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I have the following problem. Assume I have $n$ independent Pareto random variables $X_1,...,X_n$, with the CDF of $X_i$ being $Pr(X_i \leq x_i) = F(x_i) = 1 - (\frac{b_i}{x_i})^{\alpha_i}$. For convenience of notation, I use the complementary CDF $Pr(X_i \geq x_i) = G_i(x_i) = 1 - F_i(x_i) = (\frac{b_i}{x_i})^{\alpha_i}$.

Let $X = min(X_1,...,X_n)$ be the minimum of $X_1,...,X_n$ and let $H(X_1,...,X_n) = \frac{n}{\frac{1}{X_1} + ... + \frac{1}{X_n}}$ be the harmonic mean.

I want to show that the distribution $Pr(X \leq x | H(X_1,...,X_n) \geq \tau)$ follows a Pareto Distribution.


I know how to do this if we do not condition on the harmonic mean. It is straightforward to show that $X = \min(X_1,...,X_n)$ has a Pareto distribution with complementary CDF given by $$G(x) = \prod_{i=1}^n (\frac{b_i}{x})^{\alpha_i} = (\prod_{i=1}^n b_i^{\alpha_i}) \cdot (\frac{1}{x})^{\alpha_1+...\alpha_n} $$

However, when we condition on the harmonic mean being above a certain threshold, I'm not sure where to start tackling this problem.

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The correct formula defining the Pareto distribution with positive real parameters $a$ and $b$ is this: $$P(X>x)=\Big(1\wedge\frac bx\Big)^a $$ for all real $x>0$, where $u\wedge v:=\min(u,v)$.

So, if $X=X_1\wedge\cdots\wedge X_n$, where the $X_i$'s are independent random variables (r.v.'s) such that for each $i$ the r.v. $X_i$ has the Pareto distribution with parameters $a_i>0$ and $b_i>0$, and not all $b_i$'s are the same, then $X$ will not have a Pareto distribution. Also, then the conditional distribution of $X$ given $H:=H(X_1,\dots,X_n)\ge\tau$ will not be Pareto either, at least when $\tau$ is $0$ or close to $0$.


Even when all $b_i$'s are the same, the conditional distribution of $X$ given $H\ge\tau$ will not be Pareto in general. E.g., if $n=2$ and $X_1,X_2$ are iid Pareto with parameters $a=1$ and $b=1$, then $$P(X>x|H\ge2) =\begin{cases} 1 & \text{ if }x\le1, \\ 4/x-2/x^2-1 & \text{ if }1<x<2, \\ 2/x^2 & \text{ if }x\ge2. \end{cases} $$

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