Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 1 / 12

For a probability density function $P$, the mean and variance areThese formulas assume the required moments are finite; finiteness of the second moment ensures a finite mean as well.

The triangular probability density function is normalized because $2{\int }_0^1(1-x)dx=1$. It is even, so the mean integrand is odd and $\mu =0$. Its variance is consequently its second moment:

The standard normal density is even, so $x^{3}P(x)$ is odd. Since Gaussian decay dominates every power, the integrand is absolutely integrable. Therefore

For the standard normal density, $P^{\prime }(x)=-xP(x)$. Using integration by parts,The boundary term vanishes because Gaussian decay dominates every power. Since the given mean is zero and variance is one, $\mathbb{E}[X^2]=1$, yielding the Gaussian fourth moment