Past exam of the mathematics course of the University of Cambridge / 2016 / ia / Paper 2 / 11F (Probability)

Write $\mu =\mathbb{E}[X]$ and $A=\{X>\theta \mu \}$. The finite second moment implies a finite first moment, and nonnegativity with $\mathbb{E}[X^2]>0$ implies $\mu >0$. On $A^c$, one has $0\le X\le \theta \mu$. ThereforeSubtract this from $\mathbb{E}[X]=\mu$ to obtain the truncated first moment bound:The indicator function uses the strict event specified in the question; any mass at the threshold belongs to $A^c$.

For real random variables with finite second moments, the Cauchy-Schwarz inequality isThe product is integrable because $2|Y_1{Y}_2|\le Y_1^2+Y_2^2$. If $\mathbb{E}[Y_2^2]=0$, then $Y_2=0$ almost surely and the result is immediate. Otherwise the nonnegative expectation of a square gives, for every real $t$,Choose $t=\mathbb{E}[Y_1{Y}_2]/\mathbb{E}[Y_2^2]$. The resulting inequality is exactly the claimed Cauchy-Schwarz inequality. Equality holds precisely when $Y_1$ is a constant multiple of $Y_2$ almost surely in this nondegenerate case.

Apply the Cauchy-Schwarz inequality to $Y_1=X$ and the indicator variable $Y_2=1_A$. Since $1_A^2=1_A$, it givesBoth sides of the truncated first moment bound in part (a) are nonnegative, so squaring it preserves the inequality. Combining the bounds and dividing by the positive second moment provesThis is the Paley-Zygmund inequality. It includes $\theta =0$; at $\theta =1$ the lower bound is zero and remains valid.