# lognormal distribution radius to volume

How do I prove (and find the new parameters) that if the radius R of a sphere is lognormally distributed, its volume is also lognormally distributed?

All I can think of is: $$F_V(x) = F(\frac{4}{3}\pi R^3 \leq x) = F_R(\sqrt[3]{\frac{3x}{4 \pi}})$$

so $$f_R(\sqrt[3]{\frac{3x}{4 \pi}}) = \frac{1}{z \sqrt{2 \pi \sigma ^2}} exp(-\frac{lnz - \mu}{2 \sigma ^2})$$ where $$z = \sqrt[3]{\frac{3x}{4 \pi}}$$ but then I don't know to to separate the coefficients to figure out new $$\mu'$$ and $$\sigma' ^2$$ Thank you!

Observe that if $$R$$ is lognormally distributed with parameters $$\mu$$ and $$\sigma$$, then $$R = e^X$$, where $$X$$ is normally distributed with mean $$\mu$$ and standard deviation $$\sigma$$.
Therefore, for any nonzero real scalar $$k$$, the variable $$R^k = e^{kX}$$ is also lognormal because $$kX$$ is normal with mean $$k \mu$$ and standard deviation $$k \sigma$$.
$$c R^k = c e^{kX} = e^{k X + \log c},$$ so the new parameters are $$k \mu + \log c$$ and $$k \sigma$$. In your case, $$k = 3$$ and $$c = \frac{4}{3}\pi$$.
• Thank you! Does the $\frac{4}{3} \pi$ change anything? Oct 20, 2021 at 21:13