Hw3 - An Introduction to Quantitative Finance by Stephen Blyth work for Econ 3315 PDF

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An Introduction to Quantitative Finance by Stephen Blyth work for Econ 3315...

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Homework 3 Econ 5315/3315 Fall 2021 Due 10/1 1. Binomial tree:random interest rates Consider a two-step binomial tree, where a stock that pays no dividends has current price 100, and at each time step can increase by values at time

T

= 2 are thus 144, 108 and 81.

20% or decrease by 10%: The possible

now suppose that if the stock is at 120,

then the annually compounded interest rate from time T = 1 to T = 2 is

5%, not 10%.

u = 1 + 0:2 = 1:2 and down factor = 1  0:1 = 0:9: The stock prices are S0 = 100; S1 (H ) = 120; S1 (T ) = 90; S2 (H H ) = 144; S2 (H T ) = 108; S2 (T H ) = 108; and S2 (T T ) = 81:

For this question, we know the upper factor d

(a) Write down the value of the money market account

m

Does the joint tree for (S , It is easy to see

m

M

) recombine?

m

M

at all states of the tree.

1 ; 1 + r1 (H ) 1:05 1 1 M1 (T ) = ; = 1 + r1 (T ) 1:1 1 1 = ; (1 + r0 ) (1 + r1 (H )) (1:1) (1:05) M1

and

1

(H ) =

=

1 1 = 2: (1 + r0 ) (1 + r1 (T )) (1:1)

m =Mm , …nd the risk-neutral probabilities with respect to the money market numeraire Mm at each node of the tree.

(b) By using the martingale condition for

0

The risk neutral probabilities from

e =

p0

1 + r0  d u



d

S

to

=

1

1 + 0:1  0:9 0:2 2 = = 3 1:2  0:9 0:3

and

1 : 3

e =

q0

From 1 to 2 are

e (H ) =

p1

1 + r1 (H )  d u



d

are

=

and

e (H ) =

q1

Similarly,

e (T ) =

p1

and

1 + r1 (T )  d u



d

=

e (T ) =

q1

1

1 + 0:05  0:9 = 0:5 1:2  0:9 1 : 2 1 + 0:1  0:9 2 = 3 1:2  0:9 1 : 3

(c) Recall

(

2)

Z m; M

m

Z

is a martingale under the risk-neutural probability measure. Then

1

[pe1 (H ) Z (2; 2) + qe1 (H ) Z (2; 2)] 1 + r1 (H ) 1 1 1 + = 1:05 2 2 1 = 1:05

(1; 2) (H ) =

and



2 1 + Z (1; 2) (T ) = 1 + r1 (T ) 3 3 1 = : 1:1 1



Hence, Z

1 [pe0 Z (1; 2) (H ) + pe1 (H ) Z (1; 2) (T )] 1 + r0  1 2 1 1:1 1 1 + = 1:1 3 1:05 1:1 3 1:1   1 1 2 1 = 1:1 + 3 (1:1)2 3 1:05   1 1 65 1 2 1 = 1:1 + = 2 3 (1:1)2 63 (1:1) 3 1:05

(0; 2) =

as required. S (d) Recall Z (t;t2) is a martingale under the forward probability measure (using as the numeraire). Then we require

120 1 1:05

(H ) = E (S2 jS1 ) Z (1; 2) (H ) = [p S2 (H H ) + (1  p ) S2 (H T )]

=

S1

= [p  144 + (1  p )  108] : Then

126 = [144p1 (H ) + 108  108p1 (H )] = 36p1 (H ) + 108 and 

p1

Now

S0 Z

(0; 2)

= p0

(H ) = 126  108 =

1 18 = : 36 2

1 1  S1 (T ) : S1 (H ) + (1  p 0 ) Z (1 ; 2) (T ) Z (1; 2) (H ) 2

Z

(t; T )

Then

100



=

1 65 (1:1)2 63

p0

126 + (1  p0 ) 99

= p0 126 + 99  99p0 = 27p0 + 99:

Hence

100

27p0 =

1 65 (1:1)2 63

and



p0

100

1 = 27

1 65 (1:1)2 63

 99

!

 99

as required. Clearly,



p0

 e0 p

as required.

3

44 2  65 3 2 = 195 =

=

44 65...