# Conditional probability that standard gaussian random variable is larger than another

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.

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

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.

• Thanks. Is there a way to evaluate the more general case as well? I'm guessing symmetry no longer holds once $S$ is introduced. – tankerjeel Jun 5 '19 at 20:09
• 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 $0$. – tankerjeel Jun 5 '19 at 23:57