Suppose that $Y_1,...,Y_n$ is an IID sample from a uniform $U(\theta, 1)$ distribution. The method of moments estimator for $\theta$ is $\tilde \theta=2\bar Y-1$. The standard error of $\tilde \theta$ is

$$\sigma_{\tilde \theta}=\frac{1-\theta}{\sqrt{3n}}$$

Find an unbiased estimator of $\sigma_{\tilde \theta}$ and show that it is unbiased.

So normally I know how to find the estimator of a parameter of a distribution, but I don't quite understand how I'm supposed to find an estimator of the standard deviation of another estimator of a parameter of a distribution... so I really don't know how to go about this. What do I do?

  • $\begingroup$ Sorry, but those are all basically given. I don't understand how those quantities are helping me find this second estimator. $\endgroup$ – ereHsaWyhsipS Feb 9 '18 at 4:40
  • $\begingroup$ If you let $\hat{\sigma}_{\tilde{\theta}} = \frac{1-\tilde{\theta}}{\sqrt{3n}}$ and $\tilde{\theta}$ is unbiased, then $\hat{\sigma}$ is an unbiased estimator of $\sigma_{\tilde{\theta}}$. I think $\tilde{\theta} = 1-2\bar{Y}$ might be the correct method of moments estimator, though? Otherwise it's expectation is $-\theta$ $\endgroup$ – Ryan Warnick Feb 9 '18 at 19:08

My Comment (now deleted) was an attempt, perhaps not well worded, to get you to look carefully at everything again.

First verify that $E(Y_i) = \frac{\theta + 1}{2}$ and $Var(Y_i) = \frac{(1-\theta)^2}{12},$ so that $E(\bar Y) = \frac{\theta+1}{2},\,$ $Var(\bar Y) = \frac{(1-\theta)^2}{12n}$ and $SD(\bar Y) = \frac{1-\theta}{\sqrt{12n}}.$

Then, setting $\bar Y = \frac{\tilde \theta+1}{2}$ we obtain the MME of $\theta$ as $\tilde \theta = 2\bar Y - 1.$ Morover, $\tilde \theta$ is an unbiased estimator of $\theta$ because $E(2\bar Y - 1) = 2E(\bar Y) -1 = \theta.$ [Earlier typos above fixed, thanks to Comment by @ereHsaWyhsipS.]

As a reality check on the above, I use a simulation of the special case $\theta = .25,\, n = 10.$

set.seed(209)  # retain 'set.seed' to repeat same simulation, omit for fresh simulation
th = 1/4; n = 10
a = replicate(10^6, mean(runif(n, th, 1)))  # one million sample averages
th.mme = 2*a - 1
mean(th.mme);  th
## 0.2499683  # aprx E(MME of theta)
## 0.25       # exact $(MME of theta) = theta, MME unbiased

Also $Var (\tilde \theta) = Var(2\bar Y - 1) = 4Var(\bar Y) = \frac{(1-\theta)^2}{3a}$ and $SD(\tilde \theta) = SD(\tilde \theta) = \sigma^* = \frac{1-\theta}{\sqrt{3n}}.$ (This is your $\sigma_{\tilde \theta}.)$ Another reality check:

sd(th.mme);  (1 - th)/sqrt(3*n)
## 0.136847    # aprx SD(MME of theta)
## 0.1369306   # exact

A reasonable first try toward an estimator $\widetilde{\sigma^*}$ of $\sigma^*$ is simply to use $\tilde \theta$ for $\theta$ in the numerator to get $$\widetilde{\sigma^*} = \frac{1 - \tilde\theta}{\sqrt{3n}} = \frac{1 - (2\bar Y - 1)}{\sqrt{3n}} = \frac{2(1 - \bar Y)}{\sqrt{3n}}.$$ Then $E(\widetilde{\sigma^*}) = \frac{2 - (1+\theta)}{\sqrt{3n}} = \frac{1-\theta}{\sqrt{3n}},$ so that $\widetilde{\sigma^*}$ is unbiased for $\sigma^*.$

## 0.1369364     # agrees aprx with above, suggesting sg.mme unbiased
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  • 1
    $\begingroup$ This is a uniform distribution. The expectation is $\frac{1+\theta}{2}$. $\endgroup$ – ereHsaWyhsipS Feb 9 '18 at 21:34
  • $\begingroup$ Thanks. Fixing typos. Simulations OK $\endgroup$ – BruceET Feb 9 '18 at 23:32

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