Past exam of the year 1 of the computer science course of the University of Oxford / 2013 / Discrete mathematics, probability and continuous mathematics / 9

On $0<x<2$, $2x-x^2=x(2-x)>0$. A valid probability density function must integrate to one, soThus $\alpha =3/4$, which is positive and ensures nonnegativity. Its expected value isThereforeThe probability density function is symmetric around $1$, providing a second check on the expected value.

The factor $2x-x^3=x(2-x^2)$ is positive on $(0,\sqrt{2})$ and negative on $(\sqrt{2},2)$. For every nonzero $\alpha$, the proposed probability density function is therefore negative on an interval. The choice $\alpha =0$ has total integral zero. In fact,so no scaling could normalise it even without the sign obstruction. Consequently

Assume $a<b$. An increasing affine map sends the two endpoints to zero and one when $ca+d=0$ and $cb+d=1$. Solving givesIndeed, for $0\le t\le 1$, the cumulative distribution function of the transformed random variable iswhich is the cumulative distribution function of the continuous uniform distribution on $[0,1]$.

A decreasing affine map works too, by swapping the endpoints:These are the two possible choices, since the image interval must have endpoints zero and one. The case $a=b$ cannot be transformed into a nondegenerate continuous uniform distribution by deterministic constants.

Let $W=U^2$. On $0\le U\le 1$, squaring preserves order. Therefore for $0\le w\le 1$,Including the tails gives the cumulative distribution functionDifferentiating on the interior of the support gives the probability density functionThe divergence near zero is integrable: ${\int }_0^1(2\sqrt{w}{)}^{-1}dw=1$. There is no point mass at zero, and the assigned value of a probability density function at a single endpoint does not affect the probability distribution.

Complete the square in the transformation:Since $0<X<2$ almost surely, $0<Y\le 1$. For $0\le y<1$, put $r=\sqrt{1-y}$. The event $Y>y$ is equivalent to $1-r<X<1+r$. Integrating the given probability density function over both branches at once gives the survival functionThus, including the tails,The cumulative distribution function on $(0,1)$ is $1-\sqrt{1-y}$. Differentiation givesBoth inverse branches matter: $x=1\pm \sqrt{1-y}$. The endpoint divergence is integrable, and $\mathbb{P}(Y=1)=\mathbb{P}(X=1)=0$, so no endpoint atom is needed.