Past exam of the year 1 of the computer science course of the University of Oxford / 2018 / Discrete mathematics, continuous mathematics, and probability / 8 / b

For $x>1$, write the proposed density as $x^{-\alpha }(c-1/x)$. Since the first factor is positive, nonnegativity for all such $x$ requires $c\ge 1$; if $c<1$, values sufficiently close to one make it negative. With $c\ge 1$, the positive tail is integrable only if $\alpha >1$. In that case normalization requiresImposing $c\ge 1$ now gives ${\alpha }^2-\alpha -1\ge 0$. Combining this with $\alpha >1$ yieldsConversely, these conditions ensure a nonnegative, integrable function with integral one, so they are sufficient as well as necessary. The endpoint is allowed: the density vanishes at the left-hand boundary limit, not on its whole support.

Normalization gives $c=8/3$. Integrating the probability density from one to two givesThe expected value exists, since the first-moment integrals converge, and equals