Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 2 / 3F / Solution

Because $X$ is a nonnegative integer, its indicator function satisfies $1_{\{X>0\}}\le X$. Taking expected values proves the upper bound. The finite second moment also gives a finite expected value, by the Cauchy-Schwarz inequality. Applying that inequality to $X$ and $1_{\{X>0\}}$ givesThe assumed positive second moment permits division, yieldingThe lower bound is the basic second moment method; the upper bound also follows from Markov's inequality at threshold one. Integer-valuedness is essential for that upper bound, but not for the lower bound. For example, a constant $X=1/2$ would violate the upper bound.