Past exam of the year 1 of the computer science course of the University of Oxford / 2023 / Discrete mathematics, probability and continuous mathematics / 9

The cumulative distribution function is $F_X(x)=\mathbb{P}(X\le x)={\int }_{-\infty }^{x}f(u)du$. This formula uses the assumed probability density function; the density itself need not be a continuous function.

An absolutely continuous probability distribution assigns zero probability to a singleton, because its density has zero integral over that set. Thus $\mathbb{P}(X=x)=0$ for every real $x$, including points where a chosen version of $f$ is nonzero.

The events $\{X=k\}$, for $k\in \mathbb{Z}$, are pairwise disjoint. Their union is $\{X\in \mathbb{Z}\}$, and the integers form a countable set. By countable additivity,Countability is the necessary justification: one cannot apply this argument to an arbitrary uncountable union of zero-probability singletons. This is a special case of countable sets have zero probability under a density.

Normalization of the probability density function requires $1=c{\int }_1^{\infty }{x}^{-t}dx=c/(t-1)$. Hence $c=t-1$. It is positive for the given $t>2$, and the density is nonnegative, so this does define a probability distribution. It is a Pareto distribution with scale $1$ and shape $t-1$.

The expected value is finite because $t>2$:The condition on $t$ is precisely the first moment threshold for this parameterization of the Pareto distribution.

For $k\ge 1$, the event $\{\lfloor X\rfloor \ge k\}$ equals $\{X\ge k\}$. Integrating the Pareto distribution's density givesThe mass formula follows by subtracting consecutive tail probabilities, or by integrating over $[k,k+1)$. With the paper's convention $0\in \mathbb{N}$, include the separate endpoint values $\mathbb{P}(\lfloor X\rfloor \ge 0)=1$ and $\mathbb{P}(\lfloor X\rfloor =0)=0$, since $X\ge 1$ almost surely. Substituting $k=0$ into the power expression would be invalid.

The nonnegative integer-valued tail-sum formula for expectation givesThis is finite because $t-1>1$. Here $\zeta$ is the Riemann zeta function on its convergent real domain. The floor function is nonnegative on the support, so the tail-sum identity applies directly.

Part (a) gives $\mathbb{P}(X\in \mathbb{Z})=0$. Therefore the ceiling function and floor function satisfy $\lceil X\rceil =\lfloor X\rfloor +1$ almost surely. Both are integrable, since $0\le \lfloor X\rfloor \le X$ and $\lceil X\rceil \le X+1$. Thus $\mathbb{E}[\lceil X\rceil ]=\zeta (t-1)+1$.

Now take expectations in $\lfloor X\rfloor \le X\le \lceil X\rceil$:Writing $s=t-1$, which ranges over every $s>1$, and rearranging yieldsThese are the zeta bounds from rounded Pareto variables: rounding below and above converts one elementary expected value into bounds on an infinite series.