Let $X_1, ... , X_n$ be iid with common density $f(x) = ax^{a-1}$, $ 0 < x < 1$, where $a$ is an unknown parameter $a > 0$.

(a) Find the MLE for $a.$

(b) Show that $W = - \sum^n_{i=1} \ln X_i$ has a Gamma distribution with parameters $\alpha = n$, $\beta = 1/a$. Hint: Use the mgf.

(c) Show that $2aW$ has $\chi^2 (2n)$ distribution.

(d) Use (c) to find a $100(1-a)%$ confidence interval for $a$.

I'm a bit unsure about part (d). Let's say the critical values are $\chi^2_{1 - \alpha/2}$ and $\chi^2_{\alpha/2}$.

Then we have $$ (1 - \alpha) = P(\chi^2_{1-\alpha/2} < 2aW < \chi^2_{\alpha/2}) = P \left(\frac{\chi^2_{1-\alpha /2}}{2W} < a < \frac{\chi^2_{\alpha/2}}{2W} \right)$$

But I'm not sure if this works, since I'm dividing by a random variable.

So is this correct? If not, can you give me a hint?

Thanks in advance

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    $\begingroup$ A hint: What happens to $W$ in your last formula? $\endgroup$ – Adam Bailey Jan 15 '14 at 14:06
  • $\begingroup$ @AdamBailey Oh, I'm sorry I made a typo...I'll fix it now. $\endgroup$ – Artus Jan 15 '14 at 19:52
  • $\begingroup$ Wrong fix. Go back and divide by $a$ instead of $W$. $\endgroup$ – whuber Jan 15 '14 at 20:46
  • $\begingroup$ @whuber Thanks. From part (a), I got $\hat{a} = n/W$. So, we have $(1 - \alpha) = P(\chi^2_{1-\alpha/2} < 2aW < \chi^2_{\alpha/2}) = P \left(\frac{\chi^2_{1-\alpha /2}}{2a} < W < \frac{\chi^2_{\alpha/2}}{2a} \right) = P \left( \frac{1}{\frac{\chi^2_{\alpha /2}}{2a}} < \frac{1}{W} < \frac{1}{\frac{\chi^2_{1-\alpha/2}}{2a}} \right) = P \left( \frac{n}{\frac{\chi^2_{\alpha /2}}{2a}} < \frac{n}{W} < \frac{n}{\frac{\chi^2_{1-\alpha/2}}{2a}} \right) = P \left( \frac{n}{\frac{\chi^2_{\alpha /2}}{2a}} < \hat{a} < \frac{n}{\frac{\chi^2_{1-\alpha/2}}{2a}} \right)$, right? $\endgroup$ – Artus Jan 15 '14 at 22:05
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    $\begingroup$ Why are you now forming an interval for $\hat a$? You know the value of that. What is the problem with dividing by a random variable which - whatever its unknown value - is greater than zero? @whuber - doesn't the OP want an interval for the parameter $a$? Why would one divide by $a$ rather than everything else? $\endgroup$ – Glen_b -Reinstate Monica Jan 15 '14 at 23:24

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