Probability theory - multivariate random variable(computational problem)

Question

Question: Given the following covariance matrix of symmetric gaussian multivariate random variable $(X,Y)$ to be of the form:

\begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}

determine $P\{\omega \in \Omega : X(\omega)>Y(\omega)\}$

My attempts

1) I know that the probability density function for gaussian multivariate random variable $(X,Y)$ is of the form:

$f(x,y)=\frac{1}{2\pi\sigma_{X}\sigma{Y}\sqrt{1-\rho^{2}}}exp\{\frac{-1}{2(1-\rho^{2})}[\frac{(x-m_{X})^{2}}{\sigma_{X}^{2}}+\frac{(x-m_{Y})^{2}}{\sigma_{Y}^{2}}-2\rho \frac{(x-m_{X})(y-m_{Y})}{\sigma_{X}\sigma{Y}}]\}$

where: $m_{X}=E(X)$, $m_{Y}=E(Y)$, $\sigma_{X}^{2}=Var(X)$, $\sigma_{Y}^{2}=Var(Y)$, $\rho=\rho(X,Y)$ (correlation coefficient)

2) From given covariance matrix $\sigma_{X}^{2}=Var(X)=E(X^{2})-[E(X)]^{2}=E(X \times X)- E(X)\times E(X) = Cov(X,X)=2$

$\sigma_{Y}^{2}=Var(Y)=E(X^{Y})-[E(Y)]^{2}=E(Y \times Y)- E(Y)\times E(Y) = Cov(Y,Y)=1$

$\rho=\rho(X,Y)=\frac{Cov(X,Y)}{\sqrt{Var(X)Var(Y)}}=\frac{1}{\sqrt{1 \times 2}} = \frac{1}{\sqrt{2}} =\frac{\sqrt{2}}{2}$

3)Putting these values in the formula of multivariate random variable $(X,Y)$ we have:

$f(x,y)=\frac{1}{2\pi\times \sqrt{2} \times \sqrt{1} \times \sqrt{1-\frac{1}{2}}}exp\{\frac{-1}{2(1-\frac{1}{2})}[\frac{(x-m_{X})^{2}}{2}+\frac{(x-m_{Y})^{2}}{1}-2\rho \frac{(x-m_{X})(y-m_{Y})}{\sqrt{2} \times \sqrt{1}}]\} = \frac{1}{2\pi}exp[-\frac{(x-m_{X})^{2}}{2} -(y-m_{Y})^{2} + (x-m_{X})(y-m_{Y}] = \frac{1}{\sqrt{2\pi}}e^{-\frac{(x-m_{X})^{2}}{2}} \times \frac{1}{\sqrt{2\pi}}e^{-(y-m_{Y})^{2}} \times e^{(x-m_{X})(y-m_{Y})}$

4) $P(\{\omega \in \Omega : X(\omega) > Y(\omega)\}) = P(X < Y \wedge Y \in \mathbb{R}) = P(Y \in \mathbb{R} \wedge X < Y)=\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{y} f(x,y)dydx =\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{y} \frac{1}{\sqrt{2\pi}}e^{-\frac{(x-m_{X})^{2}}{2}} \times \frac{1}{\sqrt{2\pi}}e^{-(y-m_{Y})^{2}} \times e^{(x-m_{X})(y-m_{Y})} dydx $

At this point I got stuck since I do not know how to compute E(X) and E(Y)

I asked my friend for help and I was only told that final calculations for this problem are as follows:

$\frac{1}{2\pi}\int\limits_{-\infty}^{+\infty}e^{-\frac{y^{2}}{2}}\int\limits_{y}^{+\infty}e^{-\frac{(x-y)^{2}}{2}} dxdy = \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty}e^{-\frac{y^{2}}{2}} \times \frac{1}{2} dy = 1 \times \frac{1}{2} = \frac{1}{2}$

From this hint it implies that $E(X)=E(Y)=0$. But I can't see why it is the case. Besides, I can't see how to make bridge between what I did by far and the final calculations of my friend.

Help in this matter will be highly appreciated!

Answer 1

I found that in such cases it is often said that: w.l.o.g. : $E(X)=E(Y)=0$. It must be the case that in this problem such assumption was used. Using such assumption, further calculuations are straigthforward:

$$P(\{\omega \in \Omega : X(\omega) > Y(\omega)\}) = P{\omega \in \Omega : X(\omega) > Y(\omega)} = P(X < Y \wedge Y \in \mathbb{R}) = P(Y \in \mathbb{R} \wedge X < Y)=\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{y} f(x,y)dydx =\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{y} \frac{1}{\sqrt{2\pi}}e^{-\frac{(x-0)^{2}}{2}} \times \frac{1}{\sqrt{2\pi}}e^{-(y-0)^{2}} \times e^{(x-0)(y-0)} dydx = \frac{1}{2\pi}\int\limits_{-\infty}^{+\infty}e^{-\frac{y^{2}}{2}}\int\limits_{y}^{+\infty}e^{-\frac{(x-y)^{2}}{2}} dxdy = \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{+\infty}e^{-\frac{y^{2}}{2}} \times \frac{1}{2} dy = 1 \times \frac{1}{2} = \frac{1}{2}$$

Answer 2

If $\boldsymbol X \sim \mathcal N(\boldsymbol \mu, M)$, then $\boldsymbol a \cdot \boldsymbol X \sim \mathcal N(\boldsymbol a \cdot \boldsymbol \mu, \boldsymbol a^t M \boldsymbol a)$. Therefore $Y - X \sim \mathcal N(\mu_Y - \mu_X, 1)$, $$\operatorname P(Y < X) = \frac 1 2 \operatorname{erfc} \frac {\mu_Y - \mu_X} {\sqrt 2}.$$ If $\mu_X = \mu_Y$, the probability is $1/2$.