Fds312 Assignment 1 Trimester 3a 2013

FINANCIAL DERIVATIVES SECURITIES 312

Question 1 [5 marks]

“No-arbitrage” value is the absence of arbitrage opportunities, and also defined as the existence of a fair value, where a complete market requires all possible payoffs to be spanned by available securities.

2-dimensional space is spanned by primitive securities A= (1, 0) and D= (0, 1). Hence, if we can replicate A and D with the combination of Security J and Security K, then the market is complete.

Setting j as the quantity of payoff Security J, and k as the quantity of payoff Security K, to return payoff A = (1, 0) and D = (1, 0).

Let j(2, 3) + k(3, 1) = (1, 0). Thus, 2j + 3k = 1...Eq. (1) and 3j + 1k = 0…Eq. (2).

Eq. (1) – 3Eq. (2) gives -7j = 0 resulting in j = -1/7.

Substitute j = -1/7 in Eq. (1) or Eq. (2), we get k = 3/7.

Thus, (-1/7)j + (3/7)k = (1, 0)A.

Let j(2, 3) + k(3, 1) = (0, 1). Thus, 2j + 3k = 0...Eq. (1) and 3j + 1k = 1…Eq. (2).

Eq. (1) – 3Eq. (2) gives -7j = -3 resulting in j = 3/7.

Substitute j = 3/7 in Eq. (1) or Eq. (2), we get k = -2/7.

Thus, (3/7)j + (-2/7)k = (0, 1)D.

The combination of Security J and Security K can replicates A and D by

(-1/7)j + (3/7)k = (1, 0)A and (3/7)j + (-2/7)k = (0, 1)D.

Therefore, the market is complete.

Find Security L = (1, 4) as the sum of primitive securities, L=A + 4D

[i.e., (1, 4) = (1, 0) + 4(0, 1)] and find the no-arbitrage values of A = (1, 0)

and D = (0, 1).

A = (-1/7)J + (3/7)K

(1, 0) = (-1/7) (2, 3) + (3/7) (3, 1)

PA = (-1/7) PJ + (3/7) PK

PA = (-1/7) (0.08) + (3/7) (0.05)

PA = 0.01

D = (3/7)J + (-2/7)K

(0, 1) = (3/7) (2, 3) + (-2/7) (3, 1)

PD = (3/7) PJ + (-2/7) PK

PD = (3/7) (0.08) + (-2/7) (0.05)

PD = 0.02

No-arbitrage value of Security L = (1, 4), PL = PA + 4 PD

No-arbitrage value of Security L = (1, 4), PL = 0.01 + 4(0.02)

In conclusion, no-arbitrage value of Security L = (1, 4), PL = 0.09.

Question 2 [2 marks]

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