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Supplement 3:

Supplement 3:. Expectations, Conditional Expectations, Law of Iterated Expectations.

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Supplement 3:

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  1. Supplement 3: Expectations, Conditional Expectations, Law of Iterated Expectations *The ppt is a joint effort: Ms Jingwen zHANG discussed the law of iterated expectations with Dr. Ka-fu Wong on 1 March 2007; Ka-fu explained the concept with an example; Jingwen drafted the ppt; Ka-fu revised it. Use it at your own risks. Comments, if any, should be sent to kafuwong@econ.hku.hk.

  2. Joint, conditional and marginal probability, when there are two random variables. • Let (X,Y) be two random variables with a joint probability of P(X,Y). • From the joint probability, we can compute • the marginal probability PX(X) and PY(Y). PX(X=k) = ∑Y P(X=k,Y); PY(Y=k) = ∑X P(X,Y=k) • the conditional probability Px|y(X) and Py|x(Y).PX|Y=k(X) = P(X,Y=k)/ PY(Y=k) ; PY|X=k(Y) = P(X=k,Y)/ PX(X=k) • Unconditional expectation E(Y) =∑Y ∑X Y*P(X,Y) • Conditional expectations: E(Y|X) and E(X|Y) E(Y|X) = ∑Y Y*PY|X(Y)E(X|Y) = ∑XX*PX|Y(X)

  3. Conditional expectations are random variables The conditional expectation can take different values. The probability of the conditional expectation taking a particular value.

  4. Expectation of conditional expectations E[E(Y|X)] = ∑X{E(Y|X)*PX(X)} = ∑X{[∑Y Y*Py|x(Y)]*PX(X)} since E(Y|X) = ∑y Y*Py|x(Y)= ∑X{[∑Y Y* P(X,Y)/ PX(X)]*PX(X)} since PY|X=k(Y) = P(X=k,Y)/ PX(X=k) = ∑X ∑Y Y* P(X,Y) = E(Y)

  5. Let X and Y be random variables. X = education attainment (1=degree holder, 0 without degree)Y = income (only three groups for simplicity; 1, 2, 3 thousands)

  6. Let X and Y be random variables. X = education attainment (1=degree holder, 0 without degree)Y = income (only three groups for simplicity; 1, 2, 3) Expected education of a person randomly drawn from the income group Y=1. Expected income of a person randomly drawn from the education group X=1.

  7. Joint, conditional and marginal probability, when there arethree random variables. • Let (X,Y, Z) be three random variables with a joint probability of P(X,Y, Z). • From the joint probability, we can compute • The marginal probability PX(X), PY(Y), PZ(Z). PX(X=k) = ∑Y ∑Z P(X=k,Y, Z); PY(Y=k) = ∑X ∑Z P(X,Y=k, Z)PZ(Z=k) = ∑X ∑Y P(X,Y, Z=k) • The bivariate distribution of any pair of the three random variablesPXY(X,Y), PXZ(X,Z), PYZ(Y,Z) • The conditional probability PX|Y,Z(X), PY|X,Z(Y), PZ|X,Y(Z). PX|Y=k,Z=m(X) = P(X,Y=k,Z=m)/ PYZ(Y=k,Z=m) ; PY|X=k,Z=m(Y) = P(X=k,Y,Z=m)/ PXZ(X=k,Z=m) PZ|X=k,Y=m(Z) = P(X=k,Y=m,Z)/ PXY(X=k,Y=m) • The conditional bivariate probability PXY|Z(X,Y), PYZ|X(Y,Z), PXZ|Y(X,Z).PXY|Z=m(X,Y) = P(X,Y,Z=m)/ PZ(Z=m)

  8. Joint, conditional and marginal probability, when there is only three random variables. • Let (X,Y, Z) be three random variables with a joint probability of P(X,Y, Z). • Unconditional expectation E(Y) =∑y ∑x ∑Z Y*P(X,Y,Z) • Conditional expectations: E(Y|X,Z) and E(X|Y,Z) E(Y|X,Z) = ∑Y Y*Py|x,Z(Y)E(X|Y,Z) = ∑XX*PX|Y,Z(X)

  9. Conditional expectations are random variables. E[E(Y|X,Z)|Z] = ∑X{E(Y|X,Z)*PX|Z(X)} = … = E(Y|Z) E[E(Y|Z)]=E(Y)

  10. X, Y and Z random variables. X = education attainment (1=degree holder, 0 without degree)Y = income (only three groups for simplicity; 1, 2, 3 thousands)Z = gender (1= male, 2=female) E(Y|X=1,Z=1) The expected income of a person randomly drawn from the group of male degree holders. E(Y|X=1,Z=2) The expected income of a person randomly drawn from the group of female degree holders. E(Y|X=1,Z=1) - E(Y|X=1,Z=2) >0 and E(Y|X=0,Z=1) - E(Y|X=0,Z=2) >0 For the same education attainment, male’s expected income of a person is higher than female’s. Sometimes, it is interpreted as a piece of evidence of sex discrimination against female.

  11. X, Y and Z random variables. X = education attainment (1=degree holder, 0 without degree)Y = income (only three groups for simplicity; 1, 2, 3 thousands)Z = gender (1= male, 2=female) E(Y|X=1,Z=1) The expected income of a person randomly drawn from the group of male degree holders. E(Y|X=0,Z=1) The expected income of a person randomly drawn from the group of male non-degree holders. E(Y|X=1,Z=1) - E(Y|X=0,Z=1) >0 and E(Y|X=1,Z=2) - E(Y|X=0,Z=2) >0 The return to education/schooling is positive. Education/schooling thus helps to accumulate the “human capital” embodied in us.

  12. X, Y and Z random variables. X = education attainment (1=degree holder, 0 without degree)Y = income (only three groups for simplicity; 1, 2, 3 thousands)Z = gender (1= male, 2=female) E(Y | X=1,Z=1) The expected income of a person randomly drawn from the group of male degree holders. E(Y | Z=1) The expected income of a person randomly drawn from the group of male, regardless of education attainment. E(Y|Z=1)=E[E(Y|X,Z)|Z=1] = 1.5*0.4+2.5*0.6 E(Y|Z=2)=E[E(Y|X,Z)|Z=2] = 1.3*0.6+2.1*0.4 E(Y|Z)=E[E(Y|X,Z)|Z]

  13. Law of iterated expectations • Given E(e|X) = 0, find E(eX). E(eX) = E[E(eX|X)] =E[E(e|X)X] =E[0*X] =0 E(eX) =P(X=1) E(eX|X=1) + P(X=2) E(eX|X=2) =0.4*E(e|X=1)*1+0.6*E(e|X=2)*2 =0.4*0 + 0.6*0 = 0 E(eX) =0.1*(-1) + 0.1*0 + 0.1*1 + 0.2*(-4) + 0.2*(-2) + 0.3*(4) =0 + 0.

  14. Law of iterated expectations • Given E(Y|X,Z) = 0, E(XY) = 2, E(Z) =4, find E(XYZ). E(XYZ) = E[E(YXZ|X,Z)] = E[E(Y|X,Z) X Z] = E[ 0* X Z] = 0.

  15. Definition: Estimator • Estimator is a formula or a rule that takes a set of data and returns an estimate of the population quantity (also known as population parameter) we are interested in. θ(x1,x2,...,xn)

  16. Example: An estimator for the population mean • If we are interested in the population mean, a very intuitive estimator of the population mean based on a sample (x1,x2,...,xn) is θ(x1,x2,...,xn)= (x1+x2+...+xn)/n • Suppose someone suggest θ(x1,x2,...,xn)= (x1+x2+...+xn+1)/n

  17. Desired property: unbiased. That is, on average, the estimator correctly estimates the population mean. • θ(x1,x2,...,xn)= (x1+x2+...+xn)/n E [θ(x1,x2,...,xn)]= E [(x1+x2+...+xn)/n] = (1/n)*{E(x1) +E(x2)+...+E(xn)}= (1/n)*n*E(x)= E(x) • θ(x1,x2,...,xn)= (x1+x2+...+xn+1)/n E [θ(x1,x2,...,xn)]= E [(x1+x2+...+xn+1)/n] = (1/n)*{E(x1) +E(x2)+...+E(xn) + 1}= (1/n)*{n*E(x) + 1}= E(x) + 1/n Approaches zero as sample size increases. i.e., the estimator is asymptotically unbiased.

  18. Supplement 3: Expectations, Conditional Expectations, Law of Iterated Expectations - END -

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