Exercise 1.44 (Paley-Zygmund inequality)

Question

If $X$ is non-negative and square-integrable, and $0 \leq 	heta \leq 1$, establish the Paley-Zygmund inequality
$$\displaystyle {\bf P}( X > 	heta {\bf E}(X) ) \geq (1-	heta)^2 \frac{({\bf E} X)^2}{{\bf E}(|X|^2)}.$$

Elu Thingol · Accepted Answer

We have
\begin{align*}
{\bf E}X 
	&= {\bf E}X \cdot 1_{X\geq 	heta {\bf E}X} + {\bf E}X \cdot 1_{X < 	heta{\bf E}X}\
	&\leq {\bf E}X \cdot 1_{X\geq 	heta {\bf E}X} + 	heta {\bf E}X\
	&\leq \sqrt{{\bf E}|X|^2} \cdot \sqrt{{\bf E}1_{X\geq 	heta {\bf E}X}} + 	heta {\bf E}X &	ext{Cauchy-Schwarz inequality}\
	&\leq \sqrt{{\bf E}|X|^2} \cdot \sqrt{P(X > 	heta {\bf E}X)} + 	heta {\bf E}X)\
\end{align*}
Moving the right-hand term to the left
$$(1-	heta){\bf E}X \leq \sqrt{{\bf E}|X|^2} \cdot \sqrt{P(X > 	heta {\bf E}X)}$$
and squaring both sides after moving $(1-	heta)$ to the right, we obtain
$$({\bf E}X)^2 \leq \frac{1}{(1-	heta)^2}{\bf E}|X|^2 \cdot P(X > 	heta {\bf E}X)$$
as desired.

Exercise 1.44 (Paley-Zygmund inequality)

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Exercise 1.44 (Paley-Zygmund inequality)

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