# $X,Y$ are iid standard normal. what $P( Y > 3X)$

I know the answer is 1/2 but I am not sure why. I see that there are 4 cases

1. $$X >0, Y > 0$$ Unlikely that $$Y >3X$$ as they are drawn from the same distribution
2. $$X <0, Y > 0$$ Since $$X < 0, Y > 3X$$ all the time
3. $$X >0, Y < 0$$ Since $$Y < 0$$ it wont be greater than $$3X$$
4. $$X <0, Y < 0$$ X will get more negative so $$Y > 3X$$

so basically it is just $$\frac{1}{4}* 0 + \frac{1}{4}* 1 + \frac{1}{4}* 0 + \frac{1}{4}* 1 = 1/2$$

For 1. I am assuming it is unlikely but that doesn't mean it can't happen. So I am curious if my logic here is correct.

• You don't need to know much about these distributions to figure this out: the result holds for any continuous distribution symmetric about $0.$ The standard Normal distribution is symmetric about $0,$ whence $\Pr(Y\gt 3X) = \Pr(-Y\gt 3(-X))=\Pr(Y\lt 3X),$ implying (from the Law of Total Probability) that $\Pr(Y\gt 3X) = (1 - \Pr(Y=3X))/2.$ That equals $1/2$ because $(X,Y)$ is a continuous random variable. – whuber Oct 6 at 19:36

Yes, it’s 1/2, a simple way to solve it is using normal Rv rules, i.e. $$P(Y>3X)=P(Y-3X>0)$$, and $$Z=Y-3X$$ is normal RV with mean $$0$$ and variance $$10$$. Since the mean is $$0$$, independent of the variance, $$P(Z>0)$$ is $$1/2$$, since the normal curve is symmetric around $$0$$.
Your logic is not correct. We can’t say that given the variables are positive, $$P(Y>3X)$$ is $$0$$, and vice versa for both negative case.
Edit based on request: $$E[Z]=E[Y-3X]=E[Y]-3E[X]=0$$ $$\operatorname{var}(Z)=\operatorname{var}(X-3Z)=\operatorname{var}(X)+(-3)^2\operatorname{var}(Z)=10$$
• Could you expand on how you got that $Z$ has mean 0 and variance 10? – genescuba Oct 6 at 19:30