Past exam of the probability, statistics and options I course of the University of Oxford / 2011 / M / 3

The marginal probability density function is $f_Y(y)={\int }_{\mathbb{R}}{f}_{X,Y}(x,y)dx$. For $y$ with $f_Y(y)>0$, the conditional density isIf the conditional first absolute moment is finite, the conditional expectation isConditioning here uses a density, since the individual event $Y=y$ normally has probability zero; a conditional law is specified only up to marginal-null values of $y$.

Normalization of the probability density function requiresusing ${\cos }^{2}x=(1+\cos 2x)/2$. Therefore

The density is an even function on a symmetric interval, so the integrand $xf(x)$ is odd. Its integral vanishes:

A linear combination of independent normal random variables is normal. Its mean is $2\cdot 0+0=0$, and independence removes the cross-covariance term in its variance:

The identity $Y=aX+W$ forces $b=1-2a$ and $c=1-a$. Independence will follow from zero covariance, because $X,W$ are jointly Gaussian random variables. That covariance isThusThese values give $W=(-Z_1+2Z_2)/5$ with variance $1/5$ and zero covariance with $X$. Joint normality then proves the requested independence.

From the preceding decomposition, $Y=(3/5)X+W$, where $W$ is independent of $X$ and has mean zero. Taking conditional expectation gives $\mathbb{E}[Y\mid X]=(3/5)X$. The corresponding continuous version at $X=5$ is