Tail Measure Lower Bound

Result

We bound below the measure that a nonnegative measurable real-valued function exceeds some value by its integral.

Let $(X, \mathcal{A} , \mu )$ be a measure space. Let $g: X \to [0, \infty]$ be measurable and square-integrable. Then for all $t$ such that $\int t d\mu \in [0, \int g d\mu )$,

\[ \mu (\Set*{x \in X}{g(x) > t}) \geq \frac{(\int (g - t) d\mu )^2}{\int g^2d\mu }. \]

Let $t$ such that $\int t d\mu \in [0, \int g)$. We have selected $t$ so that $\int (g - t)d\mu \geq 0$. Define $h = (g - t)^+$ and $A = \Set*{x \in X}{h(x) > 0}$. Then

\[ \int (g - t) d\mu \leq \int h d\mu = \int h \chi _{A} d\mu \leq \sqrt{\int h^2 d\mu \int \chi _{A}^2 \; d\mu } \]

Now $g^2 > h^2$, so $\int g^2 d\mu \geq \int h^2 d\mu $. Also $\chi _A^2 = \chi _A$ so $\int \chi _A^2 = \mu (A)$. $h(x) > 0$ if and only if $g(x) \geq t$ for all $x$. So $A = \Set*{x \in X}{g(x) \geq t}$. Combining we have:

\[ \int (g - t)d\mu \leq \sqrt{(\int g^2 d\mu ) \mu (A)}. \]

Let $X$ be a random variable with $\E (X^2) \leq \infty$. Then for all $t \in [0, \E (X))$, we have

\[ P(X > t) \geq \frac{ (\E (X) - t)^2 }{ \E {X^2} }. \]

The above is also called the Paley-Zygmund Inequality.