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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!

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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 $k \sigma$.

In other words, because scale transformations of a normally distributed random variable are also normally distributed, and because lognormal random variables are exponential transformations of normal random variables, then nonzero powers of lognormal random variables are lognormal.

Similarly, location transformations of a normally distributed random variable are also normally distributed, so multiplication of a lognormal random variable by a positive constant also results in a lognormal random variable. That is to say,

$$c R^k = c e^{kX} = e^{k X + \log c},$$ so the new parameters are $k \mu + \log c$ and $k \sigma$. In your case, $k = 3$ and $c = \frac{4}{3}\pi$.

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  • $\begingroup$ Thank you! Does the $\frac{4}{3} \pi $ change anything? $\endgroup$ Oct 20, 2021 at 21:13
  • $\begingroup$ @sucksatmath Yes. Please see my edited answer. $\endgroup$
    – heropup
    Oct 20, 2021 at 21:15

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