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Suppose that $X$ and $Y$ are iid normally distributed and $a$ is a scalar. What is $\Pr(Y+aX<0 | X>0)$?

ABC
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  • Sorry about that -- I clarified that they're jointly normal. – ABC Jul 26 '12 at 03:58
  • Try this paper: http://etrij.etri.re.kr/Cyber/servlet/BrowseAbstract?vol=32&num=6&pg=965 – Douglas Zare Jul 26 '12 at 05:44
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    Transform $(X,Y)\rightarrow (X,Z)$, where $Z=Y+aX$; you will obtain that $(X,Z)$ has a bivariate normal distribution with a certain mean and covariance matrix. Then use the definition ${\mathbb P}(Y+aX<0\vert X>0)=\dfrac{{\mathbb P}(Y+aX<0,X>0)}{{\mathbb P}(X>0)}=\dfrac{{\mathbb P}(Z<0,X>0)}{{\mathbb P}(X>0)}$. The numerator can be calculated using the CDF of a bivariate normal and the denominator using the CDF of the univariate normal. –  Jul 26 '12 at 09:37
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    If the means are zero, this requires (virtually) *no* calculation at all. Are you interested in the general case or, perhaps, this specific one? – cardinal Jul 26 '12 at 13:12
  • @cardinal I guess the OP is interested in the general case because this expression appears in the paper mentioned in the [previous question](http://stats.stackexchange.com/q/33050/10525). I also guess he did not mean *i.i.d.*. –  Jul 26 '12 at 13:20
  • @Proc: Yes, that seems likely in light of that question (which I had seemed to have missed). Thanks. – cardinal Jul 26 '12 at 13:22
  • In addition to cardinal's observation, if we take the meaning of iid at face value (it is, after all, an edit made by the OP), there is also a simple answer requiring very little calculation for the case $a = 1$ and $\mu_x = \mu_Y \neq 0$. – Dilip Sarwate Jul 27 '12 at 03:06

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