Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 16

A probability density function must have total integral one. The numerator is an even function, soThis is positive, ensuring nonnegativity of the probability density function.

Put $N=6-2e^{-2}$. The function graph is symmetric about the vertical axis, zero outside $[-2,2]$, and has its unique maximum at $x=0$ with height $2/N$. On $0<x<2$, it decreases as $(1+e^{-x})/N$; the negative side is its reflection. The one-sided interior endpoint values are $(1+e^{-2})/N$, so the probability density function jumps to zero at both ends of its support. It also has a corner at its central maximum because of the absolute value in the exponent.

Values assigned exactly at $x=\pm 2$ do not affect any probability.

Since the probability density function is an even function, $xf(x)$ is an odd function. Its integral over the symmetric support vanishes, so

The expected value is zero, so the variance is $\mathbb{E}[X^2]$. Symmetry givesThe polynomial term contributes $8/3$. Two applications of integration by parts giveso its definite value on $[0,2]$ is $2-10e^{-2}$. Therefore

Let $F$ be the cumulative distribution function of $X$. Normalisation of the proposed probability density function requires $M={\int }_{-2}^{2}F(y)dy$. The integrated cumulative distribution function on a bounded interval identity givesFor a direct symmetry proof, the continuous symmetric probability distribution satisfies $F(-y)=1-F(y)$. HenceIn particular $g(y)=F(y)/2$ is nonnegative and has total integral one over its stated support.