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Take two independent and identically distributed random vectors $X_i$, $X_j$.

I want to find a multivariate distribution for these vectors such that the dot product $X_i^\top X_j \sim U[a, b]$.

This question actually popped up in computational research that I'm doing, rather than in a course of any kind, so it'd be sufficient to solve it in for the specific case where $X_i, X_j \in \mathcal{R}^2$. Even more specifically, for my case $a=.2, b=.8$, although that shouldn't matter too much.

Anybody have any answers? I looked around a bit and couldn't find anything straightforward.

If nobody knows the specific case for the uniform distribution, I have a second question (which I'm guessing is much easier): I need the bound $X_i^\top X_j \in [0, 1]$. How can I ensure that this bound holds when $X_i$ and $X_j$ are distributed according to some arbitrary multivariate distribution $F$?

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    $\begingroup$ I'm not even sure a uniform dot product is possible for independent vectors. $\endgroup$
    – Carlo Beenakker
    Commented Nov 9, 2021 at 15:45
  • $\begingroup$ I'm pretty sure it's not possible if you want the entries of $X_i$ and $X_j$ to all be independent vectors in the sense, that the entries of $X_i$ are also independent of each other. One can easily check, that all entries would need to have PDFs and if the dimension of the vectors is greater that two, then the distribution of the dot product will have a PDF, which is a convolution of some PDFs and thus continuous. Since $1_{[0,1]}$ is not continuous, it can not be the distribution function of the dot product. $\endgroup$
    – Ben Deitmar
    Commented Nov 9, 2021 at 22:00
  • $\begingroup$ Correction: The entries need not necessarily have PDFs, they could for example be constants. $\endgroup$
    – Ben Deitmar
    Commented Nov 9, 2021 at 22:08

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You may want to use the following two facts:

  1. Minus the logarithm of an uniform distribution on $[0,1]$ is an exponential distribution (easy).
  2. An exponential distribution is infinitely divisible as a member of the family of Gamma distributions.

Simulation of Gamma distributions is extensively discussed in the literature.

In particular, if $Y_i,Y_j$ are two i.i.d. $\textrm{Gamma}(1/2,1)$ distributions, then $e^{-Y_i}e^{-Y_j}$ is uniformly distributed on $[0,1]$. From here you can simply define: $$ X_i =( e^{-Y_i} \sqrt{(b-a)}, \ \sqrt{a}) $$ and it feels like it solves your problem.

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  • $\begingroup$ The elementary version of this is that $Y$ has pdf $f(y)=e^{-y}/\sqrt{\pi y}$ on $(0,\infty)$. So \begin{align}\int_0^t f(y)f(t-y)dy &= \frac{e^{-t}}{\pi}\int_0^t\frac{dy}{\sqrt{y(t-y)}}\\ &= \frac{-2e^{-t}}{\pi}\arccos\left(\sqrt{y/t}\right)\Big|_0^t\\ &= e^{-t}.\end{align}$$P\big[e^{-Y_1}e^{-Y_2}<u\big]=P\big[Y_1+Y_2>-\ln u\big]=\int_{-\ln u}^\infty e^{-t} dt = u$$ as desired. $\endgroup$
    – user44143
    Commented Nov 10, 2021 at 15:19

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