2
$\begingroup$

Let $X_n = X_1, X_2,..., X_n$ be a random sample of $X \sim U(0, \theta)$, where $\theta$ is an unknown parameter. Assuming confidence level $1 — \alpha$, find confidence interval for $\theta$ where:

a) n = 2

b) n $\geqslant$ 100

I've got answers for this question yet I still cannot solve it. I need an explanation.

Answers:

a) $\langle\frac{2\overline{X}_2}{2-\sqrt{\alpha}},\frac{2\overline{X}_2}{\sqrt{\alpha}}\rangle$

b) $\langle\frac{2\overline{X}_{100}}{2-\sqrt{\alpha}},\frac{2\overline{X}_{100}}{\sqrt{\alpha}}\rangle$

$\endgroup$
  • 3
    $\begingroup$ Because there are many ways to obtain confidence intervals in this situation, you might begin not by guessing what the question has in mind but instead by verifying these intervals have the properties needed of a confidence interval: namely, that they indeed will include $\theta$ with probability $1-\alpha.$ That calculation might help you see how the intervals were developed. $\endgroup$ – whuber Jun 5 '18 at 16:15
2
$\begingroup$

I am wondering how you get this confidence interval. My method could be using the pivot, but result in different interval.

Let $Q = \frac{X_{(n)}}{\theta}$, where $X_{(n)}$ is the largest order statistics of $X_1, ......, X_n$. Then,

$$P(Q \le t) = \prod_{i=1}^nP(X_i \le t\theta) = t^n$$

, so Q is a pivot (independent of the parameter $\theta$). Take $c_n = \alpha^{\frac{1}{n}}$ we obtain $P(Q \le c_n) = \alpha$ and given that $P(Q \le 1) = 1$, we have

$$1 - \alpha = P(c_n \le Q \le 1) = P(c_n \le \frac{X_{(n)}}{\theta} \le 1) =P(X_{(n)} \le \theta \le \frac{X_{(n)}}{c_n})$$

So the $1-\alpha$ confidence interval for $\theta$ is $\Big(X_{(n)}, \frac{X_{(n)}}{\alpha^{\frac{1}{n}}}\Big)$

Finally you can plug in n = 2 and n = 100 to get different answers. (But sorry it is not the same you give here.)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ This is the shortest length CI but not the one OP is interested in, which is based on a different pivot. $\endgroup$ – StubbornAtom Jan 21 '19 at 18:02
  • 1
    $\begingroup$ True, but I would think the OP should be interested in this. Just because OP has been given a pre-specified answer does not mean he should not be exposed to other techniques. $\endgroup$ – Ben - Reinstate Monica Jan 21 '19 at 21:05
  • 1
    $\begingroup$ @Ben Well, OP is long gone but absolutely agree with you. $\endgroup$ – StubbornAtom Jan 21 '19 at 21:22
2
$\begingroup$

For $n=2$, note that $X_1/\theta$ and $X_2/\theta$ are independent $U(0,1)$ variables.

This means $T=(X_1+X_2)/\theta$ has the density

$$f_T(t)=t\mathbf1_{0<t<1}+(2-t)\mathbf1_{1<t<2}$$

So here $T=\frac{2}{\theta}\overline X$ is your desired pivot for constructing a confidence interval for $\theta$.

Now there exist $t_1,t_2$ such that for all $\theta>0$, $$P_{\theta}\left(t_1<\frac{2\overline X}{\theta}<t_2\right)=1-\alpha\tag{1}$$

That is, $$P_{\theta}\left(\frac{2\overline X}{t_2}<\theta<\frac{2\overline X}{t_1}\right)=1-\alpha$$

And $t_1,t_2$ is found in terms of $\alpha$ from condition $(1)$, namely $$\int_{t_1}^{t_2} f_T(t)\,dt=1-\alpha$$

Similar reasoning should be valid for a general $n$ variables, where a pivot for $\theta$ is given by $\frac{1}{\theta}\sum\limits_{k=1}^n X_k=\frac{n}{\theta}\overline X$, having the Irwin-Hall distribution.

| cite | improve this answer | |
$\endgroup$

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.