Model-Free Static Replication · Source:Physics and Beyond : Encounters and Conversations (1971) by...
Transcript of Model-Free Static Replication · Source:Physics and Beyond : Encounters and Conversations (1971) by...
Introduction Dirac’s Delta “Function” Static Replication Applications Conclusions
Model-Free Static Replication
Christopher Ting
Christopher Ting
http://www.mysmu.edu/faculty/christophert/
k: [email protected]: 6828 0364ÿ: LKCSB 5036
April 8, 2017
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Introduction Dirac’s Delta “Function” Static Replication Applications Conclusions
Table of Contents
1 Introduction
2 Dirac’s Delta “Function”
3 Static Replication
4 Applications
5 Conclusions
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Introduction
M If a corporate client wants to buy a product that produces adesired payoff at maturity, is there a way to replicate thespecified payoff?
M As a quant, it is your job to figure out a way that satisfiesthe client’s requirements by designing a bespoke solution.
M What is the cost of replication?
M What is the price of the product?
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What is a Unit Step Function?
N Definition
1x>0 :=
1 if x > 0
0 if x ≤ 0.0
1
N Mathematical identity
1x>0 + 1x≤0 = 1.
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Integration of Step Function
N There is nothing sacrosanct about (0, 0). You can shift theorigin to a non-zero number.
N Recall that x+ := max(x, 0).
N Let λ and a be a positive real number. Then∫ λ
−∞1x>a dx = (x− a)+
∣∣∣∣λ−∞
= (λ− a)+.
N Likewise ∫ ∞λ
1x≤a dx = (a− x)+∣∣∣∣∞λ
= (a− λ)+.
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Dirac’s Delta “Function”
N Definition 1 ∫ ∞−∞
δ(x)dx = 1
δ(x) = 0 for x 6= 0.
(1)
N The most important property of δ(x) is exemplified by thefollowing equation,∫ ∞
−∞f(x)δ(x)dx = f(0), (2)
where f(x) is any continuous function of x.
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Dirac’s Delta “Function” (Cont’d)
N By making a shift of origin to a for Dirac’s δ function, wecan deduce the formula∫ ∞
−∞f(x)δ(x− a)dx = f(a). (3)
N The process of multiplying a function of x by δ(x− a) andintegrating over all x is equivalent to the process ofsubstituting a for x.
N The range of integration need not be from −∞ to∞. Anydomain, say the interval (−g2, g1) containing the criticalpoint at which δ(x) does not vanish, will do.
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Dirac’s Alternative Definition of δ(x)
N Consider the differential coefficient ε′(x) of the stepfunction ε(x) given by
1x>0 :=
ε(x) = 1 if x > 0
ε(x) = 0 if x ≤ 0.
(4)
N Substitute ε′(x) for δ(x) in the left side of (3). For positiveg1 and g2, integration by parts leads to∫ g1
−g2f(x)ε′(x) dx = f(x)ε(x)
∣∣∣∣g1−g2−∫ g1
−g2f ′(x)ε(x) dx
= f(g1)−∫ g1
0f ′(x) dx
= f(0)
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Who is Dirac?
Theoretical physicist who predicted the existence of anti-matter.
Picture source: Bubble Chamber
Dirac’s equation
i}γµ∂µψ = mcψ
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Which One is Paul Adrien Maurice Dirac?
Picture source: Paul Dirac and the religion of mathematical beauty
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Mathematics and Beauty
We must admit that religion is ajumble of false assertions, with nobasis in reality. The very idea ofGod is a product of the humanimagination.
— Dirac (1927), atheist
Source: Physics and Beyond : Encounters and Conversations
(1971) by Werner Heisenberg, pp. 85-86
God used beautiful mathematics increating the world.
— Dirac (1963), ex-atheist
Source: The Cosmic Code: Quantum Physics As The Language
Of Nature (2012) by Heinz Pagels, pp. 295
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Conversion by Mathematics
“God is a mathematician of a veryhigh order, and He used veryadvanced mathematics inconstructing the universe.”
The Evolution of the Physicist’s Picture ofNature
Scientific American, May 1963
Source: http://ysfine.com/dirac/dirac44.jpg
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Dirac’s Delta Function in QF
Ý For any payoff function f(S), since δ(x) = δ(−x), we have,for any non-negative S and λ,
f(S) =
∫ ∞0
f(K)δ(K − S)dK
=
∫ λ
0f(K)δ(K − S)dK +
∫ ∞λ
f(K)δ(S −K)dK
Ý Integrating each integral by parts results in
f(S) = f(K)1S<K
∣∣∣∣λ0
−∫ λ
0f ′(K)1S<KdK
− f(K)1S≥K
∣∣∣∣∞λ
+
∫ ∞λ
f ′(K)1S≥KdK
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Replication by Bonds and Options
Ý Integrating each integral by parts once more!
f(S) = f(λ)1S<λ − f ′(K)(K − S)+∣∣∣∣λ0
+
∫ λ
0f ′′(K)(K − S)+dK
+ f(λ)1S≥λ − f ′(K)(S −K)+∣∣∣∣∞λ
+
∫ ∞λf ′′(K)(S −K)+dK
= f(λ) + f ′(λ)[(S − λ)+ − (λ− S)+
]+
∫ λ
0f ′′(K)(K − S)+dK +
∫ ∞λf ′′(K)(S −K)+dK.
= f(λ) + f ′(λ)(S − λ)
+
∫ λ
0f ′′(K)(K − S)+dK +
∫ ∞λf ′′(K)(S −K)+dK.
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Static Replication
Proposition 1
f(ST)= f(λ) + f ′(λ)(ST − λ)
+
∫ λ
0f ′′(K)(K − ST )+dK +
∫ ∞λf ′′(K)(ST −K)+dK.
(5)Ý The payoff f(S) contingent on the outcome S at maturity T
can be replicated by• f(λ): number of risk-free discount bonds, each paying $1 atT
• f ′(λ): number of forward contracts with delivery price λ
• (K − ST )+: European put option’s payoff at T of strike K
• (ST −K)+: European call option’s payoff at T of strike K
• f ′′(λ)dK is the number of put options of all strikes K < λ,and call options of all strikes K > λ
Ý The payoff replication is static, and model-free.Christopher Ting QF 604 Week 3 April 8, 2017 15/20
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Risk Neutral Pricing
Ý The pricing formulas for options is
c0(K) = e−r0TEQ0
[(ST −K
)+]; (6)
p0(K) = e−r0TEQ0
[(K − ST
)+]. (7)
Ý Proposition 2 The price of replication is
EQ0
(f(ST )
)= f(λ)e−r0T
+
∫ λ
0f ′′(K) p0(K)dK +
∫ ∞λf ′′(K) c0(K)dK.
(8)
Note that at time 0, the price of a forward contract is zero.
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Important Example: Natural Log Contract
Ý Let f(x) = ln(x). Then f ′(x) =1
x, and f ′′(x) = − 1
x2. We
set S = ST , and λ = F0. It follows that
ln(ST ) = ln(F0) +1
F0(ST − F0)
−∫ ST
0
(K − F0)+
K2dK −
∫ ∞ST
(F0 −K)+
K2dK
Ý Accordingly,
ln
(STF0
)=
1
F0(ST − F0)−
∫ ST
0
(K − F0)+
K2dK
−∫ ∞ST
(F0 −K)+
K2dK. (9)
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Risk Neutral Expectation for Log Contract
Ý Under the risk neutral measure,
EQ0 (ST ) = F0
Ý Consequently,
EQ0
(1
F0
(ST − F0
))=
1
F0
[EQ0 (ST )− F0
]= 0.
Ý Therefore, under the risk-neutral measure Q, (9) becomes
EQ0
[ln
(STF0
)]= −er0T
∫ ∞F0
c0(K)
K2dK − er0T
∫ F0
0
p0(K)
K2dK.
(10)
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Example: Square Payoff
Ý Suppose the payoff function is f(x) = x2. Then f ′(x) = 2x,and f ′′ = 2.
Ý Moreover, we assume that λ = F0.
Ý By Proposition 1, the payoff function is
f(ST ) = F 20 + 2F0
(ST − F0
)+ 2
∫ F0
0(K − ST )+dK + 2
∫ ∞F0
(ST −K)+dK.
Ý The total cost of static replication is
F 20 e−r0T + 2
∑Ki≤F0
c0(Ki)∆Ki + 2∑
Kj≥F0
p0(Kj)∆Kj
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Example: Takeaways
9 Static replication is one-off effort and easy to do.
9 Model-free approach frees you from model risks.
9 The biggest caveat obviously is the fact that option chainsare not continuous.
9 Most of the options are not liquidly traded and liquidity riskhas to be dealt with carefully.
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