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Suppose i have a sample $x_1,...x_n$ from a random variable $X$ that is supposed to be infinitely divisible.

Since $X$ is $2$-divisible, there exists $Y$ and $Z$, independant and with the same distribution, such that $Y+Z$ has the same distribution as $X$. Namely, the laplace transform of $Y$ and $Z$ is the root square of the Laplace transform of $X$.

My goal is to construct samples $y_1,...y_n$ and $z_1,...z_n$ such that:

  1. (hard constraint) for all i, $x_i = y_i + z_i$
  2. (soft constraint, i.e. probably the loss) $y_1,...y_n$ and $z_1,...z_n$ are as i.i.d. as possible.

The second constraint could evidently never be satisfied fully and should proabbly be represented as the loss of the problem, while the first one is sctrict and must be fullfilled.

Do you know how i could do such a thing ? Is there some kind of literature on these concerns and potential algorithmic approaches, or some nomenclature that I do not have (my bad wording might be the reason I find nothing..).

Thanks !

PS: Halving a discrete random variable? seems related.

PPS: I did not precise it, but all the values i'm considering are positives reals. The distributions are all defined on $\mathbb R_+$, the available $x_i$ are in $\mathbb R_+$, and so must be the $y_i,z_i$.

Pilou l'Nrv
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    May I ask whether you only have the samples of the random variable $X$? Or do you also know the underlying distribution of it? – Peter Pang Apr 06 '22 at 16:58
  • Both cases are arising : somtimes I know that $x_1,...x_n$ were drawn from a given distribution (say a LogNormal, which is infinitely divisible, or a compound Poisson), sometimes I do not (and I assume that the underlying distribution is infinitely divisible without really knowing that it is so). Why do you ask @Peter ? PS: I added the precision that all these distributions are and must stay positives reals. – Pilou l'Nrv Apr 07 '22 at 12:29
  • Lognormal distributions are not infinitely divisible. Although you state a goal, you haven't supplied sufficient constraints. What, exactly, are your inputs?? – whuber Apr 07 '22 at 13:01
  • So let's say you can generate an arbitrary number of samples of $X$ using, e.g., MCMC, which does not require you to know the underlying distribution fully. Instead of directly sampling $x$, you draw $y$ and $z$ from the same uniform distribution and run MCMC with the acceptance $\alpha = {\rm min}(1, p_X(y^\prime + z^\prime) / p_X(y + z))$. With proper burn-in and thining, you could get i.i.d. samples. – Peter Pang Apr 07 '22 at 13:04
  • Indeed, that would produce such samples. But sadly I do not have the possibility to change the sample I already have (the one that is already drawn and given to me): x_1,...x_n are not random but constants, already drawn, and i need to deal with them. Otherwise instead of MCMC you could just compute an empirical laplace transfrom of $X$, take the root square and an inverse la place transform before re-sampling the obtained distribution (it works well), but it does not fullfill the hard constraint... – Pilou l'Nrv Apr 07 '22 at 13:21
  • @whuber Sorry i missed your comment. Yes Lognormal distributions are infinitely divisible, see Thorin 1977. My inputs are exactly the values of $x_1,...,x_n$, and my outputs should be the values of $y_1,...,y_n$ and $z_1,...,z_n$. Sufficient constraint are hard to express, I still need a computable loss that tells me "how far are ys and zs to the second constraint", which i would then minimize under the first constraint. Tell me if i'm still not clear. – Pilou l'Nrv Apr 13 '22 at 16:11
  • Because the sum of two independent lognormal distributions is not lognormal, which would contradict the usual meaning of infinitely divisible, please tell us what you mean by "infinitely divisible," – whuber Apr 13 '22 at 16:39
  • Well, you have the definition wrong. See https://en.wikipedia.org/wiki/Infinite_divisibility_(probability) for the usual definition, which is exaclty the reverse : 2-divisibilty of the lognormal distribution means that there exists another distribution, call it D, such that the sum of two inependant random variables ith distribution D is lognormally distributed. See Steutel-van harm for more details, or directly Thorin. The divisibility property is also well explained in https://en.wikipedia.org/wiki/Indecomposable_distribution if you want. – Pilou l'Nrv Apr 13 '22 at 17:21
  • Thank you -- I had the concept confused with stable distributions. It is very interesting to learn that the lognormal is infinitely divisible. – whuber Apr 14 '22 at 16:51
  • You are welcome. Any thoughts on my initial post ? – Pilou l'Nrv Apr 15 '22 at 17:30

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