Suppose there is another n × 1 random vector which is a linear transformation of vector ε:ξ = C + Dε,

INSTRUCTIONS TO CANDIDATES

ANSWER ALL QUESTIONS

1. Simple Linear Model

Consider the following simple linear model:

yi = α + xiJ β + εi, for i = 1, ..., n (1)

where y , α, ε are scalars, x= xi,1, β = β1

Recall one of the assumptions on the OLS data generating process is: Assumption 3: the error terms are exogenous to the independent vari- ables:

E[ε|x1, ..., xn] = 0

(a) Write (1) in a matrix form:

y = Xγ + ε, α where γ = β and give y, X, ε and their dimensions.β2

Show that

E[ε] = 0, E[x1Jε] = 0 E[x2Jε] = 0

Hint: Law of Iterated Expectation

(c) Unbiasedness: In statistics, an estimator γˆ of parameter γ is

unbiased if E[γˆ] = γ. Prove the OLS estimator is unbiased.

Hint: Use Assumption 3 and Law of Iterated Expectation.

(d) Suppose there is a random variable z. Its realizations are observed as {zi}, i = 1, ..., n. The sample average of z is defined as z¯ : Σnzi. As an approximation, let’s assume here that z¯ = E[z]when n > N . Use the result from (b) to show that

2. Vector Differentiation

Let f (b) be a function of a random vector b. For questions (a) to (c), we have:

A = 2 1 , b = b13 2∂f (b)

(a) Let f (b) = Ab. Calculate ∂b b2

(b) Let f (b) = bJAb. Calculate ∂f (b).∂b

(c) Let C be a 2 × 2 constant matrix, α be a constant and f (b) = (bJCb)α.Calculate ∂f (b) ∂b

Hint: Read A.8 of Appendix A of the Greene’s book and use the Chain Rule.

3. Multivariate Normal Distribution

An n × 1 random vector ε follows a multivariate normal distribution:ε ∼ N (0, σ2In), where In represents an n × n identity matrix. Suppose there is another n × 1 random vector which is a linear transformation of vector ε:ξ = C + Dε, where C is an n × 1 constant vector and D is an n × n constant orthogonal matrix with DJD = In.

(a) Give E [εi] , E [ε2] and E [εiεj] for i, j = 1, ..., n and i j.

(b) Write down the distribution of vector ξ.

Hint: The variance-covariance matrix of a m × 1 vector x is given by: V [x] = E[(x − E[x])(x − E[x])J] = E[xxJ] − E[x]E[x]J

(c) Calculate E[ξJξ].

Hint: If z is a univariate random variable and z ∼ N (0, 1), then

E(z2) = V (z) = 1.

Attachments:

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