91Ó°ÊÓ

X and Y are both random variables.

The variance of \(X\)is

\(\begin{align}Var\left( X \right) &= E\left( {{X^2}} \right) - {\left( {E\left( X \right)} \right)^2}\\ &= 10 - {\left( 3 \right)^2}\\ &= 10 - 9\\ &= 1\end{align}\)

The variance of\(Y\)is

\(\begin{align}Var\left( Y \right) &= E\left( {{Y^2}} \right) - {\left( {E\left( Y \right)} \right)^2}\\ &= 29 - {\left( 2 \right)^2}\\ &= 29 - 4\\ &= 25\end{align}\)

The covariance of\(X\)and\(Y\)is

\(\begin{align}Cov\left( {XY} \right) &= E\left( {XY} \right) - E\left( X \right)E\left( Y \right)\\ &= 0 - \left( {3 \times 2} \right)\\ &= 0 - 6\\ &= - 6\end{align}\)

The Correlation of\(X\)and\(Y\)is

\(\begin{align}\rho \left( {XY} \right) &= \frac{{Cov\left( {XY} \right)}}{{Var\left( X \right)Var\left( Y \right)}}\\ &= \frac{{ - 6}}{{\sqrt {1*25} }}\\ &= \frac{{ - 6}}{5}\\ &= - 1.2\end{align}\)

Thus,\(\rho \left( {XY} \right) \ne 0\), the given properties of X and Y are not possible.

Hence, it proved.