91Ó°ÊÓ

X is a random variable and \(E\left( {{X^2}} \right)\)is finite.

\(Var\left( X \right) \ge 0\)

a.

\(\begin{array}{l}Var\left( X \right) \ge 0\\E\left( {{X^2}} \right) - {\left[ {E\left( X \right)} \right]^2} \ge 0\\E\left( {{X^2}} \right) \ge {\left[ {E\left( X \right)} \right]^2}\end{array}\)

Hence proved.

b.

Suppose \(\Pr \left( {X = c} \right) = 1\); where \(c\) is a constant.

Then, \(E\left( X \right) = c\) and \(\Pr \left[ {{{\left( {X - c} \right)}^2} = 0} \right] = 1\)

Therefore,

\(\begin{array}{c}Var\left( X \right) = E\left[ {{{\left( {X - c} \right)}^2}} \right]\\ = 0\end{array}\)

\(Var\left( X \right) = 0\)

This implies,

\(\begin{array}{c}E\left( {{X^2}} \right) - {\left[ {E\left( X \right)} \right]^2} = 0\\E\left( {{X^2}} \right) = {\left[ {E\left( X \right)} \right]^2}\end{array}\)

Hence proved.