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


\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 $T$.


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 $2aT\sim \chi^2_{2n-2}$.

  • 1
    $\begingroup$ The first approach ignores the correlation between $\sum_{j=1}^n\ln X_j$ and $\ln X_{(1)}$ $\endgroup$
    – Xi'an
    May 12, 2018 at 20:51
  • $\begingroup$ Yes, I noticed that they are not independent. $\endgroup$ May 12, 2018 at 20:55

1 Answer 1


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.

  • $\begingroup$ I think you missed saying $x_{(1)}$ is independent of $x_i/x_{(1)}$ in the last sentence of the first paragraph. $\endgroup$ May 13, 2018 at 6:05
  • 2
    $\begingroup$ 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, 2018 at 13:57

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