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If X and Y are independent, then E(es(XY )) = E(esXesY) = E(esX)E(esY), and we conclude that the mgf of an independent sum is the product of the individual mgf’s Sometimes to stress the particular rv X, we write M X(s) Then the above independence property can be concisely expressed as M XY (s) = M X(s)M Y (s), when X and Y are independent. C x g j X Ƀv l ^ E u lj 08/7/16 C x g j X 7 ̗ W lj. C x g j X Ƀv l ^ E u lj 08/7/16 C x g j X 7 ̗ W lj.

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To Z and Y we obtain EE(ZjY)= E(Z) which is the same as EE(g(X)jY)= E(g(X)) 2 This property may seem to be more general statement than (i) in Theorem 1 The proof above shows that in fact these are equivalent statements 3 E(XYjY)=YE(XjY) Proof E(XYjY =y)=E(yXjY =y)=yE(XjY =y)(because y is a constant) Hence, E(XYjY)=. N X } X ̑f ނ ̖ Ŏg f ޏW ł B N X } X ̃C X g ځ N X } X J h ɂ p ł ܂ N X } X ̃C X g f ނR q A 炵 A h A 񍐁A Ε ̂ E A Ȃǂɂǂ. Z g T r X p B e p X ܑ C x g X ` ` ̑ ` 哹 E B h E f B X v C ͌^ W I } v j O y ѐ , Z g T r X 󒍔͈ p B e p C x g X ܑ w ʉƒ , H s.

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L O R hCD W P b g back top. Nature voice e l c ` { c x e @ @ @ r y 𒆐s ɍ y ̃p ̎ 舵 y сa q o y. You don't have to expand this It's a trick question If you go all the way from (xa) down to (xz), you include a factor of (xx) which equals 0.

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