Solution
= Solution
Pointwise,
$$
a^2\mathbf1_{\{|Z|\geq a\}}\leq Z^2.
$$
Taking <expectations> and dividing by $a^2>0$ gives the second-moment form of <Markov's inequality>:
$$
\boxed{\mathbb P(|Z|\geq a)\leq\frac{\mathbb E Z^2}{a^2}}.
$$
= Solution
Pointwise,
$$
a^2\mathbf1_{\{|Z|\geq a\}}\leq Z^2.
$$
Taking <expectations> and dividing by $a^2>0$ gives the second-moment form of <Markov's inequality>:
$$
\boxed{\mathbb P(|Z|\geq a)\leq\frac{\mathbb E Z^2}{a^2}}.
$$