[PPT] - CASE STUDY III EVALUATING MODEL RISK WITHIN THE BLACKSCHOLES PowerPoint Presentation

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CASE STUDY III EVALUATING MODEL RISK WITHIN THE BLACK–SCHOLES FRAMEWORK Limiting Model Risk by Short-Selling Constraints

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Outline for case study III

Samuelson (Black–Scholes) model
Exotic options, unlimited short positions
Mitigation of model risk

by short-selling constraints

Resulting market incompleteness,

upper hedging price

Incorporation of constraint into option price
Option price as stochastic control problem
Explicit valuation for several examples

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References for case study III

U. Schmock, S. E. Shreve, U. Wystup:

Valuation of Exotic Options under Shortselling Constraints Finance and Stochastics, Vol. 6 (2002) 143–172.

U. Schmock, S. E. Shreve, U. Wystup:

Dealing with Dangerous Digitals In: J. Hakala and U. Wystup (eds.): Foreign Exchange Risk: Models, Instruments and Strategies Risk Books, Risk Waters Group (2002) 327–348.

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Samuelson model Geometric Brownian motion for the exchange rate: dSt = (rd − r

f)St dt + σSt dWt,

S0 > 0 St price of one unit of foreign currency in domestic currency at time t ∈ [0, T] rd ∈ R risk-free domestic interest rate r

f ∈ R

risk-free foreign interest rate σ > 0 volatility (Wt)t∈[0,T ] Brownian motion (Wiener process) under the risk-neutral measure P r rd − r

f mean rate of return of the exchange rate

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Samuelson model (cont.) Equity model St stock price at time t rd ∈ R risk-free domestic interest rate r

f ∈ R continuously paid dividend rate

Solution of the SDE St = S0 exp

rt + σWt − σ2

2 t

,

t ∈ [0, T] Canonical probability space Ω = C([0, T], R) ∋ ω → Wt(ω) = ω(t) σ-algebra Ft is the P-completion of σ(Ws; s ∈ [0, t]),

i. e., {Ft}t∈[0,T ] is a Brownian filtration.

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Hedging a European call option with strike K

Pay-off at maturity T is (ST − K)+ where ST is

the price of the underlying at time T and K > 0 is the strike price.

To hedge the call, buy a fraction of the underlying

(delta-hedging). In the Samuelson model: Calculation of the fraction ∈ (0, 1) at time t by differentiation of the Black–Scholes formula w. r. t. St N log(St/K) + (r + σ2/2)(T − t) σ √ T − t

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95 100 105 110 2 4 6 8 10 12 90 50 days 10 days 1 day Option value Stock price St

Price of a European call option for three different maturities, interest rate r = 5%, volatility σ = 30%, strike price K = 100

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95 100 105 110 0.2 0.4 0.6 0.8 1 90 50 days 10 days 1 day Fraction of stock in hedging portfolio Stock price St

Hedge for a European call option for three different maturities, interest rate r = 5%, volatility σ = 30%, strike price K = 100

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Reverse up-and-out call European call option with strike K > 0 and knock-

ut barrier B > K. Pay-off at maturity T

g(S) (ST − K)+1{maxt∈[0,T ] St<B} No-arbitrage Black–Scholes price at time t ∈ [0, T] v(t, x) = Et,x e−rd(T −t)(S(T) − K)+1{maxu∈[t,T ] Su<B}

if St = x > 0 and no knock-out occurred before t.

Delta hedging: – If St is well below B: Buy a fraction of the underlying. – If St is just below B: Go short in the underlying.

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Price of the reverse up-and-out call v(t, x) = xe−r

f τ

N(b − θ+) − N(k − θ+)

+ xe−r

f τ+2bθ+

N(b + θ+) − N(2b − k + θ+)

− Ke−rdτ

N(b − θ−) − N(k − θ−)

− Ke−rdτ+2bθ−

N(b + θ−) − N(2b − k + θ−)

where N is the standard normal distribution function,

τ T − t, θ± r σ ± σ 2 √τ and b 1 σ√τ log B x , k 1 σ√τ log K x .

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100 110 120 130 140 150 10 20 30 40 50 90 50 days 10 days 1 day Option value Stock price St

ption pay-off

Price of a European call option with knock-out barrier B = 150 for three different maturities together with the option pay-off, interest rate r = 5%, volatility σ = 30%, strike price K = 100

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100 110 120 130 140 150 0.5 1 −0.5 −1 −1.5 50 days 10 days 1 day Stock price St Number of stocks in hedging portfolio

Hedge of a European call option with knock-out barrier B = 150 for three different maturities, interest rate r = 5%, volatility σ = 30%, strike price K = 100

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135 140 145 150 100 80 60 40 20

− − − − −

50 days 10 days 1 day Fraction πt of capital in the stock Stock price St

Fraction πt of capital invested in the stock to replicate a European call option with knock-out barrier B = 150 for three different maturities, interest rate r = 5%, volatility σ = 30%, strike price K = 100

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Problems with large FX positions

Large exposure for one sold barrier option
Liquidity risk and transaction costs
High model risk!

Possible solutions

Pay a rebate at maturity or at the first hitting

time of the barrier, when the option knocks out.

Modify the knock-out regulation

(soft barrier option, step option, Parisian option).

Impose constraint for the hedge portfolio.

→ incomplete market, superhedge the option

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Evolution of the hedge capital Xt πt fraction of Xt in foreign currency (adapted) 1 − πt fraction of Xt in domestic currency Ct capital consumed in [0, t] dXt = πtXt St dSt + r

fπtXt dt + rd(1−πt)Xt dt − dCt

= rdXt dt + σπtXt dWt − dCt Option pay-off Lower semi-continuous function g : C+[0, T] → [0, ∞) Short-selling constraint for foreign currency πt ≥ −α for all t ∈ [0, T] with α ≥ 0

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Upper hedging price v(0, S0; α) inf{ X0 | ∃ (π, C) with XT ≥ g(S) and πt ≥ −α ∀ t ∈ [0, T] } Dual maximization problem Theorem: (Cvitani´ c & Karatzas 1993, El Karoui & Quenez 1995) v(0, S0; α) = sup

λ∈L

Eλ

e−rdT −αλT g(S)
L contains all adapted, non-decreasing λ with λ(0) = 0,

which are Lipschitz-continuous in t, uniformly in ω. dPλ dP = exp

− 1

σ T λ′

t dWt −

1 2σ2 T (λ′

t)2 dt

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Simplification for dependence on final value Theorem: (Broadie, Cvitani´ c & Soner 1998) If g(S) = ϕ(ST ), then v(0, S0; α) = E

e−rdT

ϕα(ST )

with face-lift
ϕα(x) sup

λ≥0

e−αλϕ(xe−λ), x ≥ 0. Aim of our work

Generalization to path-dependent options by

conversion of the dual maximization problem to a stochastic control problem.

Explicit computation of the upper hedging price

for several examples.

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Idea behind Broadie–Cvitani´ c–Soner theorem (t, x) → v(t, x; α) is the smallest function which

satisfies the Black–Scholes PDE

vt + rxvx + 1 2σ2x2vxx − rdv = 0,

dominates the final pay-off, i.e. v(T, x; α) ≥ ϕ(x),
satisfies the constraint αv(t, x; α)+xvx(t, x, α) ≥ 0.
ϕα is the smallest function dominating the pay-off

and satisfying the constraint α ϕα(x) + x ϕ′

α(x) ≥ 0.

→ Solve Black–Scholes PDE with pay-off ϕα. Pleasant surprise: Solution satisfies constraint!

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Extension to path-dependent up-and-out call Observation: If v(t, x; α) solves the Black–Scholes PDE, then w αv + xvx solves the PDE, too. Strategy:

Boundary conditions for v give boundary condi-

tions for w.

Require w = 0 at the boundary where the uncon-

strained value function violates the constraint.

Solve Black–Scholes PDE for w.
Solve w αv + xvx for v.

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Formulation of the dual problem as singular stochastic control problem Theorem (Schmock/Shreve/Wystup): v(0, S0; α) = sup

λ∈C

E

e−rdT −αλT g(Se−λ)
where C {λ| λ adapted, non-decreasing,

continuous process, λ(0) = 0}. Remarks:

Maximization w. r. t. processes is easier.
Maximizing process can be found in many examples.
Maximizing processes can be singularly continuous.
Since g is lower instead of upper semi-continuous,

maximizing processes need not exist.

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Application to a European call option with strike K and knock-out barrier B > K Obligation at maturity T: g(S) (ST − K)+1{maxt∈[0,T ] St<B} Maximization problem: sup

λ∈C

E

e−rdT −αλT

ST e−λT −K +1{maxt∈[0,T ] Ste−λt<B}

Supremum unchanged for < → ≤.

Maximizing process: Ste−λt ≤ B ⇐ ⇒ λt ≥ log St − log B = ⇒ λt ≥ λ∗

t max u∈[0,t](log Su − log B)+

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Upper hedging price

v∗(t, x, α) = xe−r

f τ

N(b − θ+) − N(k − θ+) + e

1 2 s(s−2θ+) ×

esbN(−b + θ+ − s) − eskN(−k + θ+ − s)
+ sxe−r

f τ+2bθ+

s − 2θ+

N(b + θ+) − N(ℓ + θ+) + e

1 2 s(s−2θ+)

×

e(s−2θ+)bN(−b + θ+ − s) − e(s−2θ+)ℓN(−ℓ + θ+s)
− Ke−rdτ

N(b − θ−) − N (k − θ−) + e

1 2 ˜

s(˜ s−2θ−)

e˜

sbN(−b + θ− − ˜

s) − e˜

skN(−k + θ− − ˜

s)

− ˜

sKe−rdτ+2bθ− ˜ s − 2θ−

N(b + θ−) − N (ℓ + θ−) + e

1 2 ˜

s(˜ s−2θ−)

×

e(˜

s−2θ−)bN(−b + θ− − ˜

s) − e(˜

s−2θ−)ℓN(−ℓ + θ− − ˜

s)

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. . . with the abbreviations τ = T − t b = 1 σ√τ log B x s = (1 + α)σ√τ k = 1 σ√τ log K x ˜ s = ασ√τ θ± = r σ ± σ 2 √τ ℓ = 2b − k N(y) = 1 √ 2π y

−∞

exp(−u2/2) du

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100 110 120 130 140 150 10 20 30 40 50 090 50 days 10 days 1 day Stock price St Upper hedging price v∗(0, S0; α)

f the barrier option

for three different maturities

Knock-out barrier B = 150, portfolio constraint α = 10, interest rate r = 5%, volatility σ = 30%, strike price K = 100

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100 110 120 140 150 1 3 2 1 − − − 130 50 days 10 days 1 day Number of stocks in hedging portfolio Stock price St

Super-replication of a European call option with knock-out barrier B = 150, hedge-portfolio constraint α = 10, interest rate r = 5%, volatility σ = 30%, strike price K = 100

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135 140 145 150 2 4 10 8 6 4 2 − − − − − 50 days 10 days 1 day Fraction πt of capital in the stock Stock price St

Fraction πt of capital invested in the stock to super-replicate a European call option with knock-out barrier B = 150, portfolio constraint α = 10, interest rate r = 5%, volatility σ = 30%, strike price K = 100

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100 110 120 130 140 150 160 10 20 30 40 50 60 50 days 10 days 1 day Option value Price of a European call option with knock-out barrier B = 150 for three different maturities, portfolio constraint α = 10, interest rate r = 5%, volatility σ = 30%, strike price K = 100, and linear extrapolation. The dashed lines show the price without portfolio constraint but a barrier moved to B′ = B(1 + 1/α) = 165.

90

Stock price St

ption pay-off

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Stochastic impulsive control problem

0 < t1 < t2 < · · · < tI ≤ T fixed dates for impulses
R[0, T] set of non-decreasing, on [0, T]\{t1, . . . , tI}

continuous, in t1, . . . , tI right-continuous functions which start in the origin Theorem: (Schmock/Shreve/Wystup) v(0, S0; α) = sup

λ∈R

E

e−rdT −αλTg∗(S, λ)
where R { λ | λ adapted process, paths in R[0, T]},

g∗(S, λ) inf

{λn}n∈N lim inf n→∞ g(Se−λn),

Infimum over all {λn}n∈N in C with λn → λ pointwise.

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Applications of the stochastic impulsive con- trol problem European call option with knock-out barrier B > K, which is checked only at times 0 < t1 < · · · < tI ≤ T. Pay-off at maturity: g(S) =

ST − K

+ I

i=1 1{Sti<B}

Maximizing process (for < → ≤): λ∗

t =

max

{i;ti≤t}

log Sti − log B

+ Upper hedging price v(0, S0, α): E

e−rdT −αλ∗

T

ST e−λ∗

T − K

+

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2. Example: Asian put option

Pay-off at maturity T: g(S) =

A(S) − ST

+ with arithmetic average A(S) 1 T T St dt Maximizing process for α > 0: λ∗

t =

log (1 + α)ST

αA(S)

+

1{t=T } Upper hedging price v(0, S0, α): E

e−rdT −αλ∗

T

A(S) − ST e−λ∗

T +

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3. Example: Lookback put

Pay-off at maturity T: g(S) =

M(S) − ST

+ with maximal stock price M(S) max

t∈[0,T ] St

Maximizing process for α > 0: λ∗

t =

log (1 + α)ST

αM(S)

+

1{t=T } Upper hedging price v(0, S0, α): E

e−rdT −αλ∗

T

M(S) − ST e−λ∗

T +

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Further application: Pricing and hedging a book of options Advantages:

More realistic
Elimination of opposite risks
Higher value of the book

→ hedge-portfolio constraint less severe

Lower option prices, hence more competitive

pricing in the financial market Challenge: Evaluation of the stochastic impulsive control problem is more complicated.

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Example of a book: Two European call

ptions with knock-out barriers U > L > K

Decision at time t∗ T ∧ min{t ≥ 0 | ∀ s ∈ [t, T] 2vL(s, L; α) ≥ vU(s, L; α)} Maximizing process: For t ∈ [0, t∗] λ∗

t = max u∈[0,t]

log Su − log U

+ and for t ∈ [t∗, T] λ∗

t = max u∈[0,t]

log Su − log U

+1{Mt∗>L} + max

u∈[t∗,t]

log Su − log L

+1{Mt∗≤L}

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Sketch of proof for the singular stochastic control problem

1. Step: Reduce the supremum in the result of Cvi-

tani´ c & Karatzas, El Karoui & Quenez to piecewise linear processes λ ∈ L, i. e., sup

λ∈L

Eλ

e−rdT −αλT g(S)
= sup

λ∈Lpl

Eλ

e−rdT −αλT g(S)
.
2. Step: Define for every process λ ∈ Lpl a new

process λ ∈ Lpl (and vice versa), so that Girsanov’s theorem implies Eλ

e−rdT −αλT g(S)
= E
e−rdT −αλT g
Se−λ

.

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Details for the 2. Step Transformation of Trajectories: Consider λ ∈ Lpl. Then there exist m ∈ N and 0 = t0 < t1 < · · · < tm = T such that λ(t, ω) =

m−1

i=0

ai(ω)

(ti+1 ∧ t) − ti

+ for t ∈ [0, T] with FW(ti)-measurable ai: Ω → [0, ∞). Define ϕλ : Ω → Ω by ϕλ(ω)(t) ω(t) + 1 σ λ(t, ω). ϕλ is injective.

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Surjectivity of the Transformation: For surjectivity, take ω ∈ Ω, set ω(0) = 0 and inductively, for i ∈ {0, 1, . . . , m−1} and ti < t ≤ ti+1, ω(t) ω(t) − 1 σ

i

j=0

aj(ω)

(tj+1 ∧ t) − tj

+. Then ω = ϕλ(ω). Define λ(·, ω) λ(·, ϕ−1

λ (ω))

for all ω ∈ Ω. By Girsanov’s theorem: ω ∼ Pλ ⇐ ⇒ ω = ϕλ(ω) ∼ P

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3. Step: Extension of the supremum to all λ ∈ C, i. e.,

sup

λ∈Lpl

E

e−rdT −αλT g
Se−λ

= sup

λ∈C

E

e−rdT −αλT g
Se−λ

. Pointwise approximation of λ ∈ C by piecewise linear processes λn ∈ Lpl; lower semi-continuity is required here for the application of Fatou’s lemma.

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Sketch of proof for the stochastic impulsive control problem Extend to supremum from C to R, i. e., sup

λ∈C

E

e−rdT −αλT g
Se−λ

= sup

λ∈R

E

e−rdT −αλT g∗(S, λ)
.

Pointwise approximation of λ ∈ R by continuous processes λn ∈ C. Turn the jumps at t1, . . . , tI into bounded, non-negative martingales: Mi,n(t) = E[(λ(ti) − λ(ti−)) ∧ n|Ft]. {Ft}t∈[0,T ] is a Brownian filtration, hence there exist continuous modifications of these martingales.