Given the joint distribution of two random variables, compute the probability that one is less than the other? Let $X$, $Y$ have the joint density function
$$f(x,y) = \frac{1}{2\pi} e^{-(x^2+y^2)/2}$$
Compute $P(X<Y)$.
I believe that I should set up a double integral over this function, like so:
$$\int_{-\infty}^{\infty}\int_{-\infty}^y\frac{1}{2\pi} e^{-(x^2+y^2)/2} dx dy$$
However, I do not get a convergent integral like I should. My work is shown below:
When I first integrate with respect to x, I get:
$$\int_{-\infty}^{\infty} \frac {1}{(4\pi*y)} dy$$ and after solving this integral, I get $$-\frac {ln(y)}{4\pi} $$ evaluated from $-{\infty}$ to $\infty$, which doesn't give me a conclusive result. 
Thoughts?
 A: Two thoughts: separable and symmetric


*

*$\frac{1}{2\pi} e^{-(x^2+y^2)/2} = \frac{1}{\sqrt{2\pi}} e^{-(x^2)/2} \frac{1}{\sqrt{2\pi}} e^{-(y^2)/2}$

*$\frac{1}{2\pi} e^{-(x^2+y^2)/2} = \frac{1}{2\pi} e^{-(y^2+x^2)/2}$
A: By Symmetry and Seperability, the two Random Variables $X$ and $Y$ are Independent.So
$$f_X(x)=\frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}},x \in \mathbb{R}$$
$$f_Y(y)=\frac{1}{\sqrt{2\pi}}e^{\frac{-y^2}{2}},y \in \mathbb{R}$$ Now
$$P(X<Y)=\int_{y=-\infty}^{\infty}P(X<y/Y=y)f_Y(y)dy$$ $\implies$
$$P(X<Y)=\int_{y=-\infty}^{\infty}\left(\int_{x=-\infty}^{y}f_X(x)dx\right)f_Y(y)dy$$ $\implies$
$$P(X<Y)=\frac{1}{\sqrt{2\pi}}\int_{y=-\infty}^{\infty}\left(\frac{1}{\sqrt{2\pi}}\int_{x=-\infty}^{y}e^{\frac{-x^2}{2}}dx\right)e^{\frac{-y^2}{2}}dy$$ $\implies$
$$P(X<Y)=\frac{1}{\sqrt{2\pi}}\int_{y=-\infty}^{\infty}\left(\frac{1}{\sqrt{2\pi}}\int_{x=-y}^{\infty}e^{\frac{-x^2}{2}}dx\right)e^{\frac{-y^2}{2}}dy$$ $\implies$
$$P(X<Y)=\frac{1}{\sqrt{2\pi}}\int_{y=-\infty}^{\infty} Q(-y)e^{\frac{-y^2}{2}}dy$$ $\implies$
$$P(X<Y)=\frac{1}{\sqrt{2\pi}}\int_{y=-\infty}^{\infty} Q(y)e^{\frac{-y^2}{2}}dy=\frac{1}{\sqrt{2\pi}}\left(\int_{-\infty}^{0}Q(y)e^{\frac{-y^2}{2}}+\int_{0}^{\infty}Q(y)e^{\frac{-y^2}{2}}\right)=\frac{1}{\sqrt{2\pi}}\left(\int_{0}^{\infty}Q(-y)e^{\frac{-y^2}{2}}+\int_{0}^{\infty}Q(y)e^{\frac{-y^2}{2}}\right)$$ But $$Q(y)+Q(-y)=1$$ $\implies$
$$P(X<Y)=\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}e^{\frac{-y^2}{2}}=\frac{1}{2}$$
A: My first thought is that the standard bivariate normal random variable is radially symmetric about the origin.  That's a pretty big clue, since it means that the probability that a point lying below the line y = x equals the probability of lying below the line y = 0 equals 1/2.
If you want more analysis and less geometry, try converting to polar coordinates.  We have $(X,Y) \sim N_{0,1}$, and we define:
$$R^2 = X^2 + Y^2$$
and
$$\tan \Theta = \frac{Y}{X}$$
The joint density of $R^2$ and $\Theta$ is
$$\frac{1}{2\pi} \frac{1}{2} e^{-\frac{r^2}{2}}$$
for $0 \leq r^2$ and $0 \leq \Theta \leq 2\pi$.  In particular, they are independent, $R^2$ is exponentially distributed, and $\Theta$ is uniform on $[0,2\pi]$.  Marginalize on $R^2$ and integrate from $\frac{\pi}{4}$ to $\frac{5\pi}{4}$.
