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Sep 14, 2023 at 0:17 history edited Ben CC BY-SA 4.0
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Sep 13, 2023 at 22:44 history edited Zhanxiong CC BY-SA 4.0
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Sep 13, 2023 at 22:35 comment added Dilip Sarwate @whuber Zhanxiong has already answered your question regarding symmetry. I apologize for the typo in my displayed equation; that $-$ should have been a $=$. Oh well, comments can’t be edited this late.
Sep 13, 2023 at 20:53 comment added Zhanxiong @whuber I guess Dilip means $P(X > 0, Y > 0) = P(X > 0, Y > 0, X > Y) + P(X > 0, Y > 0, Y > X)$, and the probabilities on the right hand side are identical to $P(X > Y > 0)$.
Sep 13, 2023 at 20:37 comment added whuber @Dilip Which symmetry, specifically?
Sep 13, 2023 at 20:22 answer added Zhanxiong timeline score: 2
Sep 13, 2023 at 18:24 comment added Dilip Sarwate According to Abramowitz and Stegun 26.3.19, p 937, $$P(X>0, Y>0) - \frac 14 + \frac{\arcsin \rho}{2\pi}$$ and so, because of the symmetry, $P(X>Y>0)$ should be half that number.
Sep 13, 2023 at 17:53 comment added whuber I hope you will find it's equivalent to one of my two geometric descriptions ;-).
Sep 13, 2023 at 17:34 comment added Zhihao Xu Thank you! I will read it now. By the way, What do you think about what I have done using Cholesky?
Sep 13, 2023 at 17:07 comment added whuber I would like to suggest that simply looking at the plots in my post at stats.stackexchange.com/a/71303/919 will enable you to write down a correct simple formula. All you need to is reverse the transformations to find where the wedge $x\gt y\gt 0$ originated; the probability will equal the angle divided by $2\pi.$
Sep 13, 2023 at 16:49 history edited Zhihao Xu CC BY-SA 4.0
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Sep 13, 2023 at 16:48 history edited Zhihao Xu CC BY-SA 4.0
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Sep 13, 2023 at 16:38 history edited Zhihao Xu CC BY-SA 4.0
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S Sep 13, 2023 at 16:36 review First questions
Sep 13, 2023 at 17:20
S Sep 13, 2023 at 16:36 history asked Zhihao Xu CC BY-SA 4.0