91Ó°ÊÓ

\({\left( {\left( {X - {\mu _X}} \right) \pm \left( {Y - {\mu _Y}} \right)} \right)^2} \ge 0\)

\(\begin{align}{\left( {\left( {X - {\mu _X}} \right) + \left( {Y - {\mu _Y}} \right)} \right)^2} \ge 0\\ \Rightarrow \left( {X - {\mu _X}} \right)\left( {Y - {\mu _Y}} \right) \le \frac{1}{2}\left( {{{\left( {X - {\mu _X}} \right)}^2} + {{\left( {Y - {\mu _Y}} \right)}^2}} \right)\end{align}\)

Also,

\(\begin{align}{\left( {\left( {X - {\mu _X}} \right) - \left( {Y - {\mu _Y}} \right)} \right)^2} \ge 0\\ \Rightarrow - \left( {X - {\mu _X}} \right)\left( {Y - {\mu _Y}} \right) \le \frac{1}{2}\left( {{{\left( {X - {\mu _X}} \right)}^2} + {{\left( {Y - {\mu _Y}} \right)}^2}} \right)\end{align}\)

Hence,

\(\left| {\left( {X - {\mu _X}} \right)\left( {Y - {\mu _Y}} \right)} \right| \le \frac{1}{2}{\left( {\left( {X - {\mu _X}} \right) + \left( {Y - {\mu _Y}} \right)} \right)^2}\)

Hence, proved.

By taking Expectation on both side, it becomes,

\(\begin{align}E\left( {\left| {\left( {X - {\mu _X}} \right)\left( {Y - {\mu _Y}} \right)} \right|} \right) \le \frac{1}{2}\left( {Var\left( X \right) + Var\left( Y \right)} \right)\\ < \infty \end{align}\)

Since,\(E\left( {\left| {\left( {X - {\mu _X}} \right)\left( {Y - {\mu _Y}} \right)} \right|} \right)\)is finite,

Hence,\(Cov\left( {X,Y} \right) = E\left( {\left( {X - {\mu _X}} \right)\left( {Y - {\mu _Y}} \right)} \right)\)exists and finite.

Therefore,\(Cov\left( {X,Y} \right)\)exists and finite.