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If two lognormal random variables $X, \:Y$ are neither independent nor jointly normally distributed, the product $XY$ may not be lognormally distributed.

Where can I find the explanation for the 'may not be' part in the second statement?

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    $\begingroup$ Applying the logarithm converts questions about lognormal variables and products into equivalent questions about normal variables and sums. Explicit (and beautifully illustrated) examples are posted at stats.stackexchange.com/a/30205/919, where it is clear the sum needn't be Normally distributed. $\endgroup$
    – whuber
    Dec 17, 2021 at 17:15
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    $\begingroup$ Thanks @whuber for the helpful link. The examples are clear and intuitive, helped me to grasp the ideas. $\endgroup$
    – yufiP
    Dec 22, 2021 at 22:24

2 Answers 2

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Suppose $W$ and $X$ are independent and lognormally distributed, each with median $m$. Let

$$Y = \begin{cases} \min(W, m/W) \text{ if } X\le m\\ \max(W, m/W) \text{ if } X>m \end{cases}$$

Then $X$ and $Y$ are both lognormal, and either $X$ and $Y$ are both above $m$ or both below $m$. So $XY$ is bimodal and not lognormal.

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Suppose $X\sim LN(\mu, \sigma^2)$ is log-normally distributed.

Then its reciprocal $\frac{1}{X}\sim LN(-\mu, \sigma^2)$ is also log-normally distributed.

But of course $X\cdot\frac{1}{X}=1$ is a degenerate point mass at $1$ and not log-normal any more.

The problem is that $X$ and $\frac{1}{X}$ are not independent.

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    $\begingroup$ The atom at $1$ is a Lognormal distribution. It is the $LN(0,0)$ distribution (exponential of the point mass at $0$). $\endgroup$
    – whuber
    Dec 17, 2021 at 17:15

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