Knowledge Innovation Excellence
LILONGWE UNIVERSITY OF AGRICULTURE AND NATURAL
RESOURCES
CITY-CAMPUS
TO
:
DR. HENRY KAMKWAMBA
FROM
:
FELIX MOYO
ID NUMBER
:
200200102
PROGRAME
:
BABM
DEPARTMENT :
APPLIED ECONOMICS DEPARTMENT
TITLE
:
ACONOMETRICS 1 ASSIGNMENT
COURSE
:
ACONOMETRICS 1
COURSE CODE :
AAE 313
DUE DATE
26th September, 2022.
:
QEUSTION 1
I.
When 𝑆𝐴𝑇 = 800, the expected GPA will be;
E[GPA|SAT = 800] = .70 + .002×800 = 2.3
When SAT= 1400, the expected GPA will be;
E[GPA|SAT = 1,400] = .70 + .002×1,400 = 3.5
 This shows that the SAT score is correlated to GPA score. The higher the
SAT score, the higher is the GPA.
II.
If the average 𝑆𝐴𝑇 in the university is 1100, then the average 𝐺𝑃𝐴 is;
Using Law of iterated expectations;
EGPA = E(E[GPA|SAT]) = E(.70 + .002 SAT)
= .70 + .002 ESAT
= .70 + .002×1,100.
= 2.9
III.
If the SAT score is 1100, then the student will have the same GPA like the average GPA
because they are both related to E[𝐺𝑃𝐴|𝑆𝐴𝑇] = 0.70 + 0.002𝑆𝐴𝑇.
QUESTION 2
I.
Proof
LHS
= ∑𝑛𝑖=1 xi - ∑𝑛𝑖=1 x̅
𝑛
= ∑𝑖=1 xi - n x̅
𝑛
= ∑𝑖=1 xi – n(
𝑛
= ∑𝑖=1 xi -
1
∑𝑛𝑖=1 xi)
𝑛
∑𝑛𝑖=1 xi
=0
II.
∑𝑛𝑖=1(xi - x̅ )2= ∑𝑛𝑖=1 xi (xi - x̅ )
Proof
LHS
∑𝑛𝑖=1[(xi - x̅ ) (xi - x̅ )]
=∑𝑛𝑖=1( xi2 - x̅ xi - x̅ xi + x̅ 2)
=∑𝑛𝑖=1( xi2 - 2 x̅ xi + x̅ 2)
=∑𝑛
̅ ∑𝑛𝑖=1 xi + ∑𝑛𝑖=1 𝑥̅2
𝑖=1 𝑥𝑖2 – 2x
=∑𝑛𝑖=1 𝑥𝑖2 -2n x̅ 2 +nx̅ 2
=∑𝑛𝑖=1 𝑥𝑖2 - n x̅ 2
=∑𝑛𝑖=1 𝑥𝑖2 - n x̅ x̅
1
=∑𝑛𝑖=1 𝑥𝑖2 - x̅ [n( ∑𝑛𝑖=1 xi)]
𝑛
𝑛
𝑛
=∑𝑖=1 𝑥𝑖2 - x̅ ∑𝑖=1 xi
𝑛
=∑𝑖=1( xi2 - x̅ xi)
= ∑𝑛𝑖=1[ xi(xi - x̅ )]
Proved
III. ∑𝑛𝑖=1( xi - x̅ ) (Yi - 𝑌̅)= ∑𝑛𝑖=1 xi(Yi - 𝑌̅)= ∑𝑛𝑖=1 Yi(xi - x̅ )
Proof
LHS
= ∑𝑛𝑖=1( 𝑋𝑖𝑌𝑖 − 𝑋𝑖 𝑌̅ − x̅Yi + x̅ 𝑌̅)
=∑𝑛𝑖=1( 𝑋𝑖𝑌𝑖 - 𝑋𝑖 ∑𝑛𝑖=1 𝑦𝑖 - Yi ∑𝑛𝑖=1 𝑋𝑖 + x̅ 𝑌̅)
=∑𝑛𝑖=1 𝑋𝑖𝑌𝑖 - ∑𝑛𝑖=1 𝑋𝑖 ∑𝑛𝑖=1 𝑌𝑖 - ∑𝑛𝑖=1 𝑋𝑖 ∑𝑛𝑖=1 𝑌𝑖 +∑𝑛𝑖=1 x̅ 𝑌̅
=∑𝑛𝑖=1 𝑋𝑖𝑌𝑖 - 2∑𝑛𝑖=1 𝑋𝑖 ∑𝑛𝑖=1 𝑌𝑖 + ∑𝑛𝑖=1 𝑋𝑖 ∑𝑛𝑖=1 𝑌𝑖
=∑𝑛𝑖=1 𝑋𝑖𝑌𝑖 - ∑𝑛𝑖=1 𝑋𝑖 ∑𝑛𝑖=1 𝑌𝑖
Taking ∑𝑛𝑖=1 𝑌𝑖=𝑌̅
=∑𝑛𝑖=1 𝑋𝑖𝑌𝑖 - 𝑌̅ ∑𝑛𝑖=1 𝑋𝑖
=∑𝑛𝑖=1 𝑋𝑖 (Yi - 𝑌̅)
𝑛
Taking ∑𝑖=1 𝑋𝑖 = x
̅
𝑛
=∑𝑖=1 𝑌𝑖 (xi - x̅ )
QUESTION 3
I.
E[(𝑋 − E[𝑋])] = 0
Proof
LHS
E(X) =
µ
Therefore; E(X - µ)
=E(X) - µ
=µ
=0
II.
-µ
Var(𝑋) = E[𝑋(𝑋 − E[𝑋])]
Proof
RHS
=E[X(X - µ)]
=E [(X2 – X. E(𝑋)]
= E (𝑋)].E(X) - E(𝑋)]. E(𝑋)]
= µ. µ - µ. µ= µ2- µ2
=0
Var(x) = 0
III.
Cov(𝑋, 𝑌) = E[𝑋(𝑌 − E[𝑌])] = E[(𝑋 − E[𝑋])𝑌]
Proof
E[X(Y – E(Y)] = E[XY – X.E(Y)]
=E(XY) – E(X).E(Y)
But, E(XY)= E(X)E(Y)
Therefore, E(X)E(Y) – E(X)E(Y)= 0
E[(X –E(X)Y)= E(YX – YE(X)
= E(YX) – E(Y)E(X)
=E(Y)E(X) – E(Y)E(X)
=0