# Probability of standard normal greater than another standard normal conditional on truncation

$$X,Y \sim N(0,1)$$ independently. Find $$P(Y > 3X | Y > 0)$$. My attempt:

$$\begin{eqnarray*} P(Y > 3X | Y > 0) &=& P(X < Y/3 | Y > 0) \\ &=& E(1(X < Y/3)| Y > 0) \\ &=& E\big(E(1(X < y/3) | Y=y)| Y > 0\big) \\ &=& E\big(P(X < y/3) | Y>0 \big) \quad \because X \perp Y \\ &=& E(\Phi(Y/3) | Y>0) \end{eqnarray*}$$

Is there any way to get a closed form expression for this? I know how to get $$E(\Phi(aY + b))$$ but getting the expectation over a truncated normal distribution seems elusive.

• Hint: draw a picture of this event and bear in mind that $(X,Y)$ is rotationally symmetric around the origin. – whuber Sep 13 '20 at 0:17
• I believe this one is already answered on site. – Glen_b Sep 13 '20 at 8:51

Graphical comment: @whuber's picture.

## a million points for 2-3 digits of accuracy
set.seed(319)
X = rnorm(10^6); Y = rnorm(10^6)
mean(Y > 3*X & Y > 0)
[1] 0.300751
mean(Y > 3*X & Y > 0)/mean(Y > 0)
[1] 0.6027038

## Fewer points for figure
x = X[1:50000];  y= Y[1:50000]
plot(x,y, pch=".")
abline(a = 0, b = 3, col="green", lwd=2)
abline(h=0, col="red", lwd=2)


• Nice work--but because the question focuses on the condition $Y\gt 0,$ don't you think it would be better to plot only the upper half plane of this figure? – whuber Sep 14 '20 at 14:31
• @whuber. Or at least given a note to look there, which you have now done. (I guess I showed the whole thing because you mentioned rotation about the origin.) – BruceET Sep 14 '20 at 18:00