# The product of two lognormal distributions may not be a lognormal

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?

• 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.
– whuber
Commented Dec 17, 2021 at 17:15
• Thanks @whuber for the helpful link. The examples are clear and intuitive, helped me to grasp the ideas. Commented Dec 22, 2021 at 22:24

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.

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.

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