One-period Binomial Case
Consider an 1-period binomial case. Assume : T = 1 and up probability p = 0.1.
Consider a call with K = 21.
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What do we know of the option value at the expiry T = 1?
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What’s the option value today?
Is it $.1 = 0.1 × 1 + 0.9 × 0?
What about time value of the money, ie. interest rate?
payoff = 1
payoff = 0
Riskless Asset (constant interest rate)
Cash account continuously compounds at risk free rate r ≥ 0. Lending (depositing) money to a bank ⇒ buying a bond from a
Borrowing money from a bank ⇒ selling a bond.
Let β(t) denote the value of a riskless bond at time t dβ(τ) T dβ(τ) T
β(τ) = rdτ ⇒ β(τ) = rdτ tt
log(β(T )) − log(β(t)) = r(T − t)
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Discounting: β(T ) = 1 ⇒ β(t) = e−r(T −t).
Compounding: β(t) = 1 ⇒ β(T ) = er(T −t).
Back to the example: Is the fair value e−0.05 × 0.01, if r = 0.05?
Determine the fair value by trading
Determining the fair option value needs to consider trading in a financial market of bond, stock, option.
An arbitrage is a trading opportunity to make a no-risk (guaranteed) profit which is greater than that of a bank deposit which earns the interest rate r ≥ 0.
Definition. The fair value of a financial instrument is the price which does not lead to arbitrage.
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Why? Arbitrage can only occur momentarily
How? Under no arbitrage: two instruments have the same values at a future time, they must be priced at the same price today.
Constructing Arbitrage. We represent a trading strategy as a portfolio.
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Example: buy one share of stock and borrow $100 (sell bonds) today
Π0= 1 ·S0 − 100 or Π0={S0,−100}
The value of this portfolio at time t:
Πt = St − 100ert
Mathematical Characterization of an Arbitrage Strategy
• A portfolio with an initial value Π0 = 0 but ΠT > 0 (always!) • AportfoliowithΠ0 <0butΠT ≥0forsure
Put and Call Parity
Assume stock St does not pay dividend, interest rate r ≥ 0, and no arbitrage. Then at any time 0 ≤ t ≤ T , European call Ct and put Pt, with the same strike K and expiry T, on the same underlying, satisfies
Ct =Pt +St −Ke−r(T−t) Proof: At T we have
CT =max(ST −K,0)
PT =max(K−ST,0)
CT −PT =max(ST −K,0)−max(K−ST,0)≡ST −K
Ct −Pt =St −Ke−r(T−t)