Ex. 10.12

Referring to (10.49), let \(\mathcal{S}=\{1\}\) and \(\mathcal{C} = \{2\}\), with \(f(X_1, X_2)=X_1\). Assume \(X_1\) and \(X_2\) are bivariate Gaussian, each with mean zero, variance one, and \(E(X_1X_2)=\rho\). Show that \(E(f(X_1, X_2)|X_2)=\rho X_2\), even though \(f\) is not a function of \(X_2\).

Soln. 10.12

We need to show that

\[\begin{equation} E[X_1|X_2] = \rho X_2.\non \end{equation}\]

Note that

\[\begin{eqnarray} E[(X_1-\rho X_2)X_2] &=& E[X_1X_2] - \rho E[X_2^2]\non\\ &=&\rho - \rho\non\\ &=&0,\non \end{eqnarray}\]

thus \((X_1-\rho X_2)\) and \(X_2\) are uncorrelated. Note also that both \(X_1-\rho X_2\) and \(X_2\) are jointly normal, it follows that \(X_1-\rho X_2\) and \(X_2\) are independent (see, e.g., Section 4.7 in Introduction to probability).

Thus, we have

\[\begin{eqnarray} E[X_1|X_2] &=& E[X_1-\rho X_2 + \rho X_2|X_2]\non\\ &=&E[X_1-\rho X_2|X_2] +E[\rho X_2|X_2]\non\\ &=&E[X_1-\rho X_2] + \rho X_2\non\\ &=&\rho X_2.\non \end{eqnarray}\]