Question

Law of iterated expectations (LIE) The law of iterated expectations says that the unconditional expectation can be written as the unconditional expectation of the CEF: E [E[Y|X]] =E[Y] where the outer expectation uses the distribution of X. ► Proof E[E[Y\X]] = = y;|X = x;)); P(X = x;) =E[Y]X=x;] (P(Y = yi|X = x;)P(X = x;)) yi =P(Y=y,X=x;) (Y = yi, X = x;lyi S P j=1 i=1 Law of iterated expectations (LIE) Proof: E[E[Y]X]] = PY = yi,X = x;)yi DPCY = yi, X = x;) Yi =P(Y=y:) = P(Y = y;)yi = E[Y]

Can someone explain the proof step by step and how everything just simplifies to Expectation of Y

Answer 1

1 = mx | We can assume, FL assume, E[yıx] = mx by definition, E [1/x] = Ï Yi PCY=y:] X= zņ), which is a function of tj. where Mx is a function of x for all j=1, 2, ... :: E(E(\x)) = E("x) = { x: P(=x;) [= x= m-&m ] [94 is assumed, x takes the value, . x1, My... xem ] (3 4: (+4c\*= 2;)) P(x+28) . [ More is defined above] (*= z;). P(x=2;)]*: [ chameling the order of [Note {{ x:4; LE 2:) *; = {(343):] P(**i,x=2;)] i [PCA 18)= PLACE? ] = F(*). *[ Theored the probability Plz-4.)EF(=+*+,x=x;)] = E() [By Definition]