Error of the bias-corrected uniform estimator

Question

Values are uniform between 0 and an unknown ceiling $\theta$ . From $n = 6$ draws you form the bias-corrected estimator $\hat\theta = \frac{n+1}{n}M$ , where $M$ is the sample maximum. Take $\theta = 1$ .

What is the mean squared error of $\hat\theta$ ?

Accepted Answer

The idea. Scaling the maximum up by \frac{n+1}{n} makes it unbiased, so its entire MSE is variance. Scaling multiplies variance by the square of the scale factor. Variance of the maximum. For n uniform draws on (0,	heta), 	ext{Var}(M) = \frac{n}{(n+1)^2(n+2)}	heta^2. Scale up. Multiplying M by \frac{n+1}{n} multiplies its variance by \left(\frac{n+1}{n}ight)^2: 	ext{Var}(\hat	heta) = \frac{(n+1)^2}{n^2}\cdot\frac{n}{(n+1)^2(n+2)}	heta^2 = \frac{	heta^2}{n(n+2)}. Since \hat	heta is unbiased, this variance is its MSE. For n = 6, 	heta = 1: 	ext{MSE}(\hat	heta) = \frac{1}{6 \cdot 8}…

Error of the bias-corrected uniform estimator

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