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I found the following result used in this paper, but it was just claimed without proof and it seems extremely strong to me, so I would like a proof, or at least a reference, of a proof.

Let $D$ be probability distribution. For any $k\geq 1$, there exists another distribution $D'$ such that if $Y_1, ..., Y_k \sim D'$, then the distribution of $\max(Y_1, ..., Y_k)$ is $D$.

This seems very counter intuitive to me, specially that the max of a distribution of iid random variables behaves like a Gumbel distribution for large $k$, and Gumbel obviously doesn't cover all distributions.

Silverfish
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AspiringMat
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    "specially that the max of a distribution of iid random variables behaves like a Gumbel distribution for large k" ... no, this is not true in general. Asymptotically, with a suitable kind of rescaling, it's one of three distribution classes you can end up in. https://en.wikipedia.org/wiki/Fisher%E2%80%93Tippett%E2%80%93Gnedenko_theorem If you want one encompassing distribution, it's the GEV. https://en.wikipedia.org/wiki/Generalized_extreme_value_distribution. The result you discuss is not an asymptotic one, though, it's an exact finite sample result; it's consistent with the asymptotic one – Glen_b Apr 20 '23 at 04:26
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    Thanks @Glen_b, I was thrown off by the false assumption I had. In retrospect, this is easy to prove. – AspiringMat Apr 20 '23 at 04:32
  • The original title said "Approximating a distribution with the maximum of other iid distributions" but the question body doesn't seem to mention approximation at all, nor does the answer. So I have renamed it to "Is every probability distribution also the distribution of the maximum of several i.i.d. random variables?" which seems closer to the intention. Feel free to edit if you disagree with the new title. (Also I'm not sure that "iid distributions" makes sense! Hence I have used "iid random variables" instead.) – Silverfish Apr 22 '23 at 12:50

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Let $D$ be a probability distribution with CDF $F$, and fix a positive integer $k$. Consider a new CDF $F_k(y) = F(y)^{1/k}$ (you can check that it satisfies the axioms of a CDF). Then $F_k$ defines another distribution $D^\prime$. If $Y_1,\ldots,Y_k \sim D^\prime$ (i.i.d.), then $$ \begin{aligned} P(\max\{Y_1, \ldots, Y_k\} \leq x) &= P(Y_1 \leq x, \ldots, Y_k \leq x) \\ &= F_k(x)\cdots F_k(x) \\ &= F_k(x)^k \\ &= F(x), \end{aligned} $$ so $\max\{Y_1,\ldots,Y_k\}$ has distribution $D$. (See also How do you calculate the probability density function of the maximum of a sample of IID uniform random variables?)

Artem Mavrin
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    Ahh, thanks, well that was simple. I was thrown off by a false assumption (that the max of iid distributions behaves like a Gumbel) which @Glen_b corrected. – AspiringMat Apr 20 '23 at 04:30
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    When contrasting this to extreme-value convergence, one important distinction is that the distribution of $Y$ will be different for different $k$ -- if it were the same, you would end up converging to one of the extreme-value classes as $k$ increased (with suitable rescaling) – Thomas Lumley Apr 20 '23 at 22:44