's an extra 4 observations you didn't use.

The concept you want is efficiency which is a measure of the mean square error or the variance. The more efficient an estimator is the closer it's variance is to the CR bound.

Alex R.
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  • Thank you for your help! Sorry to say it only now, but what I am actually looking for is an unbiased estimator. My apologies for being so unclear. – dreamer May 29 '13 at 15:06
  • And why I think I need a sufficient estimator first, is because of the Rao-Blackwell theorem (http://en.wikipedia.org/wiki/Rao%E2%80%93Blackwell_theorem) that I learned through which I can then find an estimator which attains the CRLB. – dreamer May 29 '13 at 15:08
  • You can't apply the Rao--Blackwell theorem to get an unbiased estimator unless you've already got an unbiased estimator. The latter is in typical applications a very crude estimator; the former is typically a vastly improved one. – Michael Hardy May 30 '13 at 03:13
  • There is ALWAYS a biased estimator with the smallest possible variance, namely any estimator that always has the same value regardless of the data. What's more interesting is that sometimes a biased estimator has a smaller mean squared error (equal to the variance plus the square of the bias) than does the best unbiased estimator. In some cases there are biased estimators having a mean squared error smaller than the CR bound. The CR bound applies only to unbiased estimators, not to biased ones. – Michael Hardy May 30 '13 at 03:22