Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Discrete mathematics, continuous mathematics and probability / 9 / b

First $\mu >0$. If a nonnegative random variable had zero expected value, it would equal zero almost surely, which is impossible for a distribution having a probability density function on $[0,\infty )$: a singleton has probability zero. Thus the given division by $\mu$ is legitimate.

Because $x\ge 0$ and $f(x)\ge 0$, $g(x)=xf(x)/\mu \ge 0$. MoreoverExtend $g$ by zero for negative $x$. It is a probability density function, defining the size-biased distribution of $X$: larger observations receive weight proportional to their value.

The law of the unconscious statistician and the finite variance giveThus size bias increases the mean by ${\sigma }^2/\mu$, in accord with preferentially selecting larger values.

Write $F_X(x)=\mathbb{P}(X\le x)$ and $F_Y(x)=\mathbb{P}(Y\le x)$. For $0\le x\le \mu$, the factor $t/\mu$ is at most one throughout $0\le t\le x$, soFor $x>\mu$, compare the tails instead: $t/\mu \ge 1$ throughout $t\ge x$, givingRearranging provesThis is stochastic domination of $X$ by its size-biased distribution. The proof needs only the finite positive mean, not the finite variance assumed in (ii).