Statistical Laboratory University of Cambridge University of Cambridge Mathematics and Big Data Showcase 20 April 2016
How much is an option worth? A call option is the right, but not the obligation, to buy one share of a certain stock for a certain price (the strike price) at a certain time in the future (the maturity date).
How much is an option worth? A call option is the right, but not the obligation, to buy one share of a certain stock for a certain price (the strike price) at a certain time in the future (the maturity date). For example, the option with symbol MSFT161021C00052500 is the right to buy Microsoft (MSFT) for $ 52.50 on Friday 21 October 2016.
How much is an option worth? A call option is the right, but not the obligation, to buy one share of a certain stock for a certain price (the strike price) at a certain time in the future (the maturity date). For example, the option with symbol MSFT161021C00052500 is the right to buy Microsoft (MSFT) for $ 52.50 on Friday 21 October 2016. Microsoft costs $ 56.39 this morning. How much would you pay for the option?
First modelling issues. Notation K = strike price T = maturity date St = price of stock at time t
First modelling issues. Notation K = strike price T = maturity date St = price of stock at time t Simplification: no frictions no bid-ask spread no transaction costs no price impact
The option payout. If S T > K, then exercise buy the stock for K, immediately sell the stock to the market for S T, pocket the difference S T K
The option payout. If S T > K, then exercise buy the stock for K, immediately sell the stock to the market for S T, pocket the difference S T K If S T K, then do nothing the option expires worthless
The option payout. If S T > K, then exercise buy the stock for K, immediately sell the stock to the market for S T, pocket the difference S T K If S T K, then do nothing the option expires worthless Payout = max{s T K, 0}.
A stochastic model (with easier numbers) strike K = 6. stock 7 1/2 5 call option 1 1/2? 1/2 4 1/2 0
Key idea: hedging borrow $ 4/3, buy 1/3 shares
Key idea: hedging borrow $ 4/3, buy 1/3 shares portfolio (1/3)(7) (4/3) = 1 1/2 (1/3)(5) (4/3) = 1/3 1/2 (1/3)(4) (4/3) = 0
Key idea: hedging borrow $ 4/3, buy 1/3 shares portfolio (1/3)(7) (4/3) = 1 1/2 (1/3)(5) (4/3) = 1/3 1/2 (1/3)(4) (4/3) = 0? = 1/3
Key idea: hedging borrow $ 4/3, buy 1/3 shares portfolio (1/3)(7) (4/3) = 1 1/2 (1/3)(5) (4/3) = 1/3 1/2 (1/3)(4) (4/3) = 0? = 1/3 Note probabilities play no role!
Real data. Microsoft since October 2015. Data from http://finance.yahoo.com
Generalisations More complicated distribution of S T
Generalisations More complicated distribution of S T Dynamics of (S t ) t
Generalisations More complicated distribution of S T Dynamics of (S t ) t Famous example: Black Scholes model log(s T ) normally distributed (S t ) t a geometric Brownian motion Hedging argument requires trading at intermediate times
Black Scholes (1973): initial price of a call option is d ± = log ( S 0 /K ) Φ(x) = C BS (T, K, S 0, σ) = S 0 Φ(d + ) KΦ(d ) σ T x ± σ T 2, e u2 /2 2π du
Black Scholes (1973): initial price of a call option is d ± = log ( S 0 /K ) Φ(x) = C BS (T, K, S 0, σ) = S 0 Φ(d + ) KΦ(d ) σ T x ± σ T 2, e u2 /2 2π du σ = volatility of the stock
Assuming the truth of the Black Scholes model: Volatility can be estimated from past prices.
Assuming the truth of the Black Scholes model: Volatility can be estimated from past prices. Volatility is implied from observed option prices.
Assuming the truth of the Black Scholes model: Volatility can be estimated from past prices. Volatility is implied from observed option prices. The implied volatility σ solves C BS (T, K, S 0, σ) = C obs
Assuming the truth of the Black Scholes model: Volatility can be estimated from past prices. Volatility is implied from observed option prices. The implied volatility σ solves C BS (T, K, S 0, σ) = C obs In our example, C obs = 5.62 so the (annualised) implied volatility is 21 %
Why implied volatility? Always defined, regardless of the truth of Black Scholes Provides a dimensionless measure of option price Reasonably stable across time Popular among practitioners
The calibration problem Choose a parametrized family of processes (St θ ) t 0 to model the stock, where S0 θ agrees with the time-0 price for all θ Θ
The calibration problem Choose a parametrized family of processes (St θ ) t 0 to model the stock, where S0 θ agrees with the time-0 price for all θ Θ Find the parameter θ Θ in such that the predicted implied volatilities match the market the observed implied volatilities
The calibration problem Choose a parametrized family of processes (St θ ) t 0 to model the stock, where S0 θ agrees with the time-0 price for all θ Θ Find the parameter θ Θ in such that the predicted implied volatilities match the market the observed implied volatilities In principle, calibration can be done solving a PDE, by Monte Carlo simulation, etc. but this can be slow.
Practical goal Formulae relating model parameters θ and market parameters T, K to the predicted implied volatility Σ θ (T, K). General formulae impossible but asymptotic formulae good enough for calibration
Example: Exponential Lévy models Generalises Black Scholes independent, stationary returns not necessarily continuous trajectories not necessarily log-normal distributions
Cumulant generating function Let for all T λ(p) = 1 T log E(S p T )
Theorem (Lee, Friz, Benaim) Σ(T, K) = as K where log K T ( p p 1)(1 + o(1)) p = sup{p 1 : λ(p) < }.
Theorem (MT) as T for all K. Σ(T, K) = 8 min λ(p)(1 + o(1))
Example: Subordinated Brownian motion. Let Y be a subordinator, let W be a Brownian motion, and let σw (Yt)+ΘYt+mt S t = e where m is chosen such that S is a martingale. Hence 2 p(p 1) λ(p) = aσ + 2 0 (e (pθ+p2 σ 2 /2)x e (Θ+σ2 /2)x )µ(dx)
Variance gamma: µ(dx) = 1 νx e x/ν dx, a = 0 Calibrated parameters from Madan Carr Chang: σ = 0.1213 ν = 0.1686 Θ = 0.1436
log-moneyness vs total variance, T = 1, 5, 10 0.20 0.18 0.16 0.14 0.12 0.10 0.08 0.06 0.04 0.02 0.00 1.0 0.8 0.6 0.4 0.2 0.0 0.2 0.4 0.6 0.8 1.0
Example: Stochastic volatility model. Suppose and let ds t = S t [(r(z t ) δ(z t ))dt + σ(z t )dw t ] dz t = a(z t )dt + b(z t )[ρ(z t )dw t + 1 ρ(z t ) 2 db t ] Λ(p : f ) = principal eigenvalue of 1 2 b(z)2 2 z 2 + [a(z) + pρ(z)b(z)σ(z)] z f
Theorem Let λ(p) = Λ(p : (1 p)r + pδ + 1 2 p(1 p)σ2 ) (1 p)λ(0 : r) pλ(1 : δ) Then Σ(T, K(T )) 8 min λ(p)