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This might be a dumb question, but is there a simple way to not invoke the joint probability function and show that $P(X+aY>0|Y<0)$ is increasing or decreasing in $a$ given $X~N(0,1), Y~N(0,1), Corr(X, Y)=\rho>0$, $X$ and $Y$ jointly normal.

My proof is that since $$P(X+a^{'}Y>0|Y<0)=P(X+aY>(a-a^{'})Y|Y<0)$$ for any $a^{'}>a$, then $(a-a^{'})<0$ and $(a-a^{'})Y>0$ for $Y<0$.

Then, $$P(X+a^{'}Y>0|Y<0)=P(X+aY>(a-a^{'})Y|Y<0)<P(X+aY>0|Y<0).$$

So, $P(X+aY>0|Y<0)$ is decreasing in $a$. But I am not sure if I can claim that because $(a-a^{'})Y>0$ for $Y<0$, then $P(X+aY>(a-a^{'})Y|Y<0)<P(X+aY>0|Y<0)$.

che00c
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  • You have only to check that the event $[X+aY>0 ; Y < 0]$ decreases (in a wide sense) as $a$ increases. By the definition of conditional probabilities, the desired inequality follows. Additional assumptions on the real random variables $X$ and $Y$ are useless. – Christophe Leuridan Aug 29 '22 at 19:32
  • Thanks for your suggestion! – che00c Aug 29 '22 at 21:21

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Your argument is that if $a'>a$ and $Y<0$, then $(a-a')Y>0$, therefore $$X+aY = X+(a-a')Y+a'Y > X+a'Y.$$ So you're well on the way to proving the following:

Claim: If $a'>a$ then the following set inclusion holds: $$\{X+a'Y>0, Y<0\} \subset \{X+aY>0, Y<0\},$$ and therefore $$P(X+a'Y>0, Y<0)\le P(X+aY>0, Y<0).$$

Now divide through by $P(Y<0)$ to obtain the conclusion.

grand_chat
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  • Thanks for the detailed steps. May I ask if there is also a simple way to show that $P(X+Y>0|Y<0)$ is increasing or decreasing in the correlation coefficient $\rho$? – che00c Aug 29 '22 at 21:19
  • You can obtain a closed form in terms of $\rho$ from this result: https://math.stackexchange.com/a/3037330/215011 . You know $P(Y<0)=1/2$, so it remains to compute $P(X+Y>0, Y<0)=P(X+Y>0)-P(X+Y>0, Y>0)=1/2 - P(X+Y>0, Y>0)$. Now observe that $X+Y$ and $Y$ have a joint bivariate normal distribution. After you compute the correlation between $X+Y$ and $Y$, you can apply the result to obtain $P(X+Y>0, Y>0)$. – grand_chat Aug 29 '22 at 22:32