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Let $(X_1,X_2,\ldots,X_n)$ be a random sample drawn from a $\text{Pareto}(k,a)$ population with density $f(x)=\frac{ak^a}{x^{a+1}}\mathbf1_{x>k}$ where $a,k>0$. What is the distribution of $\sum_{j=1}^n \ln\left(\frac{X_{(j)}}{X_{(1)}}\right)$ ?

In what follows, I have used $\text{Exp}(a)$ to denote an exponential distribution with mean $1/a$ ($a>0$), and $\text{Gamma}(\alpha,p)$ to denote the density $g(t)\propto e^{-\alpha t}t^{p-1}\mathbf1_{t>0}$ where $\alpha,p>0$.

We have $S=\sum_{j=1}^n\ln\left(\frac{X_{(j)}}{X_{(1)}}\right)=\sum_{j=1}^n\ln(X_{(j)})-n\ln(X_{(1)})=\sum_{j=1}^n \ln X_j-n\ln X_{(1)}$.

Now, \begin{align}&X_j\stackrel{\text{i.i.d}}{\sim}\text{Pareto}(k,a)\,,j=1,\ldots,n\\&\implies \ln(X_j/k)\stackrel{\text{i.i.d}}{\sim}\operatorname{Exp}(a)\,,j=1,\ldots,n\\&\implies\sum_{j=1}^n\ln(X_j/k)=\sum_{j=1}^n\ln X_j-n\ln k\sim\operatorname{Gamma}(a,n)\end{align}

I could show that $X_{(1)}\sim\operatorname{Pareto}(k,na)$, so that $\ln\left(\frac{X_{(1)}}{k}\right)=\ln X_{(1)}-\ln k \sim \operatorname{Exp}(na)$.

Not sure if the last two facts help me get the exact distribution of $S$.


Since $S=\sum_{j=2}^n\ln\left(\frac{X_{(j)}}{X_{(1)}}\right) = \ln\left(\prod_{j=2}^n \left(\frac{X_{(j)}}{X_{(1)}}\right)\right)$, I proceeded to find the distribution of $\prod_{j=2}^n \left(\frac{X_{(j)}}{X_{(1)}}\right) = \prod_{j=2}^n Y_j$ (say).

Joint density of the order statistics $\left(X_{(1)}=U_1,\ldots,X_{(n)}=U_n\right)$ is $$f_{\mathbf U}(u_1,\ldots,u_n) = n!\prod_{j=1}^n\left(\frac{ak^a}{u_i^{a+1}}\right) = \frac{n!a^nk^{na}}{(\prod_{j=1}^nu_i)^{a+1}} \mathbf1_{k<u_1<\cdots<u_n}$$

Transforming $(U_1,U_2,\ldots,U_n)\to(Y_1,Y_2,\ldots,Y_n)$ such that

$Y_1=U_1$ and $Y_i=\frac{U_i}{U_1}$ for $i=2,3,\ldots,n$

The inverse solutions are $u_1=y_1$ and $u_i=y_1y_i$ for $i=2,3,\ldots,n$

And $k<u_1<u_2<\cdots<u_n\implies y_1>k\,,\quad 1<y_2<y_3<\cdots<y_n$.

The Jacobian is $J=1\times\det(\operatorname{diag}(y_1, y_1, \ldots, y_1)) = y_1^{n-1}$.

So joint density of $(Y_1,Y_2,\cdots,Y_n)$ is

\begin{align}f_{\mathbf Y}(y_1,\ldots,y_n)&=\frac{n!a^nk^{na}}{y_1^{na+n}(\prod_{i=2}^n y_i)^{a+1}}y_1^{n-1} \mathbf1_{y_1>k,\,1<y_2<\cdots<y_n} \\[8pt] &=\frac{na\,k^{na}}{y_1^{na+1}}\mathbf1_{y_1>k}\frac{(n-1)!a^{n-1}}{(\prod_{i=2}^n y_i)^{a+1}}\mathbf1_{1<y_2<y_3<\cdots<y_n}\end{align}

This implies the independence of $Y_1$ and $(Y_2,Y_3,\ldots,Y_n)$.

Now I have to find the distribution of $\prod_{j=2}^n Y_j$.

Using the change of variables $(Y_2,\ldots,Y_n)\to(W_2,\ldots,,W_n)$ where

$$W_2=\prod_{j=2}^n Y_j\,,W_3=\prod_{j=3}^n Y_j\,,\ldots,W_{n-1}=Y_{n-1}Y_n\,,W_n=Y_n,$$

I couldn't find the exact supports of the $W_i$'s. Not sure if this is the correct transformation though.

The answer given definitely shows an easier approach, but I wish to know if what I am doing leads to the desired result or not. This is indeed a lot of work but I must be able to do it the long way.

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    $\begingroup$ The first approach ignores the correlation between $\sum_{j=1}^n\ln X_j$ and $\ln X_{(1)}$ $\endgroup$ – Xi'an May 12 '18 at 20:51
  • $\begingroup$ Yes, I noticed that they are not independent. $\endgroup$ – StubbornAtom May 12 '18 at 20:55
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A simpler approach might be to use the fact that if $x \sim \text{Pareto}(k,a)$, then conditioning upon $x \geq b$ results in $x \sim \text{Pareto}(b,a)$. Consequently, $x | x_{(1)} \sim \text{Pareto}(x_{(1)}, a)$, except for the single observation corresponding to $x_{(1)}$. When we then take the ratio $x/x_{(1)}$, we are rescaling $x$ by its minimum value, and the resulting variate has a $\text{Pareto}(1, a)$ distribution, independent of $x_{(1)}$.
Therefore, if we don't pay attention to the rank of the $x_i$ in the sample, the ratios $x_i/x_{(1)} \sim \text{Pareto}(1,a)$ and are independent (except for the observation corresponding to $x_{(1)}$, which is equal to 1.)

This, combined with the fact that the log of a $\text{Pareto}(1,a)$ variate is distributed $\text{Exponential}(a)$, and the sum of $n-1$ i.i.d. variates $\sim \text{Exponential}(a)$ is $\sim \text{Gamma}(n-1,a)$, leads directly to the result that the sum

$$\sum_{j=1}^n \ln\left(\frac{X_{(j)}}{X_{(1)}}\right) \sim \text{Gamma}(n-1,a)$$

where the $n-1$ comes from the fact that exactly one of the ratios will have value $1$, hence $\log(\cdot) = 0$, leaving $n-1$ nonzero terms in the sum.

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  • $\begingroup$ I think you missed saying $x_{(1)}$ is independent of $x_i/x_{(1)}$ in the last sentence of the first paragraph. $\endgroup$ – StubbornAtom May 13 '18 at 6:05
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    $\begingroup$ That's implied by the point of the next to last sentence, admittedly not nearly as clearly stated as it might have been - the ratio and $x_{(1)}$ are independent. $\endgroup$ – jbowman May 13 '18 at 13:57

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