Expectation of an exponentiated sum of normals

eiriamjh · December 11th, 2003, 5:51 pm

I bet this will sound trivial to most people here, but I am trying to compute E( exp(aX + bY) ), where (X, Y) has a bivariate standard normal distribution with correlation rho.I laid down the integral and naively tried to do an affine change of variables, but all I got was a headache... Please help!Thanks in advance for replies,e.PS: and in the first place, isn't it the case that aX + bY should follow a normal distribution with mean 0 and variance a^2 + 2abrho + b^2?

kr · December 11th, 2003, 6:16 pm

do the one-var case first to see what the game ise^ax e^(-x^2/2) = e^(-1/2 . (x^2 - 2a + a^2 - a^2)) = e^(a^2/2) . e^(-1/2 . (x - a)^2)and since the integral is over all x's, you can dump the variable shift.

eiriamjh · December 11th, 2003, 6:27 pm

krthe one-variable case is indeed trivialsince you are so good in probability, I am convinced it will only take you a minute before you find the answer for the two-variable case... many thanks in advance for your next replye.

Yeren · December 11th, 2003, 6:35 pm

Or in a different way:Take two correlated BM B(t) and W(t) such thatdB(t) . dW(t) = rhoand conside the process:f(B, W) = exp( a*B(t) + b*W(t) )N apply the Ito's lemma and then take expectation on df(B, W), you will getE( exp(a*B(1) + b*W(1) )= exp[ (a^2 + 2 * rho * a * b + b^2 ) * t / 2 ]Please note that you can use this approach when the function a and b depend on the time t (but deterministic).Y.

kr · December 11th, 2003, 7:02 pm

ok, so no big deal in the multivariate caseyou have exp(L.x) . exp(-1/2 . (Mx) . (Mx)^t)where M is a matrix which generates the corrs and Lx is a linear form. M is orthonormal, so change vars so that you haveexp(L.M^(-1)x) . exp(-1/2 xx^t)and now that the thing is orthogonalized, you can complete the square one at a time (i.e. LM^(-1).x is still a linear form with coefficients which will turn into exp(a^2/2) factors as before

eiriamjh · December 12th, 2003, 8:52 am

Yeren, kr,Thanks for your replies.I like Yeren's trick better, but presumably kr's methodology yields the same resultswhich also answers my question: if (X,Y) is binormal, it is a gaussian vector (which I guess was kind of expected)... I presume the converse is not true, but I will leave this to you guys...txe.

Nonius · December 13th, 2003, 10:17 am

QuoteOriginally posted by: eiriamjhYeren, kr,Thanks for your replies.I like Yeren's trick better, but presumably kr's methodology yields the same resultswhich also answers my question: if (X,Y) is binormal, it is a gaussian vector (which I guess was kind of expected)... I presume the converse is not true, but I will leave this to you guys...txe.your problem has the same answer as the one dim because it really IS a one dim problem. you just need to compute the mean and variance of the linear combination and use it in the one dim formula.