I am trying to prove:
Suppose $(X, \mathcal S, \mu)$ is a measure space with $\mu(X) = 1$ and $h\in \mathcal L^1(\mu)$. Prove that
\begin{align}
\mu\left(\left\{ x\in X: \left|h(x) – \int h\; d\mu\right|\ge c
\right\}\right) \le
\frac{1}{c^2} \left(\int h^2 \; d\mu – \left(\int h \; d\mu
\right)^2\right)
\end{align}
for all $c>0$.
Attempt:
Suppose $(X, \mathcal S, \mu)$ is a measure space with $\mu(X) = 1$ and $h\in \mathcal L^1(\mu)$. Fix $c>0$.
Markov's Inequality: $\mu\left(\left\{x \in X: |f(x)|\ge c
\right\}\right) \le \frac{1}{c^2} \|f\|_1 = \frac{1}{c^2} \cdot \int\left|f\right|\; d\mu$.
Take $f(x) = h(x) – \int h \; d\mu$ in Markov's inequality:
\begin{align*}
\mu\left(\left\{x \in X: \left|h(x) – \int h \; d\mu\right|\ge c \right\}\right)
&\le \frac{1}{c^2} \int\left|h(x) – \int h \; d\mu\right|\; d\mu \\
&\le \frac{1}{c^2} \int\left(h(x) – \int h \; d\mu\right)^2\; d\mu\\
&= \frac{1}{c^2} \int\left[ h^2(x) +\left(\int h \; d\mu\right)^2 – 2h(x) \int h(x) \; d\mu\right] d\mu \\
&= \frac{1}{c^2}\left( \int h^2(x)\; d\mu +\int\left[\left(\int h \; d\mu\right)^2 – 2h(x) \int h(x) \; d\mu\right] d\mu\right)
\end{align*}
I am not sure how to proceed from here. I also know that I still have to use $\mu(X) = 1$. Intuitively, I think I should be using the property that integration is homogeneous, but $h(x)$ is not independent of $x$, of course. Can someone please provide some hints?
Best Answer
Markov's inequality says that if $X$ is a random variable (i.e. a measurable function whose domain is a probability space) and $\Pr(X\ge0)=1,$ and $\operatorname E(X)<+\infty$ (or $\int_\Omega X(\omega)\,P(d\omega)<+\infty$ if you like) then for every $x>\mu,$ we have $\Pr(X>x) \le \mu/x.$
Chebyshev's inequality says that if $Y$ is a random variable with finite variance $\sigma^2,$ and expected value $\nu,$ then for every $w>\sigma^2,$ we have $\Pr(\left| Y-\nu\right|>w) \le \sigma^2/w^2.$
You can use Chebyshev's inequality by applying Markov's inequality to the random variable $X=(Y-\nu)^2$ with $w^2$ in the role in which we put the variable $x$ in Markov's inequality, and then observing that the event $\left|Y-\nu\right|>w$ is the same as the event $(Y-\nu)^2>w^2.$