Alastair Hall ECON61001: Autumn 2020 Econometric Methods
Solutions to Problem Set for Tutorial 1
1.(a) ∂y/∂x = β0,2, and so β0,2 is the partial derivative of y with respect to x.
1.(b) Using the Chain rule: ∂y/∂x = x−1β0,2. So β0,2 gives the multiple of the proportional change
in x by which y changes.
1.(c) Using the Chain rule: ∂y/∂x = yβ0,2 and so β0,2 is the proportionate change in y that is 100β0,2 is the percentage change in y; in this case β0,2 is sometimes referred as the semi- elasticity of y with respect to x.
1.(d) Using the Chain rule: ∂y/∂x = (y/x)β0,2, and so β0,2 is (x/y)(∂y/∂x), the elasticity of y with respect to x.
1.(e) ∂y/∂x = β0,2 + 2β0,3x, and so β0,2 is ∂y/∂x at x = 0.
2.(a) By definition h(θ) = pi=1 aiθi and so ∂h(θ)/∂θi = ai. Since ∂h(θ)/∂θi is the ith element of
∂h(θ)/∂θ the result then follows immediately.
2.(b) By definition g(θ) = pi=1 pj=1 Ai,jθiθj where Ai,j is the i − jth element of A. Therefore,
pp ∂g(θ)/∂θs = Ai,sθi + As,jθj
i=1 j=1 = θ′A·s + As·θ
where As· and A·s are the sth row and column respectively of A. Now A·s = Bs· where B = A′ , and so
∂g(θ)/∂θs = (As· + Bs· )θ (1) Since ∂g(θ)/∂θs is the sth element of ∂g(θ)/∂θ, it follows from (1) that ∂g(θ)/∂θ = (A+A′)θ.
2.(c) From (1) it follows that
∂2g(θ)/∂θs∂θl = As,l + Al,s (2) Since ∂2g(θ)/∂θs∂θl is the s − lth element of ∂2g(θ)/∂θ∂θ′ the desired result follows directly
from (2).
2.(d) If A is symmetric then A = A′ and so (b) becomes ∂g(θ)/∂θ = 2Aθ, and (c) becomes ∂2g(θ)/∂θ∂θ′ = 2A.
1
3.(i) Following the hint, we have c′X′Xc = b′b for b = Xc. By construction b′b = Ti=1 b2i where bi is the ith element of b, and so it follows immediately that c′X′Xc ≥ 0. However, we must show this inequality is strict in order for X′X to be pd. Note b = Xc = ki=1 xici where xi is the ith column of X and ci the ith element of c, and so b is a linear combination of the columns of X. rank(X) = k implies that the columns of X forms a linear independent set and so it must be that b ̸= 0 and hence b′b > 0. This proves c′X′Xc > 0 for any non-zero vector c, and therefore that X′X is pd.
3.(ii) If rank(X) < k then for some choices of c we have b = Xc ̸= 0 but for others b = 0. Therefore, c′X′Xc ≥ 0 for all c, and so X′X is positive semi-definite.
4.(i) The normal equations can be written as: X′(y − XβˆT) = X′e = 0. Substituting for X we
have:
X′e=ι′T e=ι′Te=0. X 2′ X 2′ e
Theseareasystemofk×1equations,thefirstofwhichisι′Te=0. Sinceι′Te=Tt=1et,it follows immediately that e ̄ = 0.
4.(ii) Frompart(i),itfollowsthatι′T(y−XβˆT)=0whichinturnimpliesT−1ι′T(y−XβˆT)=0
(asT >0)andso
T−1ι′Ty = T−1ι′TXβˆT (3) ̄ ̄
Since the left hand side of (3) is y ̄ and the right-hand side of (3) is yˆ, it follows that y ̄ = yˆ. 5. Prove A−1 = B by showing B satisfies AB = BA = I. Now if AB = I then
A1,1B1,1 + A1,2B2,1 = I (4) A1,1B1,2 + A1,2B2,2 = 0 (5) A2,1B1,1 + A2,2B2,1 = 0 (6) A2,1B1,2 + A2,2B2,2 = I (7)
and so we now verify that the stated formulae for Bi,j satisfy these equation. Substituting B1,2 = −A−1A1,2B2,2 into the left-hand side of (5) gives
1,1
A1,1(−A1,1A1,2B2,2) + A1,2B2,2 = −A1,2B2,2 + A1,2B2,2
=0
and so B1,2 = −A−1A1,2B2,2 satisfies (5). Substituting B2,1 = −A−1A2,1B1,1 into the left-
1,1 2,2 hand side of (6) gives
A2,1B1,1 + A2,2(−A2,2A2,1B1,1) = A2,1B1,1 − A2,1B1,1 =0
A2,1(−A−1A1,2B2,2) + A2,2B2,2 = I 1,1
2
and so B2,1 = −A−1A2,1B1,1 satisfies (6). Substituting B1,2 = −A−1A1,2B2,2 into the left-
2,2 hand side of (7) gives
1,1
which implies
and so B2,2 = (A2,2 − A2,1A−1A1,2)−1 satisfies (7). Finally substituting B2,1 = −A−1A2,1B21
(A2,2 − A2,1A−1A1,2)B2,2 = I 1,1
1,1 2,2 into the left-hand side of (4) gives
A1,1B1,1 + A1,2(−A2,2A2,1B1,1) + A2,2B2,2 = I (A1,1 − A1,2A−1A2,1)B1,1 = I
which implies
and so B1,1 = (A1,1 − A1,2A−1A2,1)−1 satisfies (4).
3
2,2
2,2
SinceAandBarenonsingular,AB=I ⇒ BAB=B ⇒ BA=BB−1 =I. Therefore
B = A−1.
Notice that if we begin with BA = I then it is possible to derive alternative – but equivalent
– representations for Bi,j in terms of Ai,j for i ̸= j. These are
B1,2 = −B1,1A1,2A−1 (8)
B2,1 = −B2,2A2,1A−1 (9) 1,1
2,2
(10)