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

For the continuous uniform distribution on $[0,1]$, the probability density function isChanging the values at the endpoints does not change any probabilities, because individual points have zero probability.

The law of the unconscious statistician gives the improper integralIn particular $\mathbb{E}(X)=1/2$ and $\mathbb{E}(X^2)=1/3$, soFor $p=-1$, ${\int }_{\epsilon }^1{x}^{-1}dx=-\log \epsilon \to +\infty$; for $p<-1$, the power integral also tends to $+\infty$. Thus $\mathbb{E}(X^p)=+\infty$ for $p\le -1$, so there is no finite expected value. Negative powers need only be defined almost surely on $X>0$; the null event $X=0$ has no effect on this conclusion.

Use the covariance identity and the moments from (ii):The positive value is consistent with both $X$ and $X^2$ increasing together on $[0,1]$.

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).