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Derive mgf bivariate normal distribution

  1. Apr 25, 2015 #1
    Does anyone know the proof of joint moment generating functions for bivariate normal distributions?
    M_x,y (s,t)= E(e^(xs+yt))
     
  2. jcsd
  3. Apr 25, 2015 #2

    mathman

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    Expand the exponential in a power series. E(each term) is that moment.
     
  4. Apr 25, 2015 #3
    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-ρ2xσy)) exp [ [-1/2(1-ρ2)] [(x-μxx)2 + (y-μyy)2 - 2ρ(x-μxx)(y-μyy)}
     
  5. Apr 26, 2015 #4

    mathman

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    No. [itex]e^{(xs+yt)}[/itex] must be under the integral sign.
     
  6. Apr 26, 2015 #5
    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.
     
  7. Apr 27, 2015 #6

    mathman

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    That will depend on what [itex]f_{XY}(x,y)[/itex] is. If it is bivariate normal, then you will get its moment generating function.
     
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