Let the independent random variables $X,Y\sim N(0,1)$. Prove that $P(X/Y < 1) = 3/4. $
Could anyone help me prove this analytically? Thanks.
Progress: My first thought was to integrate the joint density function: $\dfrac{e^{-\frac{x^2}{2}-\frac{y^2}{2}}}{2\pi}$ but I'm not sure where to go from here.
It is well-known that the random variable $X/Y$ has the two-sided Cauchy density $\frac{1}{\pi(1+x^2)}$ for $-\infty<x<\infty$. Thus
$P(X/Y<1)=\int_{-\infty}^{1}\frac{1}{\pi(1+x^2)}dx=0.5+\int_{0}^{1}\frac{1}{\pi(1+x^2)}dx$
and so
$P(X/Y<1)=0.5+\frac{1}{\pi}\hbox{arctg}(1)=0.5+0.25=0.75$.
Note: A much simpler way is to consider the random vector $(X,Y)$ on the plane. Then,
$P(X/Y<1)=P(X/Y<1\mid Y>0)\times\frac12+P(X/Y<1\mid Y<0)\times\frac12\\ = (\frac12+\frac14) \times \frac12+ (\frac12+\frac14)\times\frac12=\frac34.$