# Distribution of $\sum_{j=1}^n\ln\left(\frac{X_{(j)}}{X_{(1)}}\right)$ when $X_i$'s are i.i.d Pareto variables

Let $$X_1,X_2,\ldots,X_n$$ be i.i.d variables having a Pareto distribution with density $$f(x)=\frac{a\theta^a}{x^{a+1}}1_{x>\theta}\,,$$ where $$a,\theta>0$$. What is the distribution of $$\sum\limits_{j=1}^n \ln\left(\frac{X_{(j)}}{X_{(1)}}\right)$$ ?

Suppose $$\mathsf{Gamma}(p,\alpha)$$ denotes the density $$g(t)\propto e^{-\alpha t}t^{p-1}1_{t>0}$$.

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

Now,

\begin{align}&\ln(X_j/\theta)\stackrel{\text{i.i.d}}{\sim}\mathsf{Exp}\text{ with mean }1/a\qquad,\,j=1,\ldots,n \\&\implies\sum_{j=1}^n\ln(X_j/\theta)=\sum_{j=1}^n\ln X_j-n\ln \theta\sim\mathsf{Gamma}(n,a) \end{align}

I could show that $$X_{(1)}$$ has another Pareto density, so that $$\ln\left(\frac{X_{(1)}}{\theta}\right)=\ln X_{(1)}-\ln \theta \sim \mathsf{Exp}\text{ with mean }1/(na)$$

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

Edit:

Turns out this was rather simple had I simply rewritten $$T$$ as

$$T=\sum_{j=1}^n \ln\left(\frac{X_j}{X_{(1)}}\right)=\sum_{j=1}^n(\ln X_j-\ln X_{(1)})=\sum_{j=1}^n (Y_j-Y_{(1)})\,,$$

where $$Y_j=\ln (X_j/\theta)$$. Since $$aY_j\sim\mathsf{Exp}(1)$$, using this result I have $$a T\sim \mathsf{Gamma}(n-1,1)$$.

This is equivalent to $$T\sim \mathsf{Gamma}(n-1,a)$$ or $$2aT\sim \chi^2_{2n-2}$$.

• The first approach ignores the correlation between $\sum_{j=1}^n\ln X_j$ and $\ln X_{(1)}$ – Xi'an May 12 '18 at 20:51
• Yes, I noticed that they are not independent. – StubbornAtom May 12 '18 at 20:55

A simpler approach might be to use the fact that if $$x \sim \text{Pareto}(\theta,a)$$, 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.
• I think you missed saying $x_{(1)}$ is independent of $x_i/x_{(1)}$ in the last sentence of the first paragraph. – StubbornAtom May 13 '18 at 6:05
• 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. – jbowman May 13 '18 at 13:57