Using the Laplace of the determinant of an S × S matrix, prove that

Homework Assignment 1

I don’t tell u who i am 2018.10.6

1 Chapter 2

1.1 Question 1

Using the Laplace of the determinant of an S × S matrix, prove that

X =











1 0 · · · 0

2 1 · · · 0

.. . .. . . . . .. . S S − 1 · · · 1











(1)

Has determinant equal to 1.

Answer : Expand the matrix by the first row. We have

X = 1 ×











1 0 · · · 0

2 1 · · · 0

.. . .. . . . . .. . S − 1 S − 2 · · · 1











(2)

Keep expanding the matrix and we have

X = 1 × 1 × 1 × · · · × 1 0 0 1



(3)

Expand the last matrix and we can conclude that the X has determinant to 1.

1.2 Question 2

Now repeat exercise 1 by adding to the securit structure put options on the

stock, following the same steps as explained in the text (for call options). Are

markets complete?

Answer: Let us consider a specific example in which stock value at time t = 1 are equal to the index of the state of the world: s = (1, 2, · · · , S). We can introduce S − 1 put options with payoff (k − s) ⁺ for k = 1, · · · , S − 1: we obtain the securities

c 2 = (1, 0, 0, · · · , 0, 0) ⁰ (4) c ₃ = (2, 1, 0, · · · , 0, 0) ⁰ (5)

.. . (6)

c S = (S − 1, S − 2, S − 3, · · · , 1, 0) ⁰ (7) Which together with the stock give rise to the security structure

X =











1 2 · · · S 0 1 · · · S − 1

.. . .. . . . . .. . 0 0 · · · 1











(8)

This is an upper triangular S × S matrix whose determinant is one (the product of the terms on the diagnal). It can be proved by following simple Laplace steps just like what we do in Question 1. Therefore X is full rank and markets are complete.

1.3 Question 3

Suppose there exists only a risk-free asset x ¹ = (1, 1, · · · , 1) ⁰ and a risky asset x ² 6= x ¹ and S states of the world. Let p 1 amd p 2 be the prices of these two assets. A forward contract on the stock is an agreement to pay an amount F at a future date t = T in exchange for the payment x ^j _s when the state s ∈ {1, 2, · · · , S} realizes, with no cash flow exchange at time t = 0. Assuming arbitrage opportunities are ruled out, find the fair value of F.

Answer Let x ^j be an equivalent of a × x ¹ + b × x ² . The payoff of the forward contract is x ^j _s − F . To replicate this, we buy ^F _T x ¹ and x ^j at time t = 0 and sell

F

T x ¹ , which equals to F, at time t = T . Therefore we have x ^j _s − F , and hence F will be equal to the value of the portfolio of long a x ^j and short ^F _T x ¹ today.

So,we have

F = (a + F

T ) × p ₁ + b × p ₂ (9)

This function equals to

F = aT p 1 + bT p 2

T − p ₁ (10)

where a and b fit x ^j = a × x ¹ + b × x ²

Note: In fact in this question x ^j is not a particular security (accoring to the

teacher), so the answer should be changed. But the concept is the same.

2 Chapter 3

2.1 Question 1

Determine whether the following statements are true or false. Provide a proof or a counter-example.

2.1.1 1.

Law of one price and complete markets imply no strong arbitrage.

Answer False. Let us consider a security structure that looks like:

X = 1 2 0 1



(11)

It’s a complete market. We have two portfolios h = k = (−1, 1) ⁰ . It definitely satisfies LOOP since h = k. And Xh = Xk = (1, 2) > 0. If we have p = (11, 1), then p · h = −10. So this statement is false.

2.1.2 2.

Law of one price and complete markets imply no arbitrage.

Answer False. Let us consider a security structure that looks like:

X = 1 2 0 1



(12)

It’s a complete market. We have two portfolios h = k = (−1, 1) ⁰ . It definitely satisfies LOOP since h = k. And Xh = Xk = (1, 2) > 0. If we have p = (11, 1), then p · h = −10. So this statement is false.

2.1.3 3.

No strong arbitrage and complete markets imply no arbitrage.

Answer False. Consider a complete market with a security portfolio X to be

X = 1 2 0 2



(13)

Consider h = (−2, 1) ⁰ and p = (1, 2). Thus Xh > 0, however p × h = 0. It fits NSA, but doesn’t fit NA. So this statement is false.

2.2 Question 2

Suppose there exists 3 states of the world s = 1, 2 and 2 assets x ¹ , x ² .

2.2.1 1.

Suppose x ¹ = (2, 1, 0) ⁰ and x ² = (0, 1, 0) ⁰ . Describe the asset plan. Are markets complete?

Answer The security structure looks like:

X =



 2 0 1 1 0 0



 (14)

There are three states. So whatever the portfolio is, if the state turns out to be 3, the payoff is always 0. If we remove state 3, it is clear that x ¹ and x ² are linearly independent. Hence the asset span

hXi = {z =



 a b 0



 : a, b ∈ R} (15)

So markets are incomplete.

2.2.2 2.

Suppose p ₁ = 4 and p ₂ = 3. What type of no-arbitrage requirements does this markets satisfy?

Answer It satisy NA. If we have a portfolio like h = (a, b) ⁰ , and p × h equals to 3a + 4b. Thus

Xh =



 2a a + b

0



 (16)

If Xh > 0, we have a > 0 and a + b > 0, which results in a > 0 and a > −b.

Hence 4a + 3b > 0. We can say this market satisfy NA.

2.2.3 3.

What are the restrictions on p ₁ and p ₂ such that this market satisfies LOOP, NSA and NA? (Write each restriction separately)

Answer Let’s consider LOOP first. if we define h = (a, b) ⁰ and k = (c, d) ⁰ then

Xh =



 2a a + b

0



 , Xk =



 2c c + d

0



 (17)

if Xh = Xk then a = c and a + b = c + d, which means a = c and b = d. So

p × h = p × k no matter what p 1 and p 2 is. LOOP is satisfied in any p 1 and

p 2 cases.

Let’s consider NSA then. Still, we define h = (a, b) ⁰ . When Xh ≥ 0, we have a ≥ 0 and a + b ≥ 0. If NSA is satisfied, p 1 × a + p 2 × b ≥ 0. Let it be divided by p 1 , we have a + ^p _p

1

b ≥ 0. Since a ≥ 0 and a + b ≥ 0, we can conclude that 0 ≤ ^p _p

1

≤ 1. Since p 1 ≥ 0andp 2 ≥ 0, hence p 1 ≥ p 2 is the restriction of NSA.

Let’s consider NA finally. Still, we define h = (a, b) ⁰ . When Xh > 0, we have a > 0 and a + b > 0. If NA is satisfied, p ₁ × a + p ₂ × b > 0. Let it be divided by p ₁ , we have a + ^p _p

1

b > 0. Since a > 0 and a + b > 0, we can conclude that 0 < ^p _p

1

< 1. Since p 1 > 0 and p 2 > 0, hence p 1 > p 2 is the restriction of NA.

2.2.4 4.

Repeat 1), 2) and 3) for x ¹ = (1, 1, 0) ⁰ and x ² = (0, 2, 0) ⁰ . Answer The security structure looks like:

X =



 1 0 1 2 0 0



 (18)

There are three states. So whatever the portfolio is, if the state turns out to be 3, the payoff is always 0. If we remove state 3, it is clear that x ¹ and x ² are linearly independent. Hence the asset span

hXi = {z =



 a b 0



 : a, b ∈ R} (19)

So markets are incomplete.

Now p ₁ = 4 and p ₂ = 3. Define h = (a, b) ⁰ ,we have

Xh =



 a a + 2b

0



 (20)

When Xh > 0, we have a > 0 and a + 2b > 0. So we have 1.5a + 3b > 0.Thus ph = 4a + 3b > 0. The market satisfies NA.

Let’s consider LOOP. if we define h = (a, b) ⁰ and k = (c, d) ⁰ then

Xh =



 a a + 2b

0



 , Xk =



 c c + 2d

0



 (21)

when Xh = Xk is satisfied, we have h = k, so LOOP is satisfied no matter what p 1 and p 2 are.

Let’s consider NSA. When XH ≥ 0, we have a ≥ 0 and a + 2b ≥ 0. p 1 Let × a + p 2 × b be divided by p 1 , we have a + ^p _p

1

b. If we want it to be greater than 0, we have ^p _p

1

≤ 2. So the condition for NSA is p 2 ≤ 2p 1 .

Let’s consider NA. When XH > 0, we have a > 0 and a + 2b > 0. p 1 Let × a + p 2 × b be divided by p 1 , we have a + ^p _p

1

b. If we want it to be greater than 0, we have ^p _p

1

< 2. So the condition for NA is p 2 < 2p 1 . 2.2.5 5.

Repeat 1), 2) and 3) for x ¹ = (1, 1, 0) ⁰ and x ² = (0, 2, 0) ⁰ and x ³ = (0, 1, 1) ⁰ . Answer The security structure looks like:

X =





1 0 0 1 2 1 0 0 1



 (22)

It is clear that rank(X) = 3, which means the market is full and the asset span hXi is R ³ .

There is not enough price information so we skip question 2 and jump to question 3.

Let’s consider LOOP. If we define h = (a, bc) ⁰ and k = (d, e, f ) ⁰ then

Xh =



 a a + 2b + c

c



 , Xk =



 d d + 2e + f

f



 (23)

When Xh = Xk,we have h = k, thus LOOP is satisfied no matter what p ₁ and p ₂ are.

Let’s consider NSA. When Xh ≥ 0, we have a ≥ 0, c ≥ 0 and a + 2b + c ≥ 0.

We have ph to be p 1 a + p 2 b + p 3 c. To make every a, b, c fit NSA, The condition is p 2 ≤ 2p 1 and p 2 ≤ 2p 3 .

Let’s consider NA finally. When Xh > 0, we have a > 0, c > 0 and a+2b+c >

0. We have ph to be p 1 a + p 2 b + p 3 c. To make every a, b, c fit NSA, The condition is p 2 < 2p 1 and p 2 < 2p 3 .

3 Additional Question

3.1 1

Prove that if v is a linear functional, LOOP holds.

v(z) ≡ {p · h : z = Xh} (24)

Answer We prove the contrapositive ”if LOOP doesn’t hold, v(z) is not a linear functional”.

Define h and k to be two portfolios that satisy Xh = Xk. LOOP doesn’t

hold implies that p · h 6= p · k. Assume that v(z) is a linear functional. We have

v(0) = v(Xh − Xk) = v(Xh) − v(Xk) = p · h − p · k (25)

Also,

v(0) = v(Xk − Xh) = v(Xk) − v(Xh) = p · k − p · h (26) Since p · h 6= p · k, we have v(0) 6= v(0), which is of course wrong. Thus, v(z) can’t be a linear functional. So the contrapositive proves true. The statement proves true, too.

0 = p · h − p · k (27)

So we have p · h = p · k. Then LOOP holds.