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x ~ N(a,b), y ~ N(c,d). x and y are independent Random Variables.

I am trying to compute this expectation:-

E[f(x,y)]..... x Where f(x,y) = y if (x-y)>0 f(x,y) = x, if otherwise

The way I was thinking is

E[f(x,y)] = yP((x-y)>0) + x(1-P((x-y)>0))

P((x-y)>0 is something I can compute from the distributions of x and y, but E[f(x,y)] is not giving a closed form solution it's still having randomness as I still don't know what x and y can be.

Any help is appreciated.

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  • $\begingroup$ Hint: the problem might become easier or the solution more obvious if you were to re-express it in terms of $U=X/b-Y/d$ and $V=X/b+Y/d$, because $U$ and $V$ are uncorrelated. You will have to assume $(X,Y)$ have a jointly bivariate Normal distribution, because that will imply $U$ and $V$ are independent--and therein lies the possibility of simplification. Please check your question for typos: for instance, "x ~ N(c,d)" most likely is intended to mean "$y \sim N(c,d)$". $\endgroup$
    – whuber ♦
    Commented Jul 16, 2017 at 17:15
  • $\begingroup$ @whuber - Thanks, but can you elaborate a little more with some equations if possible. I have been away from direct Mathematics for a long time so I am bit rusty with my equations. $\endgroup$
    – vicky113
    Commented Jul 16, 2017 at 17:18

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