September 23rd, 2003, 4:19 pm
Assume the world is normal and all we know is correlation…If (X,Y) is N(0,0,sX,sY,r) and N1, N2 are iid N(0,1) then we have the (distribution) equality(X,Y)=( sX*N1 , sY(r*N1 + sqrt(1-r^2)*N2) )If |r|<1, thenP(Y>0 | X>0) = P( sY(r*N1 + sqrt(1-r^2)*N2)>0 | sX*N1>0)= P( r*N1 + sqrt(1-r^2)*N2>0 | N1>0)= P( N2 > -N1* r/sqrt(1-r^2) | N1>0)=f(r)and solving for f(r)=r numerically could give r around 0.8 while the true value probably involves pi. >if the correl is 0, you cannot say anythingWell, if r=0 then P(Y>0 | X>0) = P(Y>0)=1/2SPAAGGs original statement isn’t that wrong. If r is 0.8 then P(Y>0|X>0) is approximately equal to r. Not an answer to QuietStorms question however, who should look at f(r) closer or a generalization if the means are to be non-zero. I don’t have time to look closer at f(r) but I expect that it has a similar curvature as arcsine, shifted up by 1/2. Prove me wrong.p.s. how beautiful life would be if Wilmott.com had Tex…