Problem Set 1 for Week 2: sketch
answers
1. Assets A, B and C have the following prices and dividends
Asset
A
B
C
Pit
10
100
50
Pit+1
10
105
25
Dit+1
0.5
0
27.5
(a) Calculate payo¤s and returns for each asset
Asset
A
B
C
Pit
10
100
50
Pit+1
10
105
25
Dit+1
0.5
0
27.5
Yit+1
10.5
105
52.5
Rit+1
10:5
10
105
100
52:5
50
1 = 0:05
1 = 0:05
1 = 0:05
(b) Assume all values in the table are known in period t; and investors can
buy any quantity of each asset (including fractions). Which if any of
the assets would investors prefer?
They’d be indi¤erent between all three as all o¤er the same risk-free
return.
(c) Give examples of assets with patterns of prices and dividends that might
resemble the numbers in the table.
A and C might be stocks, but C would be a stock of a company that is
probably on the verge of winding up, hence paying out a high proportion
of its value in dividends. B pays no dividends so might be gold or some
other commodity.
2 Assume X has a discrete distribution with the following probabilities
Xi
pr (X = Xi)
0
0.4
1
4
0.3
1
2
0.2
3
4
0.1
1. (a) Derive and sketch the cumulative distribution function, F (x) = pr (X < x)
over the range x = 0 to x = 1
Answer
Xi
pr (X = Xi)
F (Xi)
0
0.4
0
1
4
0.3
.4
1
2
0.2
.7
3
4
0.1
.9
1
0
1
Formally F (x) jumps up to 0.4 at x = 0; etc...
CDF for discrete case
1.1
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
(b) Calculate E (X ) and var(X ) (for the latter, use both standard formulae
as cross-check)
Answer:
Formally F (x) jumps up to 0.4 at x = 0; etc...
(b) Calculate E (X ) and var(X ) (for the latter, use both standard formulae
as cross-check)
Answer:
Xi
pr (X = Xi)
pr (X = Xi) Xi
(Xi E (X ))2
pr (X = Xi) (Xi E (X ))2
Xi2
pr (X = Xi) Xi2
hence E (X ) =
E X2 =
P
P
0
:4
0
:0625
1
:4 16
0
0
1
2
1
4
:3
:3
0
:3
:25
:2
:2
1
16
0
:2
1
16
1
4
:01875
:05
3
4
:5
1
16
pr (X = Xi) Xi = :25
pr (X = Xi) Xi2 = :125; V AR (X ) = :0625
:1
:1 :75
:25
:1 :25
9
16
:05625
1
0
F ( x) =
Z x
0
f (t) dt = 2
Z x
0
(1
"
t) dt = 2 t
#x
2
t
2 0
= 2x 1
x
2
(b) Calculate E (X ) and var(X ) (hint, use simpli…ed formula for var(X ))
Answer:
E (X ) =
Z 1
0
"
f (x) :x:dx = 2
x2
= 2
2
#1
Z 1
0
(1
x3
1
=2
3 0
2
var (X ) = E X 2
(E (X ))2
E X2
=
Z 1
0
"
f (x) :x2:dx = 2
#1
x) :x:dx = 2
Z 1
x
0
x2 :dx
1
1
=
3
3
Z 1
0
(1
x) :x2:dx = 2
x3 x4
1
1 1
=2
=
= 2
3
4 0
3 4
6
1
1 1
1 2
1
var (X ) =
=
=
= 0:05_
6
3
6 9
18
Z 1
0
x2
x3 :dx
(c) Using the CDF in this question, calculate F (Xi) for the Xi values in
question 1. Sketch the two functions on the same graph and comment
on the similarities and di¤erences. Graph the density alongside the
probabilities in the previous question. Compare the density with a
straight line that passes through each of the probabilities. Why is the
intercept for the density so much higher?
Xi
F (Xi) = 2Xi 1
Xi
2
0
0
1
4
1
2
3
4
0.4375
0.75
0.9375
CDFs
1.1
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1
1
Probablities in Discrete Case
(pr (X=Xi) )
0.45
0.4
0.35
0.3
0.25
Series2
0.2
0.15
0.1
0.05
0
0
0.25
0.5
Xi
0.75
1
Density Function for Continuous Case
2
1.5
1
0.5
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
4 Show that properties 1 to 4 of expected values as in C&W Chapter 4
Section A also hold for continuous random variables, ie, using
E (X ) =
E (g (X )) =
Z 1
1
Z 1
1
xf (x)dx
g (x) f (x)dx
Hints:
a) CW Appendix D, Section E summarises properties of integrals;
b) If you are struggling with the general case, use the particular form for
f (x) in question 3
a) CW Appendix D, Section E summarises properties of integrals;
b) If you are struggling with the general case, use the particular form for
f (x) in question 3
Answer:
(a) Property 1: E (x + a) = E (x) + a
Proof
R
E (x + a) = 11(x + a)f (x)dx
R1
R1
= 1 xf (x)dx + 1 af (x)dx
R
R
= 11 xf (x)dx + a 11 f (x)dx = E (x) + a
(a) Property 2: E (ax) = aE (x)
Proof:
R1
R1
E (ax) = 1 axf (x)dx = a 1 xf (x)dx = aE (x)
(a) Property 3: V AR(x + a) = V AR(x)
Proof : V AR(x + a) = E [(x + a
= E [(x + a
E (x))2
E (x + a))2]
a] = E [(x
E (x))2]
= V AR(x)
(Use property 1 in the second equality)
(a) Property 4: V AR(ax) = a2V AR(x)
Proof:
V AR(ax) = E [(ax
= a2E [(x
E (ax))2] = E [a2(x
E (x))2]
E (x))2] = a2V AR(x)
(Use property 2 in the second and third equalities)