# Derive probability distributions from i.i.d. Gumbel

I have a question on how to derive (if possible) the following probability distributions.

Consider 3 random variables $$(X,Y,Z)$$ mutually independent and identically distributed. Specifically, $$X$$ is distributed as a Gumbel with mean $$\mu\in \mathbb{R}$$ and location $$\beta>0$$.

1) What is the probability distribution of $$Y-X$$

As suggested in a comment below the answer to this is the third bullet point here:

$$Y-X\sim Logistic(0,\beta)$$

It seems that $$X\perp Y$$ is not necessary, correct?

2) What is the probability distribution of $$\begin{pmatrix} Y-X\\ Z-X \end{pmatrix}$$ Is it a bivariate logistic? Which are the parameters governing it? This is another question on this forum about the bivariate logistic.

3) Can we conclude that $$\begin{pmatrix} Y-X\\ Z-X \end{pmatrix} \sim \begin{pmatrix} X-Z\\ Y-Z \end{pmatrix} \sim \begin{pmatrix} X-Y\\ Z-Y \end{pmatrix}$$

• The answer to the first question is on the Wikipedia link you provided. Read the section "Related distributions" carefully. – COOLSerdash Nov 1 '18 at 12:17
• @COOLSerdash Thanks a lot, apologies, didn't read it carefully. – user3285148 Nov 1 '18 at 12:20
• The answer to the third question is an immediate "yes" because independence (as you assume) implies exchangeability, which appears to be expressed in the equations (they merely permute the variables). Given the answer to the first question, the answer to the second is that it obviously is some kind of bivariate logistic distribution. What, then, are you referring to by "bivariate logistic"? Do you have some particular such distribution in mind? What is it? – whuber Nov 1 '18 at 12:55
• Yes, the bivariate normal is one kind of bivariate distribution with normal marginals--but it is by no means the only one. See stats.stackexchange.com/questions/30159 for examples. Similar concepts apply to all bivariate distributions: marginal distributions never completely determine the joint distributions. – whuber Nov 1 '18 at 13:14
• Exchangeability of variables implies equidistribution of permutations of any function of those variables. Specifically, let $X=(X_1, X_2, \ldots, X_n)$ be an exchangeable random variable and let $f:\mathbb{R}^n\to\mathbb{R}^m$ be a measurable function. By the definition of exchangeability of $X$, the distributions of $X$ and $X^\sigma = (X_{\sigma(1)}, \ldots, X_{\sigma(n)})$ are identical for any permutation $\sigma\in\mathfrak{S}_n.$ A fortiori, $f(X)$ and $f(X^\sigma)$ have the same distribution. In your case, $n=3,$ $m=2,$ and $f(X_1,X_2,X_3)=(X_2-X_1,X_3-X_1).$ – whuber Nov 1 '18 at 13:52