A pure math problemif x, y are jointly normal (correlated), what is the expectation of exp(x+y) conditional upon x>a, y>b ie. E[exp(x+y)|(x>a,y>b)]?

Assuming that X is a multivariate normal with mean vector m and covariance matrix C, I thinkE[ exp( q' X ) 1(X > x) ] = exp( q' C q / 2 + q' m ) P(X > x - C q)where "X > a" is a cone defined by entry-wise inequalities. In your case, q = (1,1)' and P(X > a)becomes a bivariate normal prob.

I feel that I haven't quite finished answering the question, so I'm adding a line or two.E[ Y 1(X > a) | X > a ] P(X > a) + E[ Y 1(X > a) | X <= a ] P(X <= a) = E[ Y 1(X > a) ].The left is E[ Y | X > a ] P(X > a) and the right is as in my previous posting.

You could also do it numerically via monte carlo if a and b are numerically defined. All the tools that you would need are readily available. You basically generate independent normal variables. You know that you can make them corelated via:normal2 = normal1*rho + (1-rho)*z1where normal1 and z1 are independent (use Box-Muller for instance). Then reject those samles for which either z1>a or normal2 >b. Using the rest, evaluate exp(z1+normal2). You can use all sorts of variance reduction methods, like antithetic or control variates. If you must find an analytical solution then you can use the laws of conditional expectation.You basically want [ \integral_{a^infity} \integral_{b^infity} exp(x+y) . f(x,y) dy dx ] / P(x>a, x>b)f(x,y) is the joint distribution of x and y, where they are correlated.I'm sure that you can use some properties of bi-variate normal distribution to somehow evaluate this. Regards,Sam

Hi HA,I am trying to understand what you have tried to do and I hope you will forgive my ignorance...The bit that I do not agree with is your term:E[ exp( q' X ) 1(X > x) ] Are you implying that this is the same asE [ exp(q' X) | (X>x) ]?If so then I SUSPECT that this does not hold. You have calculated the expectation of a new rv whereas what is required is the conditional expectation. Maybe you could ellaborate a little bit more?Thanks,Sam

Sam,I was trying to say E[ Y | X > x ] = E[ Y 1(X > x) ] / P(X > x) provided that P(X > x) is positive.(See my second posting. Sorry if I made you confused.)

Cool... I think this is consistent to what I was saying.Regards,Sam

Thanks HA and Sam,That helps!