The joint moment generating function (mgf) for bivariate normal distributions is defined as MX,Y(s,t) = E(exs + yt). The correct approach involves expanding the exponential in a power series and integrating under the integral sign. The bivariate normal density function fXY(x,y) is given by (1/(2π√(1-ρ2σxσy))) exp[-(1/2(1-ρ2))((x-μx/σx)2 + (y-μy/σy)2 - 2ρ(x-μx/σx)(y-μy/σy))]. This method leads to the derivation of the mgf for bivariate normal distributions.
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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.