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Consider the two independent and identically distributed standard normal random variables $X,Y∼N(0,1)$. What is the probability: $P(0<X<Y)$?

Intuitively, I would think the answer is $1/8$, because $X$ has $1/2$ chance of being positive and, if positive, would have an expected value at the $75th$ percentile of the standard normal curve. $Y$ would then have $1/4$ chance of being greater than $X$. $1/2*1/4 = 1/8$

I guess otherwise I can set up integrals and change to polar coordinates to solve? A bit confused...

mathstudent12321
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  • You might find this useful: http://math.stackexchange.com/questions/261073/finding-probability-pxy – 3x89g2 Oct 20 '15 at 04:32
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    Hint: By symmetry, $P(0 < X < Y) = P(0 < Y < X)$. And these two events are disjoint - what do you get if you add these probabilities? – Nate Eldredge Oct 20 '15 at 04:34
  • Thanks guys. @NateEldredge, I assume that adding the probabilities would equal 1/4, because both X and Y would have to be positive with 1/2 probability each. Right? – mathstudent12321 Oct 20 '15 at 05:12

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