Variance of Estimator / Variance Algebra

Question

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

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 — James, May 22 '14 at 08:09

score 0 · Accepted Answer · answered May 22 '14 at 08:13

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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'.

answered May 22 '14 at 08:13

user21820

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Yes n is the sample size, thanks very much. Appreciated – James May 22 '14 at 08:14

score 0 · Answer 2 · answered May 22 '14 at 08:23

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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.

answered May 22 '14 at 08:23

JPi

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Variance of Estimator / Variance Algebra

2 Answers2