The hedging strategy att 0is toshortsell 2units of the asset

  

5α+β= 0

2α+β= 6,

with solution   

α=−2

β= +10.

The hedging strategy att= 0is toshortsell−α= +2units of the asset SpricedS0= 4, and to putβ= $10on the savings account. The price V0=αS0+βof the initial portfolio at timet= 0is

V0=αS0+β=−2×4 + 10 = $2,

which yields the price of the claim at timet= 0. In order to hedge then

option, one should:

i) At timet= 0,

• Charge the $2 option price.
• Shortsell−α= +2units of the stock pricedS0= 4, which yields $8.
• Putβ= $8 + $2 = $10on the savings account.

ii) At timet= 1,

• IfS1= $5, spend $10 from savings to buy back−α= +2stocks.
• IfS1= $2, spend $4 from savings to buy back−α= +2stocks, and

deliver a $10 - $4 = $6 payoff.Pricing the option by the expected value IE ∗ [C]yields the equality

$2 =IE

∗ [C]

= 0×P

∗

(C= 0) + 6×P

∗

(C= 6)

= 0×P

∗

(S1= 2) + 6×P

∗

(S1= 5)

= 6×q ∗ , 1 Solutions Manual for Introduction to Stochastic Finance with Market Examples, 2e by Nicolas Privault (All Chapters) Chapter 1 Exercise 1.1 The payoff C is that of a put option with strike price K = $3.Exercise 1.2 Each of the two possible scenarios yields one equation: 1 / 4

Solutions Manual hence the risk-neutral probability measureP ∗ is given by p ∗ =P ∗

(S1= 5) =

2 3 andq ∗ =P ∗

(S1= 2) =

1 3 .Exercise<> 1.3 a)<> Each of the stated conditions yields one equation, i.e.  

4α+β= 1

5α+β= 3,

with solution    α= 2

β=−7.

Therefore, the portfolio allocation att= 0consists to purchaseα= 2 unit of the assetSpricedS0= 4, and to borrow−β= $7in cash.We can check that the priceV0=αS0+βof the initial portfolio at timet= 0is

V0=αS0+β= 2×4−7 = $1.

b)<> This loss is expressed as

α×$2 +β= 2×2−7 =−$3.

Note that the $1 received when selling the option is not counted here be- cause it has already been fully invested into the portfolio.Exercise<> 1.4 a)<> i)<> Does this model allow for arbitrage? Yes|✓ No| ii)<> If this model allows for arbitrage opportunities, how can they be real- ized? By shortselling|By borrowing on savings|✓ N.A.| b)<> i)<> Does this model allow for arbitrage? Yes| No|✓ ii)<> If this model allows for arbitrage opportunities, how can they be real- ized? By shortselling| By borrowing on savings| N.A.|✓ c)<> i)<> Does this model allow for arbitrage? Yes|✓ No| ii)<> If this model allows for arbitrage opportunities, how can they be real- ized? By shortselling|✓ By borrowing on savings| N.A.| Exercise<> 1.5

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Introduction to Stochastic Finance with Market Examples, Second Edition a)<> We need to search for possible risk-neutral probability measure(s)P ∗ such that IE ∗ ˆ S (1) 1 ˜ = (1 +r)S (1)

. Letting                              p ∗ =P ∗ ` S (1) 1 =S (1)

(1 +a) ´ =P ∗

S (1) 1 −S (1)

S (1)

=a !, θ ∗ =P ∗ ` S (1) 1 =S (1)

(1 +b) ´ =P ∗

S (1) 1 −S (1)

S (1)

=b !, q ∗ =P ∗ ` S (1) 1 = (1 +c)S (1)

´ =P ∗

S (1) 1 −S (1)

S (1)

=c !, We have    (1 +a)p ∗ S (1)

• (1 +b)θ

∗ S (1)

• (1 +c)q

∗ S (1)

= (1 +r)S (1)

p ∗ +θ ∗ +q ∗ = 1, from which we obtain    p ∗ a+θ ∗ b+q ∗ c=r, p ∗ +θ ∗ +q ∗ = 1.=⇒          p ∗ =

(1−θ

∗ )c+θ ∗ b−r c−a

∈(0,1),

q ∗ = r−(1−θ ∗ )a−θ ∗ b c−a

∈(0,1).

In order forp ∗ andq ∗ to belong to the interval(0,1)we should have    0<(1−θ ∗ )c+θ ∗ b−r < c−a, 0< r−(1−θ ∗ )a−θ ∗ b < c−a, i.e.       r−a b−c < θ ∗ < c−r c−b , r−c b−a < θ ∗ < r−a b−a .Therefore there exists an infinity of risk-neutral probability measures de- pending on the value of θ ∗ ∈ „ max „ r−a b−c , r−c b−a « ,min „ c−r c−b , r−a b−a «« , in which case the market is without arbitrage but not complete."3 September 15, 2022 3 / 4

Solutions Manual b)<> Hedging a claim with possible payoff valuesCa, Cb, Ccwould require to solve              (1 +a)αS (1)

• (1 +r)βS

(0)

=Ca (1 +b)αS (1)

• (1 +r)βS

(0)

=Cb (1 +c)αS (1)

• (1 +r)βS

(0)

=Cc, forαandβ, which is not possible in general due to the existence of three conditions with only two unknowns.Exercise<> 1.6 a)<> The risk-neutral condition IE ∗ [R1] = 0reads bP ∗ (R1=b) + 0×P ∗ (R1= 0) + (−b)×(R1=−b) =bp ∗ −bq ∗ = 0, hence p ∗ =q ∗ =

1−θ

∗ 2 , sincep ∗ +q ∗ +θ ∗ = 1.b)<> We have Var ∗ " S (1) 1 −S (1)

S (1)

# =IE ∗ ˆ R 2 1 ˜

−(IE

∗ [R1]) 2 =IE ∗ ˆ R 2 1 ˜ =b 2 P ∗ (R1=b) + 0 2 ×P ∗ (R1= 0) + (−b) 2 ×(R1=−b) =b 2 (p ∗ +q ∗ ) =b 2

(1−θ

∗ ) =σ 2 , henceθ ∗

= 1−σ

2 /b 2 , and therefore p ∗ =q ∗ =

1−θ

∗ 2 = σ 2 2b 2 , provided thatσ 2 ≤2b 2 .Exercise<> 1.7 a)<> We denote the risk-neutral measure byp ∗ =P ∗ (S1= 2),q ∗ =P ∗

(S1= 1).

i)<> Yes| No|✓Comment: No loss is possible, while a 100%

profit is possible with non-zero probability1/3.

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