The Pareto distribution $P(a,c)$, with positive parameters $a$ and $c$, has density function $$ p(x;a,c) = \frac{ac^a}{x^{a+1}} $$ for $x \geq c$. Then, if $X_1, \dots, X_n$ is a random sample from the above distribution, we are asked to find the MLEs of $a$ and $c$. The likelihood function is $$\mathcal{L}(a,c;x_1,\dots,x_n) = \prod_{i=1}^n \frac{ac^a}{x_i^{a+1}} = a^n c^{an} \prod_{i=1}^n x_i^{-(a+1)},$$ and so, the log-likelihood is $$ \log\mathcal{L} = n \log a + an \log c - (a+1)\sum_{i=1}^n \log x_i. $$ This is a strictly increasing function in $c$, so we see that the MLE for $c$ is $$ \hat{c} = \min_i x_i = x_{(1)}. $$ Taking partial with respect to $a$ gives us that $$ \frac{\partial \log\mathcal{L}}{\partial a} = \frac{n}{a} + n\log c - \sum_{i=1}^n \log x_i. $$ Note that the second partial is always negative. Hence, setting the above expression to zero and solving for $a$ gives us the MLE for $a$: $$ \hat{a} = \frac{n}{\sum x_i - n \log \hat{c}} = \frac{n}{\sum_{i=1}^n \left(x_i/x_{(1)}\right)}. $$

Now, we are asked to find the distribution of $\hat{c}$ or the distribution of $2na/\hat{a}$. Finding the distribution of $\hat{c}$ was pretty straightforward. However, I have no idea how to find the distribution of $2na/\hat{a}$: $$ P\left( \frac{2na}{\hat{a}} \leq x \right) = P \left( 2a\sum \log\left(\frac{x_i}{x_{(1)}} \right) \right) = \cdots $$ and I have no idea where to go from here nor am I sure of my work this far.

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    $\begingroup$ Check en.wikipedia.org/wiki/… $\endgroup$ – Alecos Papadopoulos Feb 2 '15 at 1:16
  • $\begingroup$ self-study tag, please! $\endgroup$ – Xi'an Feb 2 '15 at 9:36
  • $\begingroup$ @AlecosPapadopoulos Thank you for the link. So basically, I'm looking at $2na/\hat{a}$ as the sum of $\log(X_i/X_{(1)})$, which should be the exponential distribution with parameter $a$, according to wikipedia. And $n$ sums of iid exponential distributions is a gamma distribution. However, I am a bit confused since $X_{(1)}$ is a random variable also and not the minimum they speak of in the article. $\endgroup$ – symmetricuser Feb 3 '15 at 0:46
  • $\begingroup$ When you look at a distribution, then you look at the random variables whose realizations you have in the sample. So you have realizations for $X_{(1)}$, $x_{(1)}$ but also for the other $X_i$'s -and all are random variables as regards their distribution. Please post your findings as an answer here, so that your question does not remain in the "unanswered" queue. $\endgroup$ – Alecos Papadopoulos Feb 3 '15 at 0:55

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