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How do I calculate $P(X > 0 | X+Y > 0)$ if X and Y are standard normal?

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    $\begingroup$ Hint: draw a picture. Divide the $(X,Y)$ plane into eight equiprobable sectors along the lines $X=0$, $Y=0$, $X+Y=0$, and $X-Y=0$. How many of those correspond to $X+Y \gt 0$? How many out of those correspond to $X \gt 0$? Draw your conclusion. $\endgroup$
    – whuber
    Sep 18, 2017 at 21:55
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    $\begingroup$ Thanks. I tried this, I got 3/4, which makes sense visually. I am wondering if there is an analytical way to solve this. $\endgroup$ Sep 18, 2017 at 21:59
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    $\begingroup$ Sure--but it's hard to imagine it would be any easier! See stats.stackexchange.com/search?q=normal+conditional+probability for our posts about conditional Normal probabilities. One way is to put the two integrals into polar coordinates. The polar and angular integrals separate and the polar integrals cancel in the ratio that forms the conditional probability. That reduces the problem to considering ranges of angles--which is precisely what you have done visually. $\endgroup$
    – whuber
    Sep 18, 2017 at 22:02
  • $\begingroup$ This question probably should have the self study tag. $\endgroup$ Sep 19, 2017 at 0:19
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    $\begingroup$ What's the joint distribution? You can't answer without knowing that. Commenters (and the present answer) seem to be assuming independence (or at least joint normality) but there's nothing to say so in the question $\endgroup$
    – Glen_b
    Sep 19, 2017 at 0:35

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If the $X$ and $Y$ distributions are uncorrelated and independent and "standard normal":

$$I_1 = \int\limits_{x = -\infty}^0 dx \int\limits_{y=x}^\infty dy {\cal N}(0,1;x) {\cal N}(0,1;y)$$

$$I_2 = \int\limits_{x = 0}^\infty dx \int\limits_{y=0}^\infty dy {\cal N}(0,1;x) {\cal N}(0,1;y) = 2 I_1$$

$$I_3 = \int\limits_{x = 0}^\infty dx \int\limits_{y=-x}^\infty dy {\cal N}(0,1;x) {\cal N}(0,1;y) = I_1$$

$$P(X>0| X+Y >0) = {I_1 + I_2 \over I_1 + I_2 + I_3} = {3 \over 4} ,$$

which conforms to the OP's comment.

If however the distributions are correlated, then one defines the covariance matrix ${\bf \Sigma} = {1\ \rho \choose \rho\ 1}$ and perform the above integrals with the above limits but the joint density:

$$p(X,Y) = {1 \over 2 \pi} e^{-{1 \over 2} {\bf x}^t {\bf \Sigma}^{-1} {\bf x}} .$$

(Somehow I don't think the OP is asking this more general question, but I could be wrong.)

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    $\begingroup$ You seem to be assuming independence. The question doesn't mention independence (it doesn't even mention joint normality) $\endgroup$
    – Glen_b
    Sep 19, 2017 at 0:36
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    $\begingroup$ Yes... I'm assuming independence and it comports with the OP's result that the conditional probability is indeed $3/4$. I suspect the OP (with reputation = 1) is asking a simple question and just not experienced enough to put in all the conditions. But of course I might be wrong. $\endgroup$ Sep 19, 2017 at 0:38
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    $\begingroup$ Yes; on the other hand an inexperienced poster may well omit critical information from their own attempt not realizing its import. I expect your assumption will turn out to be correct, but it would be careless to add such a big assumption* without any comment at all. $\:$ *(it has a clear impact considering the possible answers that you can get without it) $\endgroup$
    – Glen_b
    Sep 19, 2017 at 0:53
  • $\begingroup$ The approach suggested in the comments works almost without change when the distribution is binormal: the answer is still a ratio of angles. They are computed in the Mahalanobis metric rather than the Euclidean one. Going back to the case of independence, this answer currently doesn't show anything: it merely asserts $I_2=2I_1$ and $I_3=I_1$. It would become an actual answer if it would explain or justify these assertions. $\endgroup$
    – whuber
    Sep 19, 2017 at 17:06
  • $\begingroup$ You should at least mention explicitly that there are cases where the sum of two normals is nonnormal. $\endgroup$ Oct 1, 2019 at 19:25

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