Is there any probability distribution that $\max(X_1, X_2)$ and $\min(X_1, X_2)$ has similar "shape" to $X_1$ and $X_2$?

Question

When I was thinking about probability-based algorithm, I came up with this problem:

Question: Can we construct a probability distribution such that when two independent random variables $X_1$ and $X_2$ follows this distribution, $\max(X_1, X_2)$ and $\min(X_1, X_2)$ also follows this distribution?

1. Background - Reproductive Property

There are many probability distribution which preserves shape by addition and subtraction.

One example is normal distribution: let $X_1 = N(\mu_1, \sigma_1^2)$ and $X_2 = N(\mu_2, \sigma_2^2)$ which are independent variables. Then, if we know the basics of statistics, we can say that the distribution of $X_1 + X_2$ is equivalent to $N(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2)$ - which can also be expressed by normal distribution. (The figure below depicts $N(1, 1^2) + N(2, 2^2) = N(3, \sqrt{5}^2)$.)

Another example is Poisson distribution: when $X_1 = Po(\lambda_1)$ and $X_2 = Po(\lambda_2)$, then $X_1 + X_2 = Po(\lambda_1 + \lambda_2)$. There are many distribution like this example, like binomial distribution for a fixed $p$, and chi-square distribution.

So, in many cases, given two distributions of a certain type, their sum is also a distribution of this type. This kind of property is called reproductive property. Especially for normal distribution, the same thing also holds for subtraction, because $X_1 - X_2 = N(\mu_1 - \mu_2, \sigma_1^2 + \sigma_2^2)$.

2. Generalizing to Maximum/Minimum?

Generalizing this, can $\max(X_1, X_2)$ or $\min(X_1, X_2)$ be written by the same type of distribution? For normal distribution, it's no, and is not even symmetric. Uniform, Poisson, binomial,... also doesn't have reproductive property for $\max$.

However, some distributions have such property. One example is the distribution which:

• probability density function (PDF) is $f(x; \beta) = \frac{1}{\beta x^2} e^{-1/(\beta x)} \ (x > 0)$, therefore
• cumulative distribution function (CDF) is $F(x; \beta) = e^{-1/(\beta x)}$.

So, if we have two such distributions $X_1, X_2$ with parameter $\beta_1$ and $\beta_2$, the CDF of $\max(X_1, X_2)$ will be:

$$F(x; \beta_1) \cdot F(x; \beta_2) = e^{-1/(\beta_1 x)} \cdot e^{-1/(\beta_2 x)}= e^{-\left(1/\beta_1 + 1/\beta_2\right) x}$$

which means that $\max(X_1, X_2)$ has the same type of distribution with $\beta = \frac{1}{1/\beta_1 + 1/\beta_2}$. (The figure below is for $\beta_1 = 2$ in red, $\beta_2 = 3$ in blue, and $\beta = 1.2$ in green depicting $\max$.)

Symmetrically, some distributions preserves the shape by minimum. However, unfortunately, these distribution have reproductive property not for "both maximum and minimum".

3. The Question

Now, I want to construct a distribution which have "reproductive property" for both maximum and minimum! In other words, when we have distribution $X_1$ and $X_2$ of a certain type, the distributions $\max(X_1, X_2)$ and $\min(X_1, X_2)$ also can be written by distributions of the same type.

Let's write it more rigorously by means of mathematics: think about any probability distribution $Dist(\theta)$, which PDF is $f(x; \theta_1, \theta_2, \dots, \theta_k)$ (the parameter is $\theta = (\theta_1, \theta_2, \dots, \theta_k) \in D$, where $k$ is finite and all $\theta_i$ are real values). I want to find the probability distribution which satisfies all following properties.

1. There is a function $f: D \times D \to D$ where, letting $X_1 = Dist(\theta_1)$ and $X_2 = Dist(\theta_2)$, the $\max(X_1, X_2)$ follows the distribution $Dist(f(\theta_1, \theta_2))$.
2. There is a function $g: D \times D \to D$ where, letting $X_1 = Dist(\theta_1)$ and $X_2 = Dist(\theta_2)$, the $\min(X_1, X_2)$ follows the distribution $Dist(g(\theta_1, \theta_2))$.
3. $Dist(\theta)$ is a continuous probabilistic distribution (not discrete).

Please note that, without Property 3, the construction of answer is easy: if the random variable is constant, the maximum and minimum is also constant, which technically satisfies Property 1 and 2. Also, if the distribution is discrete and takes $m$ values, we can express maximum and minimum using $m$ parameters for $\theta$.

4. Reformation of the Question

I think this question can be easier when we think about CDF. Let's say CDF can be expressed by $F(x; \theta)$. Then:

• the CDF of $\max(Dist(\theta_1), Dist(\theta_2))$ is $F(x; \theta_1) F(x; \theta_2)$.
• the CDF of $\min(Dist(\theta_1), Dist(\theta_2))$ is $1 - (1 - F(x; \theta_1)) (1 - F(x; \theta_2))$.

So, the question can be paraphrased for finding a function $F$, which is:

• $F(x; \theta)$ is a continuous function for all $\theta \in D$.
• $\lim_{x \to -\infty} F(x; \theta) = 0$ for all $\theta \in D$.
• $\lim_{x \to +\infty} F(x; \theta) = 1$ for all $\theta \in D$.
• For all $\theta_1, \theta_2 \in D$, $F(x; \theta_1) F(x; \theta_2) = F(x; \theta)$ for some $\theta \in D$.
• For all $\theta_1, \theta_2 \in D$, $1 - (1 - F(x; \theta_1)) (1 - F(x; \theta_2)) = F(x; \theta)$ for some $\theta \in D$.

I thought about hours but still cannot find the answer, so I would like to know the distribution with such properties, or even whether such thing exists or not.