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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.

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    $\begingroup$ 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. $\endgroup$ – whuber Oct 6 at 19:36
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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: $$E[Z]=E[Y-3X]=E[Y]-3E[X]=0$$ $$\operatorname{var}(Z)=\operatorname{var}(X-3Z)=\operatorname{var}(X)+(-3)^2\operatorname{var}(Z)=10$$

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  • $\begingroup$ Could you expand on how you got that $Z$ has mean 0 and variance 10? $\endgroup$ – genescuba Oct 6 at 19:30
  • $\begingroup$ @genescuba I've added some explanation. $\endgroup$ – gunes Oct 6 at 19:33

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