# Minimum of Pareto Random Variables given Harmonic Mean

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.

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