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Reading the wiki I saw the statemeent that: "The distribution of the product of two random variables which have lognormal distributions is again lognormal. This is itself a special case of a more general set of results where the logarithm of the product can be written as the sum of the logarithms"

This enters in contadiction with my previous undernstanding( mainly the need of uspposing a bivariate lognormal distribution in order to obtain such a result)

Does anybody have some insight on the rational for such an assumption on the wiki?

Thank you a lot in advance,

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    $\begingroup$ The cited assertion is true provided those random variables have a jointly lognormal distribution (which includes the case where they are independent). $\endgroup$
    – whuber
    Apr 18, 2022 at 13:52

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If $X \sim YZ$, then $\log X \sim \log Y + \log Z$. If Y and Z are both log-normal, the two random variables $\log Y$ and $\log Z$ are both normal, hence $\log X$ is normal.

Edit: as pointed out by @whuber, $\log Y$ and $\log Z$ must also be jointly normal for their sum to be normal; one simple reason that they might be jointly normal is if they are independent, which implies joint normality. Joint normality of two distributions is not equivalent to those distributions being independent and separately normal -- a simple exception would be the dependent variables $W$ and $Q=W$, whose sum is just $2W$ and is also normal. However, the definition of being jointly normal is simply that that $a \log Y + b \log Z$ is normal for all $a,b \in \mathbb{R}$, and so proving joint normality in full generality is no easier than proving the necessary special case where $a=1$ and $b=1$.

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    $\begingroup$ Re "hence:" it might be worth emphasizing that $(\log Y,\log Z)$ must be jointly Normal. See stats.stackexchange.com/questions/557457/…. $\endgroup$
    – whuber
    Apr 18, 2022 at 13:53
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    $\begingroup$ Thanks @whuber, I've included this in answer and tried to give some (hopefully helpful) commentary. $\endgroup$
    – jwimberley
    Apr 18, 2022 at 16:16

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