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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3$\begingroup$ 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 ♦Commented Dec 17, 2021 at 17:15
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1$\begingroup$ Thanks @whuber for the helpful link. The examples are clear and intuitive, helped me to grasp the ideas. $\endgroup$– yufiPCommented Dec 22, 2021 at 22:24
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2 Answers
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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.
answered Dec 17, 2021 at 14:05
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2$\begingroup$ The atom at $1$ is a Lognormal distribution. It is the $LN(0,0)$ distribution (exponential of the point mass at $0$). $\endgroup$– whuber ♦Commented Dec 17, 2021 at 17:15