The discussion revolves around the challenge of calculating the expected value E(XY) for a bivariate normal distribution. Participants explore methods for evaluating the integral involved in this calculation, including potential transformations and the implications of the bivariate normal density function.
Participants do not reach a consensus on the best method to evaluate the integral for E(XY), and multiple approaches are discussed without resolution.
Participants express uncertainty regarding the specific form of the integral they are trying to evaluate, and the discussion includes various assumptions about the parameters of the bivariate normal distribution.
I think that this would be your integral: ∫ ∫ xy f(x,y) dy dx; that is, an iterated integral. Because of the presence of ##e^{-x^2 - y^2}## (I'm really simplifying here), a standard trick is to change the integral to polar coordinates instead of Cartesian coordinates, although the presence of xy in the integrand might make the integral doable without changing to polar coordinates.Ashton said:My main difficulty is finding E(XY), where if we find it by integrating, we get ∫xy f(x,y) dy dx where f(x,y) is the bivariate normal density function. This integral is a nightmare... are there any tricks to evaluating this integral, or finding E(XY) for that matter?
It's hard to say without seeing the integral. The trick I suggested was the standard one for evaluating ##\int_{-\infty}^{\infty} e^{-x^2}dx##. In Cartesian form, this integral can't be done, but by looking at the related iterated integral, and converting to polar form, the integration is straightforward.Ashton said:I apologize, that was indeed what I meant by my integral. There is also an e^(-xy) term in the bivariate normal density, do you think it's still possible without transforming the integral to polar coordinates?