91Ó°ÊÓ

The distribution of X is symmetric about 0. All moments of X exist. And, \(E\left( {\left. Y \right|X} \right) = aX + b\)where a and b are given constants.

Since Xis symmetric about 0, every odd ordered moment of Xis 0, that is, every \(E\left( {{X^k}} \right) = 0\;\forall k = 1,3,5,...\)

But \(E\left( {{X^{2m}}} \right)\)is an even ordered moment and hence is not equal to 0.

Now,\(E\left( Y \right)\) can also be written as,

Also, \(E\left( {{X^{2m}}Y} \right)\)can be written as,

Now,

\(\begin{align}Cov\left( {{X^{2m}},Y} \right) &= E\left( {{X^{2m}}Y} \right) - E\left( {{X^{2m}}} \right)E\left( Y \right)\\ &= bE\left( {{X^{2m}}} \right) - E\left( {{X^{2m}}} \right)b\\ &= 0\end{align}\)

Define the correlation coefficient of \({X^{2m}}\) and Y as follows,

\(\begin{align}{\rho _{{X^{2m}},Y}} &= \frac{{Cov\left( {{X^{2m}},Y} \right)}}{{{\sigma _{{X^{2m}}}}{\sigma _Y}}}\\ &= 0\;\left( {{\rm{since covariance is zero}}} \right)\end{align}\)

Thus, it has been proved that \({X^{2m}}\) and Y is uncorrelated as their correlation coefficient is equal to zero.