Proof of Chebychev inequality - help with integrals and a chain of (in)equalities

Question

I need help in deriving the following chain of (in)equalities as a part of proof of Chebychev inequality (from these lecture notes, 7.3): $$\begin{align} &\ge\int_{-\infty}^{\theta-a}(w-\theta)^2f_w(w)dw+\int_{\theta+a}^{+\infty}(w-\theta)^2f_w(w)dw\\ &\ge a^2\int_{-\infty}^{\theta-a}f_w(w)dw+a^2\int_{\theta+a}^{+\infty}f_w(w)dw=a^2P(|W-\theta|>a) \end{align}$$

Answer 1

I assume $f_w$ is the density of a probability distribution over $\mathbb{R}$, and $a > 0$. (Also, the integrals are well-defined, i.e. $X$ is a random variable with density $f_w$ which is square integrable.)

• The second equality is the simplest: starting with the RHS, $$\begin{align} \mathbb{P}\{\lvert X-\theta\rvert > a\} &= \mathbb{P}\{X-\theta > a\} + \mathbb{P}\{\theta - X > a\}\\ &= \mathbb{P}\{X > \theta+a\} + \mathbb{P}\{ X < \theta -a\}\\ &= \int_{\theta+a}^\infty f_w(w)dw + \int_{\infty}^{\theta -a} f_w(w)dw. \end{align}$$

• For the first inequality: note that (i) when $w \leq \theta-a$, then $\theta-w \geq a > 0$ and $(\theta-w)^2 \geq a^2$; and (ii) when $w \geq \theta+a$, then $w-\theta \geq a$ and $(w-\theta)^2 \geq a^2$ as well. This gives that $$\begin{align} \int_{-\infty}^{\theta-a}(w-\theta)^2f_w(w)dw&+\int_{\theta+a}^{+\infty}(w-\theta)^2f_w(w)dw\\ &\geq \int_{-\infty}^{\theta-a} a^2 f_w(w)dw+ \int_{\theta+a}^{+\infty} a^2 f_w(w)dw \\ &=a^2 \int_{-\infty}^{\theta-a}f_w(w)dw+a^2 \int_{\theta+a}^{+\infty}f_w(w)dw \end{align}$$ as wanted, since $f_w\geq 0$ (it's a density function).