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Say there are two standard Gaussian random variables $X$ and $Y$. I am trying to evaluate the probability that the larger of the two is selected, given that it is known whether $X$ is positive or negative (the strategy is selecting $X$ if $X$ is positive and selecting $Y$ if $X$ is negative). In equation form this is $$Pr(X-Y>0|X>0) + Pr(X-Y<0|X<0)$$

How can this expression be evaluated? Numerically it appears to be $\frac{3}{4}$, and intuitively this makes sense.

I am also interested in this probability in the more general case, where the strategy involves selecting $X$ if $X-S>0$ and $Y$ otherwise, where $S$ is another independent Gaussian random variable.

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  • $\begingroup$ I don't understand your question. Did you mean $\text{Pr}(X-Y>0|X>0)\text{Pr}(X>0)+\text{Pr}(X-Y<0|X<0)\text{Pr}(X<0)$? $\endgroup$ – Angela Pretorius Jun 5 '19 at 4:32
  • $\begingroup$ Yes, that's right! $\endgroup$ – tankerjeel Jun 5 '19 at 20:07
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The following assumes $X$ and $Y$ are independent.

As Angela Richardson pointed out, you probably actually want to compute $$P(X-Y > 0 \mid X > 0) P(X>0) + P(X-Y <0 \mid X< 0) P(X<0)$$ which is $3/4$. (The quantity in your post is not $3/4$.)


For the first term, it suffices to compute $P(X-Y > 0, X>0)$. (Why?)

Consider the region of the plane that contains $(x,y)$ pairs satisfying $x-y>0$ and $x>0$. Then use rotational symmetry of the vector $(X,Y)$ to compute the probability.

The other term can be handled similarly.

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  • $\begingroup$ Thanks. Is there a way to evaluate the more general case as well? I'm guessing symmetry no longer holds once $S$ is introduced. $\endgroup$ – tankerjeel Jun 5 '19 at 20:09
  • $\begingroup$ In case anyone is curious, the stackexchange post in this link gives the solution as to how to compute the more general case. When $S$ is the standard normal, the probability turns out to be exactly $\frac{2}{3}$, and approaches $\frac{3}{4}$ as the variance of $S$ approaches $0$. $\endgroup$ – tankerjeel Jun 5 '19 at 23:57

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