Bigger Models General stock model The market consists of a riskless - - PowerPoint PPT Presentation

Bigger Models General stock model

• The market consists of a riskless cash bond {Bt}t≥0 and a

single risky asset with price process {St}t≥0 governed by dBt = rtBtdt, B0 = 1, and dSt = µtStdt + σtStdWt where {Wt}t≥0 is a P-Brownian motion generating the filtra- tion {Ft}t≥0 and {rt}t≥0, {µt}t≥0 and {σt}t≥0 are {Ft}t≥0- predictable processes.

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• The real meaning of these SDEs is

Bt = exp

t

0 rsds

• ,

and St = S0 exp

t

• µs − 1

2σ2

t

• ds +

t

0 σsdWs

• .
• For these integrals to be well defined, we must assume that

t

0 |rs|ds,

t

0 |µs|ds, and

t

0 σ2 s ds

are all finite with P-probability 1.

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The risk-neutral measure

• The discounted stock price

˜ St = B−1

t

St satisfies d˜ St = (µt − rt)˜ Stdt + σt ˜ StdWt.

• If { ˜

Wt}t≥0 is defined by d ˜ Wt = dWt + µt − rt σt dt, then d˜ St = σt ˜ Std ˜ Wt.

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• By Girsanov’s theorem, if Q is defined by

dQ dP

• Ft

= exp

• −

t

0 γsdWs − 1

2

t

0 γ2 s ds

• with

γt = µt − rt σt , then { ˜ Wt}t≥0 is Q-Brownian motion and {˜ St}t≥0 is a

• Q, {Ft}t≥0
• martingale.
• So Q is the risk-neutral (equivalent martingale) measure.

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• More restrictions:

– For Girsanov’s theorem to apply, we need EP

• exp

t

1 2γ2

s ds

• < ∞;

– To ensure that {˜ St}t≥0 is a

• Q, {Ft}t≥0
• martingale, and

not just a local martingale, we need Novikov’s condition: EQ

• exp

t

1 2σ2

s ds

• < ∞.

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Replicating a claim

• If CT is FT-measurable, define the
• Q, {Ft}t≥0
• martingale

{Mt}t≥0 by Mt = EQ B−1

T CT

• Ft
• .
• By the martingale representation theorem, there exists a pre-

dictable process {θt}t≥0 such that Mt − M0 =

t

0 θsd ˜

Ws =

t

0 φsd˜

Ss where φt = θt σt ˜ St .

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• If

ψt = Mt − φt ˜ St, the portfolio (ψ, φ) is self-financing with value at time t Vt = φtSt + ψtBt = BtMt and in particular VT = BTMT = CT.

• So the self-financing portfolio (ψ, φ) replicates the claim.

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Generalized Feynman-Kac

• Special case: rt and σt (but not necessarily µt) depend on
• nly t and St.
• Then

Vt = F(t, St), where F satisfies ∂F ∂t (t, x)+1 2σ(t, x)2x2∂2F ∂x2 (t, x)+r(t, x)x∂F ∂x (t, x)−r(t, x)F(t, x) = 0.

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• In the yet more special case where rt and σt are functions of
• nly t, and CT = f(ST) is a European claim, the value is the

same as in the constant r and σ case, with: – r replaced by the average ¯ r = 1 T − t

T

t

r(s)ds; – σ2 replaced by the average ¯ σ2 = 1 T − t

T

t

σ(s)2ds.

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