1
$\begingroup$

I think ive got this correct, but i keep second guessing myself. I have an estimator of an ordered statistic, assuming distributed Uniform[0, 1]

The estimator is defined as:

$$\hat \theta = X_{(1)} - \frac{1}{n+1} $$

Im solving for the variance of this estimator, is the following correct?

$$Var(\hat \theta) = Var(X_{(1)} - \frac{1}{n+1} )$$

Can that just be simplified to:

$$Var(\hat \theta) = Var(X_{(1)} )$$

ie just drop the 1 / (n+1) term as a constant - or can this not be considered a constant.

Thanks

$\endgroup$
  • $\begingroup$ Sorry further information, the distribution is uniform between $[\theta and \theta + 1]$ ive used a transformation for calculating other variables ie E(theta) to make it uniform between 0 and 1 which is much easier. Just not sure about the variance $\endgroup$ – James May 22 '14 at 8:09
0
$\begingroup$

Well it depends on what $n$ is of course. It looks like a constant, in particular the size of your sample, in which case yes you can discard a constant when dealing with variance; a translation does not affect the 'spread'.

$\endgroup$
  • $\begingroup$ Yes n is the sample size, thanks very much. Appreciated $\endgroup$ – James May 22 '14 at 8:14
0
$\begingroup$

All is right. Also, note that the estimator of the upper bounds of a uniform minus the true upper bound has after multiplying by the sample size approximately an exponential distribution on the negative half line.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.