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

The cumulative distribution function isFor an absolutely continuous probability distribution, the probability density function is a nonnegative integrable function $f_X$ withThus interval probabilities are integrals of $f_X$, and $F_X^{\prime }=f_X$ almost everywhere, in particular at continuity points of $f_X$.

The word “continuous” in the source is understood in this density sense. A merely continuous cumulative distribution function need not possess a probability density function; it can instead describe a singular continuous distribution.

For an absolutely continuous probability distribution, its expectation isThis is finite when ${\int }_{-\infty }^{\infty }|x|f_X(x)dx<\infty$. More generally it is an extended real value if only one of the positive and negative parts diverges; it is undefined if both diverge. For nonnegative $X$ it always exists in $[0,\infty ]$.

Let $S(x)=\mathbb{P}(X>x)=1-F_X(x)$ for $x\ge 0$. Since $f_X=0$ on the negative half-line, $X\ge 0$ almost surely. We have $S^{\prime }=-f_X$ almost everywhere. On a finite interval, integration by parts givesThe lower endpoint contributes zero. The stated hypothesis gives $RS(R)\to 0$ as $R\to \infty$. Taking the increasing limits of the two nonnegative integrals proves the tail-integral identityThe equality allows the value $+\infty$; the boundary hypothesis alone does not force a finite expectation. In fact the formula for nonnegative $X$ also follows directly from Tonelli theorem, without that additional boundary hypothesis, by integrating $x={\int }_0^{\infty }{1}_{\{t<x\}}dt$ against the probability density function.

Translate the inequality inside the cumulative distribution function:Differentiating with the chain rule gives the probability density functionalmost everywhere. Translation shifts the probability distribution without changing its shape or its total probability.

Reversing sign reverses the inequality. Since $X$ has a probability density function, it has no point masses, soDifferentiating the cumulative distribution function givesalmost everywhere: the minus sign outside $F_X$ cancels the minus sign from the chain rule. If point masses were allowed, the cumulative distribution function would instead require the left limit $1-F_X((-y{)}^{-})$.

Since a square is nonnegative, $F_Y(y)=0$ for $y<0$. For $y\ge 0$, the event $X^2\le y$ is $-\sqrt{y}\le X\le \sqrt{y}$. As $X$ has no point masses,Differentiating for $y>0$ gives the contributions of the two inverse branches:The value assigned to the probability density function at $y=0$ is arbitrary. It can be unbounded near $0$, but there is no atom there because $\mathbb{P}(X=0)=0$. Both positive and negative preimages must be included; keeping just one would generally lose probability.

For any real $y$, put $k=\lfloor y\rfloor$. Since $Y$ is integer-valued,Continuity of the cumulative distribution function of $X$ therefore givesThe strict inequality $X<k+1$ matters before using the absence of point masses. The resulting probability distribution is discrete, with probability mass functionThus $Y$ has no ordinary probability density function with respect to length on the real line. Its cumulative distribution function is a step function; its derivative is zero almost everywhere and does not recover its probabilities. The floor function has converted a density into masses on the integers.