A random variable is said to have probability density function $$f_X(x)=\frac{\alpha k^\alpha}{x^{\alpha +1}},\quad \alpha , k>0 \; \text{ and }\; x>k.$$

1. Compute the MLE estimators $\widehat \alpha $ and $ \widehat k$ for $\alpha$ and k.

I figured they are $$\widehat \alpha = \frac{n}{\sum_i \text{ln}\frac{x_i}{\widehat k}} \quad\text{ while } \quad \widehat k=\text{min}_{i} \; x_i$$Correct me if I'm wrong.

  1. Derive an 95% confidence interval for $k$. Hint: Consider the distribution of the random variable $k/􏰁\widehat k$.

I don't get the point of the hint. Is it to figure out the critical value $(CV)$ to substitute in the expression $$k= \widehat k\pm CV se(\widehat k) \quad \text{where} \quad se(\widehat k)=\frac{1}{\sqrt{I_n(\widehat k)}}\quad \text{where }\quad I_n{(\widehat k)\; \text{is the Fisher Information at}\; \widehat k.}$$

If so, how do I answer this question.I feel I'll be able to get $I_n(\widehat k)$ but how do i get $CV$. Or what are other alternative methods.

  • $\begingroup$ In your MLE for $\alpha$, you need $\hat k$ where you have $k$. $\endgroup$ – Michael Hardy May 25 '15 at 3:59
  • $\begingroup$ I've now made the change. $\endgroup$ – user538762 May 25 '15 at 4:00
  • $\begingroup$ Have you found the distribution of $\hat k$? ${}\qquad{}$ $\endgroup$ – Michael Hardy May 25 '15 at 4:09

$$ \hat k >x \text{ if and only if }[X_1>x\ \&\ \cdots\ \&\ X_n>x] $$ and the probability of that is $\Big(\Pr(X_1>x)\Big)^n$. $$ \Pr(X_1>x) = \int_x^\infty \frac{\alpha k^\alpha}{u^{\alpha +1}} \, du = \left(\frac k x\right)^\alpha, $$ so $$ \Pr(\hat k>x) = \left(\frac k x\right)^{n\alpha}. $$ Thus $$ \Pr(x_1<\hat k < x_2) = \left(\frac k {x_1}\right)^{n\alpha}- \left(\frac k {x_2} \right)^{n\alpha}. $$ $$ \Pr(Ak < \hat k < Bk) = A^{-n\alpha} - B^{-n\alpha}. $$ $$ \Pr\left(A< \frac{\hat k} k < B\right) = \cdots $$ $$ \Pr\left(\frac 1 B < \frac k {\hat k} < \frac 1 A\right) = \cdots $$ $$ \Pr\left(\frac{\hat k}B<k<\frac{\hat k} A\right) = \cdots $$ This gives you a confidence interval for $k$ if $\alpha$ is known. Since $\alpha$ is not known, there is more work to do. (You have to choose $A$ and $B$ to get you the probablities that you want.)


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