The discussion revolves around deriving the joint moment generating function (mgf) for bivariate normal distributions. Participants explore various methods and steps involved in the derivation process, including the use of power series expansions and integration techniques.
Participants do not reach a consensus on the best method for deriving the mgf, and there are competing views on the placement of terms in the integral and the overall approach to the derivation.
Some participants express uncertainty about specific steps in the derivation, particularly regarding the integration and the form of the joint probability density function.
I'm actually suppose to do it likemathman said:Expand the exponential in a power series. E(each term) is that moment.
great ideamathman said:Expand the exponential in a power series. E(each term) is that moment.
No. e^{(xs+yt)} must be under the integral sign.Lewis7879 said:I'm actually suppose to do it like
great idea
it is possible to do it this way ?
its a very long step.
M(s,t)= e(xs+yt) ∫∫ fXY (x,y) dxdy
= e(xs+yt) ∫∫ (1/(2π√(1-ρ2)σxσy)) exp [ [-1/2(1-ρ2)] [(x-μx/σx)2 + (y-μy/σy)2 - 2ρ(x-μx/σx)(y-μy/σy)}
I'm actually suppose to do it likemathman said:Expand the exponential in a power series. E(each term) is that moment.
yes I know that but will I get the solution of mgf of bivariate normal distribution ? I'm stuck current trying to derive it.mathman said:No. e^{(xs+yt)} must be under the integral sign.
That will depend on what f_{XY}(x,y) is. If it is bivariate normal, then you will get its moment generating function.Lewis7879 said:I'm actually suppose to do it like
M(s,t)=
yes I know that but will I get the solution of mgf of bivariate normal distribution ? I'm stuck current trying to derive it.