Roman C. answered 04/30/16
Tutor
5.0
(851)
Masters of Education Graduate with Mathematics Expertise
I will rewrite the conditions in clearer formatting:
(i) ε ∼ N(0 , σ2)
(ii) X ∼ N(μ , τ2)
(iii) ε ⊥ X
(iv) Y = α + βX + ε
Solution:
Using condition 4 we get:
E(Y|X) = E(α + βX + ε|X)
= α + E(βX|X) + E(ε) ← Linearity of expectation
= α + βX + E(ε) ← Taking out what's known
= α + βX + ε
Condition 3 says that the residual ε has the same distribution independent of X.
Finally we have
Y|X = E(Y|X) + ε ~ E(Y|X) + N(0 , σ2) = N(α + βX + ε , σ2)