Suggested Answers
Problem Set 4
Economics 206
Beomsoo Kim
Spring 2016
1.
Notice that x and y are linearly related – for every 1 unit increase in x, y falls by 1 unit.
It is easy to show that
y=20-2x. Looking at a graph of the points, x and y are perfectly negatively correlated – for any value of x one
knows exactly the value of y. Therefore, in this case, _ˆxy =-1.
2a.
Start with the definition of the correlation coefficient --- ρfc = σfc/(σfσc)
Note that Var(F) = E[(F - μf)2] = σ2f and looking at the value for C,
0.556E[F] = -17.78 + 0.556 μf. So C - μc = 0.556(F - μf)
E[C] = μc = E[-17.78 + 0.556F] = -17.78 +
Therefore: σfc = E[(F - μf)(C - μc)] = E[(F - μf)(0.556F - 0.556μf)] = 0.556 E[(F
- μf)2] = 0.556σ2f
Notice as well that Var(C) = E[(C - μc)2] = E[(0.556F - 0.556 μc)2] = E[(0.556)2 (F
0.5562 σ2f
Therefore: ρfc = σfc/(σfσc)
2b.
=
- μf)2] =0.5562 E[(F - μf)2] =
0.556 σ2f/(σf (0.556 σf )) = σ2f/σ2f = 1.
Start with what you know: ρiP = σiP/σiσp
What is ρkc = σkc/σkσc where K = 0.45P and C=2.54I?
The answer can be generated by re-writing σkc σk and σc as a function of P and I.
Note that:
E[K] = μk = E[0.45P] = 0.45E[P] = 0.45μp
E[C] = μc
= E[2.54I] ] 2.54E[I] = 2.54μi
σ2k = E[(K - μk)2] = E[(0.45P - 0.45μp)2] = E[0.452(P - μp)2] = 0.452 E[(P - μp)2] = 0.452 σ2p
And σk = 0.45σp
σ2c = E[(c - μc)2] = E[(2.54I - 2.54μi)2] = E[2.542(I - μi)2] = 2.542 E[(I - μi)2] = 2.542 σ2i
And σc = 2.54 σi
σik = E[(K - μk)(C -
μc)]
= E[(0.45P - 0.45μp)(2.54I - 2.54μi)] =
= 0.45(2.54) E[(P -
E[0.45(2.54)(P -
μp)(I - μi)]
μp)(I - μi)] = 0.45 (2.54) ρpi
So ρkc = σkc/σkσc = 0.45(2.54)σip/[2.54σi(0.45σp)] = σiP/σiσp = ρiP =0.5
3a.
T=M+V. If we write T as T=a + bM + cV, them a=0, b=1 and c=1. We know that μt = a + bμm + cμv, so μt =
μm + μf = 510 + 475 = 985. We are given the correlation coefficient ρmv , and since ρmv = σmv/(σmσv), σmv =
σmσv ρmv = 7500.5 6100.5 0.4 = 270.6.
We know that σ2t = b2σ2m + c2σ2v +2bcσmv = 750 + 610 + 2(270.6) = 1901.2, so σt = 43.6
3b.
G = 0.25M + 0.75F. If we write G as G=a + bM + cF, then a=0, b=0.25 and c=0.75.
bμm + cμf, so μg = (0.25)μm + (0.75)μf = (0.25)71 + (0.75)69 = 69.5
σ2 t
We know that μg = a +
= (0.25)2 σ2m + (0.75)2 σ2f + 2(0.25)(0.75)(19)σmσfρmf
= (0.25)2 192 + (0.75)2 232 + 2(0.25)(0.75)(19)(23)(0.5)
= 402
4.
Let D1 be the value from the 1st die and D2 be the value on the second. T = D1 + D2.
Three things to note
E[D1]=E[D2] = 3.5
Var(D1) = Var(D2) = 1.712 =σ2
The values D1 and D2 are independent, so σ12 = 0
T = a + bD1 + cD2 where a=0, b=c=1
E[T] = E[D1] + E[D2] = 3.5 + 3.5 = 7
Var[T] = b2σ2 + c2σ2 + 2bcσ12 = σ2 + σ2 = 2σ2 = 2(1.71)2 = 5.85
Standard deviation for a pair of dice is then the square root of 5.85 or 2.42
5.
Given a linear combination of random variables, Z = a + bY1 + cY2, we showed in class that Var(Z) = b2Var(Y1) +
c2Var(Y2) + 2bcCov(Y1,Y2). In this case Z = Y1 + Y2 so a=0, b=1 and c=1. We are also given the fact that Y1
and Y2 have the same variance σ2y plus Y1 and Y2 are independent so we also know that Cov(Y1,Y2) = σ12 = 0.
Plugging all these values into the equation above, Var(Z) = 2 σ2y – the same answer as in problem 4.
6.
Given the results above, if Z = Y1 + Y2 + Y3 + ..... Yn and each value Yi is independent of Yj, then
Var(Y1 + Y2 + Y3 + ..... Yn) = Var(Y1) + Var(Y2) + ..... Var(Yn) = nσ2y
7.
Note that Var((1/n)Y1) = (1/n)2 Var(Y1) = σ2/n2, so
Var(Z) = Var((1/n)(Y1 + Y2 + Y3 + ..... Yn))
= Var((1/n)Y2 + (1/n)Y2 + (1/n)Y3 + ...(1/n)Yn)
And because each of the Y’s are independent (as in problem 6)
=(1/n2)Var(Y1) + (1/n2)Var(Y2) + (1/n2)Var(Y3) + ...... (1/n2)Var(Yn)
= (1/n2) σ2y + (1/n2) σ2y + (1/ n2) σ2y + ..... (1/n2) σ2y = nσ2/n2 = σ2/n
Var(Z) =