Past exam of the mathematical methods II course of the University of Oxford / 2021 / 2

The probability density function is normalized because ${\int }_1^{\infty }3x^{-4}dx=1$. Its expected value and second moment areBoth improper integrals converge, so the variance is

For a fixed $x\in [0,1]$, the allowed $y$ interval is $[0,\sqrt{x}]$. Integrating the joint probability density over it gives the marginal densityFor a fixed $y\in [0,1]$, the inequality $y\le \sqrt{x}$ is equivalent to $x\ge y^2$. The other marginal density is thereforeAs a normalization check, ${\int }_0^{1}3x^{2}dx=1$ and ${\int }_0^{1}3y(1-y^4)dy=1$.