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Let $X_1, \dots, X_n$ denote a random sample of size n from the probability distribution with pdf:

$$ f_X(x|\theta_1, \theta_2) = \frac{1}{\theta_2 - \theta_1} \ I(x)_{[\theta_1,\theta_2]} \ I(\theta_1)_{(-\infty,\theta_2)} \ I(\theta_2)_{(\theta_1,\infty)}\;.$$

(1) Find a pair of sufficient statistics for $(\theta_1, \theta_2)$.

$\bf{My \ thoughts:}$ This wasn't too bad. I got $(X_{(1)}, X_{(n)})$ for this part

(2) Find the maximum likelihood estimator $(\hat{\theta_1}, \hat{\theta_2})$ for $(\theta_1, \theta_2)$.

$\bf{My \ thoughts:}$ Thinking I need to use monotone functions since it has 2 parameters and the variables are part of the interval. I believe that $\frac{X_{(1)} + X_{(n)}}{2}$ will become one of my estimators.

(3) Show that $\frac{X_{(1)} + X_{(n)}}{2}$ is an unbiased estimator for $\frac{\theta_1 + \theta_2}{2}$.

$\bf{My \ thoughts:}$ I think I will need to use Cramer-Rao Lower Bound in some form but not quite sure if that is right .

(4) Construct an unbiased estimator for $\theta_2 - \theta_1$.

$\bf{My \ thoughts:}$ Very stuck on this part, but I think I can use some information from previous parts to help me.

Any help is greatly appreciated.

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    $\begingroup$ (2) What does the midrange have to do with either of the interval endpoints? You ought to rethink this one. (3) is trivial because the distribution of the midrange is symmetric under the transformation $x\to \theta_1+\theta_2-x$: this transformation would negate any bias but it does not change the distribution of the midrange, showing its bias is equal to its negative. (4) What is the expectation of the sample range? If it is a function of $n$ and $\theta_2-\theta_1$, you should be able to adjust the sample range to have zero bias. $\endgroup$ – whuber Mar 11 '13 at 23:32
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(2)

I believe that $\frac{X_{(1)} + X_{(n)}}{2}$ will become one of my estimators.

Why do you believe that, rather than something more directly related to your answer to (1)?

What would you use to just estimate the first parameter? What would you use to just estimate the second?

(3)

My thoughts: I think I will need to use Cramer-Rao Lower Bound in some form but not quite sure if that is right .

The question relates to expectation, rather than variance.

(4)

I suggest you use the sufficient statistics to construct an estimator with good properties, and then find its bias. Then figure out what simple modification to that estimator will have bias 0.

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Note that for the mle, the likelihood $(\theta_2-\theta_1)^{-n}$ always decreasing in $\theta_2$ and increasing in $\theta_1$, but we must have $\theta_1\leq X_{(1)}\leq X_{(n)}\leq\theta_2$.

Also for (3) you need the distribution for each sufficient statistic. As a hint, if the minimum is greater than $t$ then all values are greater than $t$. Similarly, if the maximum is less than $r$ then all values are less than $r$. This will give you the CDF - differentiate to get the pdf and then you can work out the expectations and hence the bias.

For (4) if you define $R=\theta_2-\theta_1$ and note that $R$ must be at least as big as the range you have observed $R\geq X_{(n)}-X_{(1)}$. Also $ X_{(n)}-X_{(1)}$ is a sufficient statistic for $R$ so the best unbiased estimator must be a function of $r_n=X_{(n)}-X_{(1)}$. The bias you work out in (3) can be used to find the bias for $r_n$

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