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I'm hunting for a probability distribution with the following properties:

  1. The support is $(0,\infty)$.
  2. Denote by $F(x)$ the CDF of this distribution.
  3. If $X_1, X_2,...$ are independent random variables following this distribution then for each $n$ the CDF of $X_1+...+X_n$ is equal to $F(x)^n$.

Does it exist?

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  • $\begingroup$ You are asking in the wrong forum. Nice question, anyway. $\endgroup$
    – Gerald Edgar
    Commented Jul 15, 2022 at 12:31
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    $\begingroup$ There is no such distribution if $n \geq 2$. If you want to show it yourself, start with some $x_0 \in (0,\infty)$ with $0 < F(x_0) = p < 1$ and estimate $\mathbb{P}(X_1 + \ldots + X_n \leq x_0)$. $\endgroup$
    – Dieter Kadelka
    Commented Jul 15, 2022 at 12:36
  • $\begingroup$ @Dieter Kadelka. Thank you very much for your response. Silly me. Indeed it's quite trivial. $P(X_1+X_2\leq x_0)\leq P(X_1\leq x_0)P(X_2\leq x_0)=p^2$. What I'm actually after is not necessarily $F(n,x)=F(1,X)^n$ (by $F(n,x)$ I denote the CDF of $X_1+...+X_n$), but sth that satisfies $F(n,x)=g(x)h(x)^n$ for some $g(x)$ and $h(x)$ or even more generally $F(n,x)=\sum_{i=1}^kg_i(x)h_i(x)^n$. $\endgroup$
    – Luka74
    Commented Jul 15, 2022 at 16:17

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