Question

Suppose that the radius X of a spherical particle has pdf f(x)= 3x^2 for x in [0,1] and 0 otherwise. Find the pdf of its volume V (V=4/3piX^3)

Answer 1

There are at least three different methods you can use, but I have a preference for the transformation method.

You're given \(V\) as a function of \(X\), call it \(v(x)=\dfrac{4}{3}\pi x^3\). Rewriting as a function of \(v\), we get \(x(v)=\sqrt[3]{\dfrac{3}{4\pi}v}\).

Then, we can find the pdf of \(V\) as follows:
\[\begin{align*}f_V(v)&=f_X(x(v))\cdot\left|\frac{dx(v)}{dv}\right|\\
&=3\left(\frac{3}{4\pi}v\right)^{2/3}\left|\frac{1}{3}\left(\frac{3}{4\pi}v\right)^{-2/3}\cdot\frac{3}{4\pi}\right|\\
&=\frac{3}{4\pi}&\text{for }v\in\left[0,\dfrac{4\pi}{3}\right]\\
\end{align*}\]
You can find plenty of info and examples here if you have trouble following: http://www2.econ.iastate.edu/classes/econ671/hallam/documents/Transformations.pdf

Answer 2

ok thank you! that is the answer that i got but I was unsure if it was correct or not!