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Everyone makes a distinction between what are called stable distributions and infinitely divisible distributions. Stable distributions act as an 'attractor' of sorts: the sum of a large number of iid random variables will, by the central limit theorem (CLT), converge to a normal distribution independent of the original distribution.

Infinitely divisible distributions are a bit weaker, it seems.

But suppose that we are not adding iid RVs, but multiplying them (and for the sake of simplicity, let's assume the RVs in question have only positive support): $$Y_n=\Pi_0^n X_i$$If we take the logarithm, then$$logY_n=\Sigma_0^nlogX_i$$ Now, assuming the original $X_i$ are iid, it seems by the CLT that the sum on the left also is the sum of iid RVs; it, too, must converge (using the CLT again) to a normal distribution as n becomes large.This seems to imply that multiplying (strictly positive) RVs must converge, in the limit, to a lognormal distribution since the "underlying distribution" must converge to a normal distribution - and,when we "go back" by exponentiating, it seems we should get: $$e^{Y_n}$$which should converge to a lognormal distribution as $n\rightarrow\infty$, since $Y_n$ is converging to a normal distribution.

I am sure there is a mathematical wormhole here, but I can't spot it.

eSurfsnake
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  • I see no connection between your proposed multiplicative CLT and the question of stability. Could you clarify? – Kevin Carlson Oct 27 '19 at 15:14
  • It may be simply verbiage. The normal distribution is often motivated as the sum of many iid RVs from other distributions, so it is a sort of "attractor" and is called "stable". Is it that the term "infinitely divisible" is just the same ide for a multiple of many iid RVs, or is somehow a "stable" distribution stronger, in some sense, than an infinitely divisible one? – eSurfsnake Dec 01 '19 at 23:45
  • These words have precise definitions. An infinitely divisible random variable is one which may be written as a sum of arbitrarily many iids. A stable distribution is one from a family invariant under linear combinations, so in particular is infinitely divisible by using scaled copies of itself. It is not accurate, in particular, that the normal is called “stable” because it obeys the central limit theorem. Neither concept has anything to do with multiplication. – Kevin Carlson Dec 02 '19 at 02:00
  • But the lognormal, I don't believe, is invariant under linear combinations, is it? – eSurfsnake Dec 03 '19 at 21:14
  • That is correct. The lognormal is not stable. – Kevin Carlson Dec 04 '19 at 00:19

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It seems to me that you're correct that the product of independent RVs tends towards a Lognormal distribution. This post seems to support it. Do you have evidence to the contrary?

$$ Y := e^Z \sim \mathrm{Lognormal} \iff Z \sim \mathrm{Normal} $$

So by the laws of the exponent and CLT, we can see that multiplying IID Lognormal distributions will give us a Lognormal distribution:

$$ Y_1 Y_2 Y_3 \cdots = e^{Z_1}e^{Z_2}e^{Z_3} \cdots = e^{Z_1+Z_2+Z_3+\cdots} = e^{Z_\text{sum}} $$

Adding Lognormals would not result in a Lognormal (remember Lognormals have the form $\exp(Z)$):

$$ Y_1 + Y_2 + Y_3 + \cdots = e^{Z_1} + e^{Z_2} + e^{Z_3} + \cdots = \exp(\ln(e^{Z_1} + e^{Z_2} + e^{Z_3} + \cdots)) $$

In this case, our would-be $Z$ is $\ln(e^{Z_1} + e^{Z_2} + e^{Z_3} + \cdots)$ which does not simplify so, by the CLT, becomes $\ln(Z_*)$. Therefore:

$$ Y_1 + Y_2 + Y_3 + \cdots = \exp(\ln(Z_*)) = Z_* \sim \mathrm{Normal} $$

Zaz
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