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Guided Tour Inside the Random Walk with

Declensions of Black, Scholes & MertonFrederic Siboulet

fsiboulet(at)nyu(dot)edu

28 March 2016 1

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Objective• The objective of this presentation is to compare and

contrast some of the theories and techniques that may be used to demonstrate either the Black-Scholes-Merton partial differential equation (BSME), or the Black-Scholes-Merton option pricing formulae (BSMF) – or both.

• Considering the extensive work of many scholars, there are at least 12 methods that can lead to the BSME or the BSMF. We limit our presentation to 5 methods, and refer the reader to specialized text books on the matter if required (see for example Paul Wilmott’s Frequently Asked Questions in Quantitative Finance or CQF classes).

• This work borrows extensively from many research papers, finance classes and other conversations. In particular, it is largely inspired by Riaz Ahmad, Peter Carr, Raphael Douady, Randeep Gug, Seb Leo, Fabio Mercurio, Zari Rachev, Paul Wilmott – and many more, all of whom I am grateful to.

28 March 2016 2

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TOC

1. Partial Differential Equation2. Binary Tree and Cox Ross Rubinstein3.1 Martingales and Probabilities3.2 Martingales and Change of Numeraire4. Kolmogorov and Fokker Planck Equations

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General Introduction• BSM: there are at least 12 methods (see Paul Wilmott’s

FAQQF) expressed either in– Equation (Partial Differential Equation, infinitesimal) or– Formulae (Option Pricing, integral)BSM Equation is more general than BSM Formulae

• For derivatives pricing on traded (Equities, Ccies, Futures) or non-traded underlying assets (Interest Rate, Credit Spread)

• Generic or specific formulations (i.e. may or may not be easily generalized)

• In continuous time or discrete time• With the Gaussian distribution assumed - or not • With or without incremental or friction costs, such as bid∆ask

spread, dividend, carry, transaction cost, liquidity premium• With constant, deterministic or stochastic parameters (e.g.

volatility)• Some solutions focus solving for the drift μ, others for the

diffusion σ.28 March 2016 4

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1/ Partial Differential Equation:Classic BSM Equation and Formulae

28 March 2016 5

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Our Assumptions (see Black and Scholes, or Markowitz)• Short selling, Fractional trading always possible• No market friction (such as BΔA spread, transaction fee,

dividend, tax, illiquidity)• Continuous trading and dynamic hedging assumed• At this stage

– Constant rate r for the RFA At=ert

– Constant stock return μ– Constant volatility σ

• The only derivative payoff is at the horizon T (V(T), Euro-style) with no path dependency

• Underlying stock asset (S) follows a GBM and assumed to be traded. In that complete market, any such contingent claim Vt (i.e. derivative) can be replicated @t with a replicating portfolio of 1.Stock (St) & 2.Risk Free Asset (At).

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1/ Partial Differential Equation

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1/ Partial Differential EquationThe risky term (not to be confused with credit risk) of the

PDE is the Brownian Motion (BM), aka Weiner Process (WP). A WP is a Random Process (RP), aka a time dependent Random Variable (RV) iff it has the four following properties:

[ An alternative definition, the Levy Characterization, states that a RP is a WP iff it is an almost surely continuous martingale, with zero origin and quadratic variation [W]t =t ]

is continuous

2*

2*

(1) X(0) 0

=1Xis continuous in time

(2) / w/proba one , ,

0,

(3) , 0, X's Variance is the time interval

,(4) , independent

' 0,

t X tt T

t s X N t s

t sX t s X t

t t

Markovian

of 'property

X t

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1/ Partial Differential Equation1. Assume a Stock S following a GBM – S is the time dependent

RV and the underlying of the derivative, where X is the WP:

2. Construct the portfolio Π, long 1xOption and short Δx stocks –∆ tbd:

3. Move forward in time, from t to t+δt, assuming Δ constant:

4. Itô Lemma (or Taylor expansion order o(dt), with δS2~δt~dt):

28 March 2016 8

, , ,S t V S t S S t

, ,t t t tS t V S t S

Sdt dXdS S

2

2

2 2 2

22

22

2

2 2

2

2

2

dt

1 1 d2

t2

t1

d

2 2

Tay

d

lo

tX

r

do

V Vdt dt

dS

dV dS o

dS

V dSt

S

dSSV dSS

dt S dt

V Vdtt SV S Vd

t

t dtt S

S

S

dX

o

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1/ Partial Differential Equation5. Neglecting the high order terms (HOT), dΠ is s.t. (w/Δ=cte):

6. Now, pick Δ so that Π is risk-free over dt (dynamic hedging argument). There should be no arbitrage, therefore the return of that portfolio over dt should be that of the RFR:

7. And therefore over dt, (5)=(6) gives:

28 March 2016 9

d r dt r V S dt

32 2 2

22

2dV

d dV dS OV d V SS dSS

Vdt dS

ttt

risk free, determini

2 2 2

2

2 2 2

2

2 2 2

ti

2

s

2

2

2

dS dS

V S Vdt dt r V S dtt S

V S Vdt Sdt dt Sdt r V S dtt S

V S V r V S S S dtt S

SdX SdX

V d

S

SV

S

S dS

V

risky, random termc term

0VS dXS

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1/ Partial Differential Equation8. According to the risk-free argument the random term is

null, eliminating the random term dX; therefore write:

9. Plug the new-found value of ∆ into the deterministic term:

10.Which gives the Black Scholes Merton Equation:

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2 2 2

2risk discountfree termdiffusiondrift

BSME2

V S V VrS rVt S S

: 00

SdXS dXV V

S S

2 2 2

2

0

20V S V Vr V S S S

t S SV VS S

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To go from the BSM “Equation” to the BSM “Formulae” for option pricing, solve the BSM PDE:11. Formulate a function U which is the forward value @T of

the present value @t of the derivative V(S,t):

12.Derive V by t and by S and express V as a function of U:

13.Replace V by its expression of U in the LHS of the BSME :

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1/ Partial Differential Equation

"numeraire" @T "numeraire" @t simplified

notation w/o T

, , , ,r T t r T tTU S t e V S t V S t e U S t

22

2 2

,,

, ,, ,

r T t r T t

r T t

r T t r T t

U S tV re U S t et tV S t e U S t

U S t U S tV Ve eS S S S

2 2 2, ,

2

2 2 2

2

BSME2

Backward Kolmogorov EquationBSME 0 BKE

identical to BSME w/o discount 2

r T tV S t e U S tr T t r T t

rV

U S U Ure U e rS rVt S S

U S U UrSt S S

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14. First change of variable, on time t:

15.With ̃ replacing t, the BKE becomes the FKE (or Forward Kolmogorov Equation, aka Fokker Planck Equation):

16.Second change of variable, on the stock S:

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1/ Partial Differential Equation

chain rule

limitsvariable

. . .

;

0parameter

t T

t tt T T t

2 2 2

2

Notice the sign of BSME 0 FKE(1)

the time derivation2

T t U S U UrSS S

chain rule

first order

2 2variable

2 2 2 2second order

. . 1 .

ln. 1 . 1 . 1 . 1 . .

S S SS S

S S S S S S S

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17.With ς replacing S, the FKE(1) becomes FKE(2):

18. Third change of variables:

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1/ Partial Differential Equation

2

2

2 2 2ln

2 2

2 2 2

2

1 1 UBSME 02

Notice the change to the UBSME 0 FKE(2)drift (& diffusion) terms2 2

S

T t

U UUt SS

U S U U rSS S

U U r

2

2

2 2 2 2

2 2

. . . . .2

, . . . .2

. . . .

x rx x

x r xx x

x x x

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19.Replace the old with the new variables into FKE(2) to get W:

20. Therefore finally:

with:

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2

2

2

2 2 2 2, ,t

2

U

,2

02 2 2

FKE(2)

W x U S

UU

x r

W W W Wr rx x x

2

2 2 2 2 2, ,

2 2ln

2(classic) Black Scholes Merton Equation Heat Equation

BKE FKE0

2 2

rtV S t e W x

x S r

T t

V S V V W WrS rVt S S x

1/ Partial Differential Equation

2

2

, , derivative's value @t

correspondence of variables

r

x r

V S t e W x

S e t T

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To value the derivative V(S,t) is equivalent to solve the Heat Equation. Therefore with the fundamental solution, seek the pdf of the Gaussian RP W for the alternate set of variables xand τ.21.May solve the heat equation by similarity reduction, i.e.

transform the PDE (two variables) by an ODE (single variable) by the appropriate change of variable

22. For that new variable z, assume α, β and y three parameters tbd and write:

23.Derive the variable z once wrt τ and once wrt x:

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1/ Partial Differential Equation

3,

, , y / , ]0, ]W x f z

x T x yz

1

1

z x yx yz

zx

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24.Derive the function W once wrt τ and twice wrt x:

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1

11

1

1

2

2

2

'

, '

'1 '

''

''

f zW z f zz

x y f z f z

W x f z zf z f zx yz f zW z

x z xz x yf z

z f zW zx

x z xzf zx

f z

1/ Partial Differential Equation

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25.Go back to the Heat equation and substitute the above .:

26.By similarity reduction, seek to reduce dimensionality – it is sufficient to pick β=1/2 to eliminate τ, hence:

27.And then:

28 March 2016 17

2

2

2 2

2

21 1 2

22

2

21 2

02

' ' '' 02

''

' '' 02

W Wx

W Wx

W zf z f z zf z f z f z

W f zx

f z zf z f z

1

2 221 2' '' 0 ' '' 0

2 2 2zf z zf z f z f z f z f z

1,2

,W x f zx y x y x yz z W x f

1/ Partial Differential Equation

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28.The values of α and β are selected so that, at any time, ταf(z) is pdf if the Normal distribution (since W is the fundamental solution of the heat equation), and therefore:

28 March 2016 18

1/ Partial Differential Equation

12

12

must be constant vav always therefore this term can independentonly also be equal to 1from

10,2

0, , 1

0, , 1

0, , 1

1

1 02

x yz

dx dz

T

T

x yT f dx

T f z dz

12

and is the pdf of the1

normal distribut

α is necessarily

equal to 1/

o

2

i n f

f z dz

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29. From there, f is defined by the following ODE:

28 March 2016 19

2 * by -22

productrule

2

anti-derivative

vav z2

boundaryconditions

2

0' 0

1 ' '' 0 ' '' 02 2 2

'' 0

' Constant

' 0z

z

f zf z

zf z f z f z f z zf z f z

d zf z f zdz

zf z f z

zf z f z

1/ Partial Differential Equation

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30.Continuing the resolution of the ODE:

31.And the constant C2 is such that:

28 March 2016 20

2

2

2 2

2

2

2

2

12

' 0

1

ln 1 C1 constant2

2* 2z

C

df zzf z f z zf z

dzdf z dzfdf zdzf

zf C

f C e C e

2

2

2

2

2

2

11 2* 1 2z

f z dz C e dz C

e d

1/ Partial Differential Equation

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32.By simple change of variable:

33. Therefore, we get the formulation of f:

From here W is the Gaussian distribution of mean y and variance .

28 March 2016 21

2

2

2

2

2 2

2

2

2

112 2 222

212

2

z

e dzC e d

Cz dz d

C

22

2 2

1,

2 22 2

1 1,2 2

yW x f z x yz

yx yzf z e W x e

1/ Partial Differential Equation

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34.At any time between t=0 (inception, τ=T) and t=T (expiration, τ=0), write for the pdf and CDF of that Gaussian RP:

35.Therefore at the τ-zero-plus limit, when the option expires, the three conditions above show that the pdf Wy tends to a y-Dirac delta function:

28 March 2016 22

0

20

Gaussian pdf

, 0 1, 0, ,, 2, ,

0, , , 1 3

y

yy

y

x y W xx Tx y W xW x N y

T W x dx

2

222 0

1, 0, , , lim ,2

0

where, by definition, y-Dirac is st:

1

x y

y y y

y

y y

y

x T W x e W x x

x y x

x x y y

x dx

1/ Partial Differential Equation

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36.According to the sifting (aka sampling) property of Dirac-delta, for a continuous function g(x):

37. Therefore write (note that x and y are interchangeable):

38. For t<T, Wy above is the Gaussian pdf of mean y and variance . At the limit τ=0+ (i.e. t=T -), with g the payoff of the

derivative V, the integral’s limit gives the expected value of the derivative V at expiration, aka that payoff, and:

28 March 2016 23

1/ Partial Differential Equation

g -continous

y-Dirac delta yy

g x x dx g y

2

2

0

2

2

22 , 0

Gaussian pdf N ,

or N x,

1, , lim ,2 x x

x y x y

y x xW y y

y

W x W y e W y g y dy g x

0

2 0

00 ln2 T Tx r S S e

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39.At the boundary for τ=0+:

40. Prior that, for τ>0, integrate a Green function (Gaussian transition pdf) multiplied by the derivative payoff (boundary condition) over all possible values of the underlying stock (state variable) to get the derivative value (solution) @t<T:

28 March 2016 24

1/ Partial Differential Equation

0 0

0

2 0 00

0

aka payoff @t=Taka derivative's value @t=T

aka boundary condition @t=T

lnx ln

,0 , , g2

, ,

, ,

tx

T Tt T x

y T T

r T tt t

t y

T t SS e e S

x r W x U S T V S T e

V S t e U S t

U S t W x

forward derivative's (Green solution) Gaussian pdf derivative's payoff @Tvalue t T value@t (Green function) (boundary condition)

0, , , , ,

0,

r T t yt t y xt T U S t e V S t W x W y g e dy

t

2

222

2

, ,2,

ln2

y xrr y

t y

t

eV S t e W x e g e dyT

x r S T t

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41.Now return to observable variables for the pricing of the derivative between inception and expiration, first write for payoff:

42. Then:

28 March 2016 25

ln y

y

ddy dy g e dy Ge

2

2

22

2

*

22

2

*

22

ln ln2

2

2

ln2

2

2risk free rate

present valuationexp

,2

e2

1 e2

t

t

x yry

t

S r T tr T t

T t

S r T t

r T t T t

eV S t e g e dy

e dGT t

de GT t

present value of the expected payoff

ected value of the payoff wrt lognormal transition probability density function

1/ Partial Differential Equation

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43.Pricing a Euro-Call; for K the strike and Υ the value of the underlying @T, the Call payoff is:

44.Hence write for V(S,t) @t<T, with :

G K

22

2

*

22

2

2

ln2

2

2

ln2... 0

2

2for a Call

ln2

2ln

, e2

e2

e2

t

tK

yt

Sr T t

r T tT t

t

Sr T t

r T tT t

K

e S rd dy r T t

K y K

e dV S t KT t

e dKT t

eT t

2

22

ln

T t y

T t y

Ke K dy

1/ Partial Differential Equation

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45.Split the integral in two parts:

46. For S, t and T known, introduce the variable uε ( 1) s.t.:

47.And the integral boundaries s.t.:

28 March 2016 27

22

2

22

2

ln2

2

2 ln

ln2

2

2 ln

1 e2

, 1 2

2 e2

t

t

S r T t yr T t

T t y

K

tS r T t y

r T tT t

K

e e dyT t

V S t

Ke dyT t

2

ln2tS r T t y

dyu du dy T tduT t T t

2 2

ln ln2 2

ln

t tS Sr T t r T tK K

y K u uT t T t

y u

1/ Partial Differential Equation

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48.Rewrite the second integral, with the variable u-:

Which is the result for the second integral.

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22

2

2

22

2

ln2

2

2 ln

ln

22

22

Gaussian CDF to =d

2 e2

e2

1 e2

t

y

S r T t yr T t

T t

K

K ur T t udy T tdu

uu

u u dur T t r T t

u

Ke dyT t

Ke T tduT t

e K du e KN d

1/ Partial Differential Equation

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49.Now rewrite the first integral:

50.With that first integral, we notice with the numerator of the exponent that (see tedious but straightforward algebra next):

28 March 2016 29

22

2

22

2

222

2

ln2

2

2 ln

ln2

2

2 ln

ln 22

2

2 ln

1 e2

1 e2

1 e2

t

t

t

S r T t yr T t

T t y

K

S r T t y

y r T tT t

K

S r T t y T t y r T t

T t

K

e e dyT t

dyT t

dyT t

222

222

ln 22

ln 2 ln2

t

t t

S r T t y T t y r T t

S r T t y T t S

1/ Partial Differential Equation

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51.Below, brief demonstration of the result above; the intuition is to change the square of a difference in d2 (=u-) to the square of a sum in d1 (=u+) , while brushing lightly over the terms that are identical between d1 to d2:

28 March 2016 30

22

22

22 2keep together

remarkable square identity with diffrence = n

2 2

2

2

l

ln2

ln2

2

2 ln2

t

t

S y r T

t

t

t

S r T t y

S y

T t y r T

r T t S y r T t

t

2 2 24 2

2 2

22 2

22 2

22 22 2

42 22

2

0 0

ln 2 l

2 2

n4 2

ln 2 ln4

2

2

2 22t t

t t

S y r

y

r T t S y r T t

S y r T t S

r r

T t r T t

y T t r T t

r yr T t r

22

remarkable square ident

22

ity

22

with s

222

um = ln2

2

22

2 l

ln 2 ln2

n2

2

2 2

t y r

t

S

t

T t

t

T t

S y S y

S y T t

r

y T t r

r t

t

t T

T

T

20 2

22

ln

22

2 2

2

2 222

ln 2 l2

2 l

n

2ntT t S

t

t tS r T t y

r T t yr T t S

T t

ty t TT

S

1/ Partial Differential Equation

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52.Back to the first integral:

28 March 2016 31

222

2

222

2222

ln 22

2

2 ln

ln 22

lnln 2 ln 2

2

2

11 e2

1 e2

t

t

tt t

S r T t y T t y r T t

T t

K

S r T t y T t y r T t

S r T t yS r T t y T t S

dyT t

T t

22

2

22

2

22

2

2 ln

2

ln

ln2

ln2

2 ln

ln2

2

2 ln

1 e2

e2

t

t

t

t

T t S

T t

K

S r T t y

ST t

K

S r T t y

T ttK

dy

dyT t

S dyT t

1/ Partial Differential Equation

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53. From there now write, with the variable u+:

54. Therefore the BSM formulae for an Euro-Call (same approach for an Euro-Put):

28 March 2016 32

22

2

2

21

1

ln2

2

2 ln

ln

22 u

u2

1

Gaussian CDF to u =d

1 e2

e2

1 e2

t

y

S r T t y

T ttK

K uudy T tdu

t

u

u u d

t t

S dyT t

S T tduT t

S du S N d

1 2, 1 2 , r T tt t tV S t V S t S N d e KN d BSMF

1/ Partial Differential Equation

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55.By solving the PDE, with the final condition aka the Euro-Option Payoff @T, for the Black and Scholes Formulae, get for Euro-Calls and Euro-Puts:

28 March 2016 33

1 2

@ ( )

2

1

2@

2 1

, t

1ln1 2

,

ln2

r T tt t

OptionValuet T

t

T T

Option PayofftT

V S S N d Ke N d

Call S r T tPut Kd

V S T S K T tS r T tK

d d T tT t

1/ Partial Differential Equation

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56.Remarks:• At the infinitesimal level, the BSME is a linear parabolic

differential equation– First order in time– Second order in underlying asset

• At the integrated level, solving BSME requires boundary conditions; one may use:– One time condition (e.g. payoff at T)– Two asset boundary condition (e.g. V for S=0 and S=∞)

• The drift μ of the underlying GBM of S is not a factor in the pricing of the derivative V – only r is. Hence “risk neutral pricing” - a direct consequence of the dynamic hedging in the risk neutral portfolio.

• Another reading of the dynamic hedging argument is that –in a complete market – the value of the derivative can be replicated with cash and ∆ stocks, where ∆ is the sensitivity of the derivative to the underlying asset.

28 March 2016 34

1/ Partial Differential Equation

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57.Variations of the BSME:

28 March 2016 35

1/ Partial Differential Equation

Stocks S with dividends s2

Currency spot C with domestic rate (rd) and foreign rate (rf). C expressed with foreign numeraire and domestic base

2

Commodity Q with cost of carry q 2

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58.Variation of the BSMF (ε=±1, Euro Call and Put)

Stocks S with dividends s

Currency spot C, domestic rate (rd) and foreign rate (rf)

Commodity Q with cost of carry q

2

, t

1 ln2

s T t r T tt t

t

V S S e N d Ke N d

Sd r s T tKT t

1/ Partial Differential Equation

2

, t

1 ln2

f dr T t r T tt t

td f

V S C e N d Ke N d

Cd r r T tKT t

2

, t

1 ln2

q T t r T tt t

t

V S Q e N d Ke N d

Qd r q T tKT t

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2/ Binomial Tree

28 March 2016 37

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2/ Binomial Tree and Time Limit1. At t, the Stock price is St ; move in time from t to t+δt.2. At t+δt, the Stock price may move up to uS or down to vS,

with 0<v<1<u; probability to move up is p, to move down is 1-p (hence “binomial”):

3. Chose u, v and p (not unique) by using the drift μ and the diffusion σ of S (note: that choice will later lead to coincide at the continuous limit δt→0 with the GBM dS=μSdt+σSdX):

28 March 2016 38

1 for t small enough:1 2 2 0<p<11 1 0<1-p<11

2 2

tpu t

v t tp

t t+δt→S

S+=uS

S-=vS

p

1-p

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4. Look at the Option price at V(t+δt); no assumption yet on Option value V(t). Option value at t+δt is either V+ if S(t+δt)=uS(t), or V- if S(t+δt)=vS(t) (note: with S going up, a Call V+ goes up, a Put V+ goes down):

5. At t, construct a portfolio Π(t), long 1xOption and short a fixed value ΔS of the Stock S; go from t to t+δt:

2/ Binomial Tree and Time Limit

28 March 2016 39

t t+δt

VV+

V-

p

1-p

t t+δt→Π

Π+

Π-

p

1-p

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2/ Binomial Tree and Time Limit6. Write the portfolio Π(t) and Π(t+δt) :

7. Chose ∆ so that Π+(t+δt) =Π-(t+δt), i.e. the value of the portfolio at t+δt is indifferent of whether the stock goes up or down between t and t+δt (risk neutral argument):

8. General remark: S(t) is such that S+(t+δt)>S(t)>S-(t+δt) , and:– for a Call, V+(t+δt)>V-(t+δt)therefore Δ>0– for a Put ,V+(t+δt)<V-(t+δt), therefore Δ<0 40

t t t t t V t t S t tt V t S t

t t V t t S t t

t t t t V t t S t t V t t S t t

V t t V t tS t t S t t

V t t V t tu v S t

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2/ Binomial Tree and Time Limit9. The choice of Δ is such that the return of Π is risk-free (again,

the risk neutral argument); therefore:

10. For instance, in the S+ up case Π+(t+δt) on the RHS (or equally Π-=Π+ in the down case) and substituting in Π(t) on the LHS

11.And replacing the value and S+(t+δt)=uS(t), and solving for V(t):

28 March 2016 41

value of portfolio at t+ twith up move of the underlying S

risk free rate portfolio growth over t value at t value of portfolio at t+ t

with down move of the

1

t t

r t tt t

real world expectation of the value of portfolio at t+ t

underlying S

1p t t p t t

1 or resp. =

1 or resp. w/

r t t t t t t

r t V t S t V t t S t t S

1

1

r t V t S t V t t uS t

V t t uS tV t S t

r t

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2/ Binomial Tree and Time Limit12.Replace ∆ in the previous equation with its prior found

value:

13.To lighten up the notation, now remove the explicit reference to the time steps:

28 March 2016 42

1

V t t V t tV t t uS t

V t t V t t u v S tV t S t

u v S t r t

1

11

11

V VV uSu v SV VV S

u v S r t

V V V VV uu v r t u v

V V uV vVu v r t u v

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2/ Binomial Tree and Time Limit14.Replace u and v with the original values:

28 March 2016 43

present

valuation

11

1 1112 2

1 1 111 2

11 2

1 11 2 2

q

V V uV vVV tu v r t u v

t V t VV Vr tt t

r t V V t V t V

r t t

r t t V r t t V

r t t

r tr t

1

risk neutral expectation of the value of thederivative V at t+ t (no drift μ, replaced by r)

1 12 2 1 1

q

qV t tr tV Vr t q V t t

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2/ Binomial Tree and Time Limit15.Where is Black and Scholes? Look at V(S,t), and perform a

Taylor expansion between t and t+δt, denoting ε=±1:

16.Side remark, the expression of a familiar term:

28 March 2016 44

uS or vS, depending on ε=±1

2Taylor2

2

2 2 2

2

2

2

2

2

2

2

21

1 ,

1 12 2

,2

,2

,2

V t S t t

V V VV VS t S t SS t S

V V S VV S t S t t tS t SV V SV S t S t t

Vt S t o tt Sdt

t

S t

o

2

V tS

o t

0

2

2 2t

VS tV V VSStS tS

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2/ Binomial Tree and Time Limit17.Combine risk neutral Option value with Option value Taylor

expansion:

28 March 2016 45

2 2 2

21

2 2 2

2

1 1 1,1 2 2 2 2

, ,2

11 , ,2 2 2

S S t

V

r t r tV S t V Vr t

V V S VV S S t t V S t S t t t o tS t S

r t V V S Vr t V S t V S t S t t tS t S

2 2 2

2

2 2 2

2

δt 0

1 ,2 2 2

1 , , 22 2

V

r t V V S VV S t S t t t o tS t S

V S V r t Vr t V S t V S t t t S t o tt S S

Vt

2 2 2

2 BSME2S V VrS rV

S S

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18.Practical remarks, when constructing trees:– The underlying S is computed forward in time, from S=S0

@t=0– The derivative V is computed backward in time, from the

payoff, aka the final boundary condition– The portfolio Π is computed forward in time, from t=0

19.The construction of the risk neutral portfolio transitions the “Real World” probability p to the “Risk Neutral” probability q.

this is called a “change of measure” (see martingale chapter). Also see Cox Ross Rubinstein next for other possible formulations of p and q.

28 March 2016 46

2/ Binomial Tree and Time Limit

asset numerairedrift drift

-measure, Real World -measure, Risk Neutral

1 12 2 2 2

t r tp q

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2.1/ Binomial Tree withCox-Ross-Rubinstein

28 March 2016 47

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2.1/ Cox-Ross-Rubinstein20.For CRR, go back to the beginning of the tree presentation,

and take the following assumptions on 1/ , and the discount factor D – with RFR r and Stock Vol σ. Make the following Taylor approximations (arbitrary for u and v):

21.Similar steps as the previous tree example (in 2/) result in the LHS below, which writes equivalently as the RHS below:

28 March 2016 48

2

2

1 discount factor on backward t

up probability for S on forward t1

rounded @ o t2

down probability for S on forward t1

rounded @ o t2

r tD e r t o t

u t t

v t t

1 1 1V V t

V V V t t

V V uV vVV D u v V v V u Vu v u v D D D

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22.By δt-order Taylor expansion, write:

23. For the up or down stock movements from t to t+δt write:

24. Lightening up notations as below and with Taylor:

28 March 2016 49

2

2 δt small

22

2

11 1 12

1 22 12

212

o to t

o tTaylor

D r t o t r tv t r tD

Du t t u v t

u t r tu v tDv t t

2.1/ Cox-Ross-Rubinstein

2

22 δt small2 2

1 12

o t o t

Taylor

t tS t t S t S S t t

t t t t

212

2

1,2

,

S t t t SV V VV S t V V V S t S o tS t S

V S S t t V

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25.Take the following expression:

and plug the Taylor expansion on RHS below (ε=±1) in the above :

Which gives on the next page…

28 March 2016 50

2

2

2 2 2

22

2

2 22 2 22

2 2 2 2

2

12

12 2

2 2

S S t t o t

S S t o t

V V VV V S t S o tS t S

V V VV V t t S t S t o tS t S

V V S V S VV V S t t o tS t S S

2.1/ Cox-Ross-Rubinstein

1 1 1u v V v V u VD D D

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26.The three terms below:

28 March 2016 51

1

2

2 2 2 2

21

1 2 01

1 22 2

1

o t

o t

o t

u vD

vD

V

r t t Vu v V

D

VV S tS

t r tV S V S Vv V tD t S S

u VD

2

2 2 2 2

21

1

12

2 2o t

o t

uD

V

VV S tS

t r tV S V S V tt S S

2.1/ Cox-Ross-Rubinstein

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27.And further, with ε=±1:

28 March 2016 52

2.1/ Cox-Ross-Rubinstein

32

2 2 2 2 2

2

22

2 2 2 2 2 32 22

2 2 2

2

2 2 2

o tV V S V S Vt r t V S t tS t S S

VV t r V S tS

V V S V S Vr S tS t S S

V t

32

22

3 3 3 2 2 32

2

2 3 2 2 32 22

2

2 2 2

2 2

1 1 2 2o t

Vr V S tS

V S V V S V S Vr S tS S t S S

V V V S VV t r V S t r S tS t S S

VV t

3 2 2 32

22V S Vr S t

t S S

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28.Therefore, get the BSME as below:

28 March 2016 53

3 32 2

32

3 2 2 32

2

2 2 2

2

2 2 2

1 1 1 0 1 1

1 2 2 22

12

2

o t o t

o t

o t

o t

u v V v V u VD D D

V V S Vr t t V V t r S tt S S

V V S Vr t V V rS tt S S

V V S Vr tV rSt S

2

2 2 20

22

t

tS

V V S VrS rV BSMEt S S

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29.Also note the following results:

30.By writing q as below, interpret q as a probability where the asset drift μ in the p formulation is replaced by RFR r in q:

28 March 2016 54

1/

for δt small enough0<q<1, a probability

1 11

1/ 1/

D vqu v

V V uV vVV Du v u v

qV t tvD uDV V V t Du v u v q V t t

D v u DD V Vu v u v

2

-measure, Real World-measure, Risk Neutral

21/ 1 12 2 4 2 22

o tt r t o t

D v r tq q t pu v t o t

2.1/ Cox-Ross-Rubinstein

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31.Therefore, V(t) is the risk free present value of the risk neutral expectation of V(t+δt) – i.e. the RHS independent of μ:

32.Derivative (backward): in a discrete formulation V is expressed with :

28 March 2016 55

risk free risk neutral probabilityvalue of the value of the derivatdiscount of up move ofderivative @t over t the underlying Swith value of the

between t and t+ tunderlying of S @t

, ,V S t D q V S S t t

ive @t+ t risk neutralprobability value of the derivative @t+ twith up move of the underlying S of down move of with

between t and t+ t the underlying Sbetween t and t+ t

1 ,q V S S t t

down move of the underlying Sbetween t and t+ t

1

1 / 12 2

o t o tr t D v r tD e q S S t

u v

1

for any back time step j the previous time step's value for any node level ifrom N-1 to 0 of the derivative, expected at back time step j

from following time stefrom 0 to j

, , ,N jj i V i j one step risk neutral following time step risk neutralrisk free probability and up node value probabilitdiscount to come p's of the derivative

back downvalue of the derivative

1, 1 1-D q V i j q

following time stepy and down node value

to come of the derivativeback up

, 1V i j

2.1/ Cox-Ross-Rubinstein

1 /j jt t t T t N

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33.Underlying (forward): From origin 0 to any time index j(i.e. at time j*δt=j*(T-t)/N, going forward in time), write by hypothesis for the underlying:

34.At the boundary (final time step N), write the payoff value Vfor time step N (taking the Call case for the demonstration):

35. The discrete formulation is:

36. From N, roll V back in time to the previous time step N-1:

28 March 2016 56

0

any time step j any node level iat time step j

, , , j i j iN jj i S i S u v

0, , N , N i N iNi V i S i K S u v K

1(j 1), , N 1 , 1, N 1- , NNN i V i D qV i q V i

1, , , 1, 1 1- , 1N jj i V i j D qV i j q V i j

2.1/ Cox-Ross-Rubinstein

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37. From N-1, roll back V to the previous time step N-2:

… a pattern which suggest an induction.

28 March 2016 57

2

1,N 1

,N 1

2

2

2

1, N 1(j 2), , N 2 ,

1- , N 1

2, N 1- 1, N

1- 1, N 1- , N

2, N

2 1- 1, N

1- , N

N

V i

V i

qV iN i V i D

q V i

q D qV i q V i

Dq D qV i q V i

q V i

D q q V i

q V i

2

222

0

1 , Nkk k

k

D C q q V i k

2.1/ Cox-Ross-Rubinstein

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38. For the Induction, assume that for time step M-h:

39. From there, prove that (induction):

40. First, change the indexation in the discrete formulation:

41.Above is the starting point of the next development:

28 March 2016 58

0

: , ,

, N 1 , N

N h

hh kh k k

hk

j N h assumed i

V i h D C q q V i k

1

111

10

1 : , ,

, N 1 1 , N

N h

hh kh k k

hk

j N h to prove i

V i h D C q q V i k

1

for any back time step j value of the for any node level iderivative @ i and jand at back time step j

1 1

, , , 1, 1 1- , 1

, , , N 1 1, N h 1-

N j

N N h

j i V i j D qV i j q V i j

h i V i h D qV i q V

, N hi

2.1/ Cox-Ross-Rubinstein

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0 0

1,N h ,N h

1 1

, N (h 1) 1, N 1- , N

induction1 1, N 1- 1 , N

assumption

1

h hh k h kh k k h k k

h hk k

V i V i

hh k kh

V i D qV i h q V i h

D q D C q q V i k q D C q q V i k

D C q q

1

0 0

11 11 1

1

1

0

1

1

factorisation1, N 1 , N

of D, q, 1-q

index boundaries1 , N 1 , N

change

1

on the

,

H

N

1

L S

h hk h kk k

hk k

h hh k h kh k k k k

h

h h

k

kh

hk

kh

h

V i k C q q V i k

D C q q

C q V i h

V i k C q

C q

q V i k

D

1 10

1

1

1

1

0

1 isolate 0 on RHSand h+1 on LH

1 , N ,S

1

, N

N

k k h

k k

hh k

h

h

hh k

hkh

k

k

khC q q V i k C q V i

V

D

q i k

1

1

01

1

10

1

11

1

1

1 0

111 11

1

1, N

1N

N

1 , ,

,

Nk

h

h

h

h

h hh kh k k

hh kk k k

h hk

h h

h

C

h

C

h

C

kD C q q V

C

C C q q V i k i k

V

h

q i

C q V i

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42. Induction’s conclusion:• For h=1:

• And for h=2 (optional):

• And from h to h+1, it was demonstrated that:

• Conclusion: for an Euro-Call (resp. []- payoff for an Euro-Put):

28 March 2016 60

1(j 1), , N 1 , 1, N 1- , NNN i V i D qV i q V i

2

222 2

0(j 2), , N 2 1 , Nkk k

Nk

N i V i D C q q V i k

0

111

10

, N 1 , N,

, N (h 1) 1 , N

hh kh k k

hk

N h hh kh k k

hk

V i h D C q q V i ki

V i D C q q V i k

*

risk freefor any back time step for any node i at a value of the derivative discount backfrom the expiration given time step N-h at node i, time step N-h from ma

, , N hN N hh i V i h D

0 binomial risk neutral boundary value of theprobability mass function derivative at time step Nturity to and at node i+ktime step N-h

1 , Nh

h kk kh

kC q q V i k

2.1/ Cox-Ross-Rubinstein

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43.Now write at the present (h=N) value V(k,N) of a derivative with defined Payoff at the boundary for any k level :

44.With the following interpretation:

the formulation above gives the value of the derivative as the RFR present value of the future RN expected payoff cash flows, aka Fundamental Asset Pricing Formulae, aka Feynman-Kac.

28 March 2016 61

00

1 , NN

N kN k kN

kV D C q q V k

0present valuationvalue of thee binomial risk neutral with risk free ratederivative now probability mass function, iefrom expiration to now

, 1(N period discount)

1

N kk kN

N kN k kN

f B N q k C q q

V D C q q

risk free present value of the derivative's risk neutral ex

0 k-payoff at time step N

of the derivative

derivative's risk neutral expected payoff, ie ,N

, NN

k

V k

V k

pected payoff

2.1/ Cox-Ross-Rubinstein

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45.Now develop the expression as follow:

28 March 2016 62

00

0

00

0 ,N0

0 ,N ,N0

1 , N

1 , N Euro-Call option case

1

1

1 1

NN kN k k

Nk

NN kN k k

Nk

NN kN k k k N k

Nk

NN kN k k k N k

N S k Kk

NN k N kN k k k N k k k

N NS k K S kk

V D C q q V k

D C q q S k K

D C q q S u v K

D C q q S u v K

D C q q S u v K C q q

1

1 1

0

0 ,N ,N0 0

1 2

1 1

N

Kk

N NN kk N kk N k kN NS k K S k K

k kS C Dqu D q v D K C q q

1 1

2.1/ Cox-Ross-Rubinstein

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46.The second term is:

since the expectation of the indicator function is the probability of the indicator condition. Here, the probability is q, under the risk neutral -measure.47. For the first term, notice that:

48. Therefore the first term is under the new -measure:

28 March 2016 63

,N

0 0,N0underlying binomial risk neutral

S, @t=0 probability mass function, ie,D 1

1 1 , N

S k K

N kkkN

N N kkkN S k K

k

f B N qu k C Dqu D q v

S C Dqu D q v S S k K

1

1

1/ 1/1 1 1

1 1 another probability measure,

u Duv Duv vD v D v u vDqu D q v Du Dvu v u v u v u v

D q v Dqu

,N

,Ncallrisk free 0 binomial risk neutral optionpresent probability mass function, iestrikevaluation

, 1

risk neutral expectation,

2 1

S k K

N kk kN

NN kN k k

N S k Kk

f B N q k C q q

D K C q q

1

1 , NND K S k K

2.1/ Cox-Ross-Rubinstein

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49. From there write:

50. From the previous expressions of D, q and u, infer the probability measure implied in o( ) approximation by these results:

Form above, note that the probabilities in the respective measures and the impact of an increase in the asset volatility 28 March 2016 64

0 0 ,N ,N0 0

1 2

0 0

value of value ofcall option underlying

@

binomial cu

t=0 S,

mulativ

t=0

e

@

1 1

, N

N NN kk N kk N k kN NS k K S k K

k kV S C Dqu D q v D K C q q

V S S k K

1 1

Eurodiscount atCallthe risk free rate

striketo t= density cumulative density

with the measure with the me0 asure

, NND K S k K

2.1/ Cox-Ross-Rubinstein

Small δt  Dqu D(1‐q)v q (1‐q) p, (1‐p)

Taylor approx.12 2 2

12 2 2

12 2 2

12 2 2

12 2

Measures ‐measure ‐measure ‐measure

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2.1/ Cox-Ross-Rubinstein51.And from the CLT on the Binomial at the limit for → ∞,

convergence to the normal distribution, from which we get the BSMF:

51.The -measure is the risk neutral measure under the currency numeraire, and N(d2) the probability of exercise of the option under the -measure.

52. The -measure is the risk neutral measure under the asset numeraire, and N(d1) the probability of exercise of the option under the -measure.

28 March 2016 65

0 0 1 2

BSMF for theN

Call Optionr T tV S d e KN d

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3/ Risk Premium andEquivalent Martingale Measures

28 March 2016 66

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3/ Risk Premium & Equivalent -Martingale1. Assume a stock following a GBM:

2. Assume a risk free asset (RFA) such as money market account with constant RFR r:

3. Discount the asset process to t=0 (removing the drift r of the RFA, or time value of money, from the Stock’s drift), and normalize by the initial stock value at t=0. Get the normalized Underlying Security Excess Return (USER) υt:

28 March 2016 67

2

200, ,

tt Xt

t tt

dSt T dt dX S S eS

1 1drift only

no diffusion

0, , 1t

rtt

tt

dA rA dt

t T A ed dA rA dt

A

0 0

is the normalized Underlying 10, , Security Excess Return (USER)

rt t tt

t

S St T eS S A

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4. Definitions and Goals:• A martingale is a driftless Random Process Mt such that:

• Equivalent measures: are two measures which are absolutely continuous v.a.v. each other - or equivalently, for which no impossible event under one measure may be possible under the other measure

• Adapted (Measurable) Process: the Random Process Mt is said to be adapted to the filtration (or measurable wrt t), iff Mt is known given the information set t

• Goal: seek a -equivalent measure , under which υ is a martingale - in other terms, seek an equivalent martingale measure for the USER, υ. We use: 1. Radon-Nikodym Derivative 2. Doléans Exponential and 3. Girsanov’ s theorem. 68

2 1 1 2

denoting thereafter: 0, , 0, ,

t

t t t tt s t t

Mt T s T t

M MM M

3/ Risk Premium & Equivalent -Martingale

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5. The USER follows the RFR adjusted GBM, removing the time value of money (apply Itô):

– Note: if μ=r, then υ is a -martingale, and reciprocally – If μ-r>0 (the natural case - investors expect positive excess

returns) then υ is a -submartingale i.e. ℙ υ υ– If μ-r<0 then υ is a -supermartingale i.e. ℙ υ υ– Assume μ r going forward (i.e. υ is no -martingale)

6. Radon-Nikodym states: if and are equivalent measures, then there exists a RP Λ(t) such that (see RHS below):

2 2

2 2

0

Ito on ln( )

t tt X r t Xrt rtt

t

tt

t

Se e e eS

d r dt dX

such that for anythere exists a random set A of events

process over , , we have

, , , ,A

dA A

3/ Risk Premium & Equivalent -Martingale

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7. The Novikov condition on a RP θt and horizon T is such that:

8. Introduce the Doléans Exponential ε of a RP θt, defined as below:

9. With the Novikov condition, the Doléans exponential is a martingale – actually an exponential martingale:

28 March 2016 70

20 0

12

00, ,

t ts s st ds dX

s st T dX e

20

2

0

12

12which can be stated

Ts

T

s

ds

ds

e

e d

20

12

0 0

martingale conditionNovikov condition

0, ,T

s ds T t

t s s s se t T dX dX

3/ Risk Premium & Equivalent -Martingale

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10.The Girsanov theorem states that:

11.And in that case, the following corollary:

• Radon-Nikodym states the existence of the relationship in the measures. Girsanov expresses the value and the explicit correspondence between the two measures – given θ. In practice, one needs to find an appropriate θ.28 March 2016 71

0thenif there exists a

such that its Doélans'random process θexpectation under is one

for any random process X

, ,

, ,

,

/ 1

/

,

T

s sdX

X

P

0

the Doléans exponential is the Radon Nikodym Derivative

there exists a such-equiv that

ameasure at any ti

t

0

nm

lee

, 0, ,

t

s s

T

t s s

d dX andd

t T

dX

and the Doleans exponential is a martingal

0

e

t

s sdX

0

0, , : driftless ABM on , , a martinga, let

t t s tt T X X ds X

3/ Risk Premium & Equivalent -Martingale

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12. From the -GBM υ in (5) and the -ABM X in (11) write:

13. For υ to be a -martingale, it is sufficient to cancel its drift under . That sufficient condition also gives an explicit formulation for θ:

28 March 2016 72

0

t t t t

t

t t s t t t

tt t

t

tt t

t

d r dt dX

X X ds dX dX dt

d r dt dX dt

d r dt dX

market price of riskaka risk premiumaka the Sharpe ratio

t t t trd dX

3/ Risk Premium & Equivalent -Martingale

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14.Therefore with RND (the Novikov condition is also valid):

15.By changing measure from to :

28 March 2016 73

2

0 0

2

0

Doléans exponential

12

12

0, ,

t tt

t

t

s

RNDr

r rdt dX

r rt X

d rt T t t dXd

e

e

220

expression of X with

220

is a -martingale(no drift, diffusion only)

0, rt tt

tt

t t

ABM

Ser t XSt X

t t t t t

rX t X

t T

d dX e S S e

3/ Risk Premium & Equivalent -Martingale

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16.Remarks: from to , and vice-versa:

17.The Market Price of Risk, aka Risk Premium, aka Sharpe Ratio: – Specifies the Doléans exponential as the Radon Nikodym

Derivative, and – Enables the measure transition from Real-World to Risk-

Neutral, and vice-versa from Risk-Neutral to Real-World with its inverse.

28 March 2016 74

2 2

0 0

2 2

RND, Doléans RND, Doléans

Real RiskWorld Neutral

t tt t t t

t t

t t s t t s

t X t X

X X ds X X ds

d de ed d

r

3/ Risk Premium & Equivalent -Martingale

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18.Replicate our derivative Vt with (1) stock S and (2) risk-free asset At=ert through a self financing portfolio Wt s.t.:

19. The Excess Returns of the Replicating Portfolio ω (RPER) and that of the Derivative ν (DER) are:

28 March 2016 75

qty qty priceprice

Stock RFA FRAStock

Ito integral Riemann integral

0 0 0change in value change in valuebetween 0 and t

portfolio composition @t0, ,

S At t t t t

t tS At t t t t

W S At T

W W dS dA

in S or A

or between S and A - only

no consumption or contribution

0 0

0 0

1 normalized RPER0, ,

1 normalized DER

rt t tt

t

rt t tt

t

W WeW W A

t TV VeV V A

3/ Risk Premium & Equivalent -Martingale

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20. W is the replicating and self financing portfolio, and through the Itô product rule, simplified here with constant r:

28 March 2016 76

1 1

1 10

0

1 1

, replication self financingcondition condition

1

t tt t

t t

tt t t t t t

t

dA rA dt dA rA dt

S A S At t t t t t t t t t

W dW

Wd d W d dA W A dWW A

rA dt S A A dS rA dt

0

1 1

1 1

1

10

delta hedging

0 0

0 0

dynamic hedging driftless martingale

tt

t

tt

t

S St t t t t t

St t t t t

SS A

S St t t t t

d dX

S St t t t t t t

rA dt S A dS

dA S A dS

d A S d S

S Sd d d dXW W

3/ Risk Premium & Equivalent -Martingale

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21.Consider the following Risk-Neutral results:• -Martingale: ωt is a -martingale and Wt is given below:

• -Expectation: further with the Itô integral:

• -Conditional Expectation: by definition of a martingale :

28 March 2016 77

0 00 0

0 0Ito integral

expectation of Ito integral = 0

1 1

0, , 1 constant RN Expectation

t tS St t t t t t t

t

S SdX dXW W

t T

0, , Martingale propertyT t t T tt T

0 0

0 00 0

0 0Ito integral

11

00 00 0

0 deterministic stochastic

RPER is 0, ,d

Martingale

1

tt

t

rtt

t t tS St t t t t t t t

WW At tS rt rt S

t t t t t t t tA e

S St T dX d dXW W

S dX W e W e S dXW

replicating

portfolio

3/ Risk Premium & Equivalent -Martingale

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22.Completeness: the replication condition gives the identity between the replicating portfolio and the derivative, as well as their respective excess returns, including boundaries, over the time period:

23.At t=T, VT=payoff=derivative value=replicating port value WT

28 March 2016 78

0 00 0

1

0, ,payoff

t

t t t t t T

T T T T

W Vt T W V

W V

0

0

is a martingale

replication condition

10, ,

V simplifies

0, ,

derivative's option payoff @T

discount to value @t =

t T t

t

t t

tt

t

t t t T t T

t t T

t t T

Vt TV A

Tt t t

T

t T

VV AA

origin 0

Feynman-Kac

3/ Risk Premium & Equivalent -Martingale

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QUANT SHEET:

79

2

200

0

0

0

1

W is replicating portfolioV is Euro-derivative

1

1

filtration & adaptation

tr t Xt

t t s t

tt

t

rtt

tt

t

r

tt

t

tt

t

t

X X ds S S e

dS dt dX dS

A eS

S A

WW A

VV A

2

2

underlying security excess return

0 00

0 0

dynamichedging replicating portfolio excess return

(complete) r

1

tt X

t t t t

tS St t t t t t t

t t

dX e

S Sd d dXW W

eplication condition

0

Radom Nikodym Derivativeaka Doléans Exponential

RPER Expectation, constant,1

no drift under

RPER Conditional Expectationdiffusion o

t

s s

t

t T t

dt dXd

nly, martingale

Derivative Value @t - (Feynman-Kac)Warning: Euro-style only

Tt t t

T

VV AA

‐m

artin

gales

3/ Risk Premium & Equivalent -Martingale

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24. Now, price derivative V: 1. Equivalent Martingale Measures and

Probability Calculations2. Equivalent Martingale Measures and Change

of Numeraire

28 March 2016 80

3/ Risk Premium & Equivalent -Martingale

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3.1/ Risk PremiumEquivalent Martingale Measures

and Probability Calculations

28 March 2016 81

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25. From Feynman-Kac in (23):

25. From (1) the GBM and (15) change of measure:

26.Therefore, Yt;T follows a normal distribution with:

28 March 2016 82

2

2

2ln2

;

the ln is a normal RandomProces with boundary @T

0, ,

ln ,2

t t

T

T tt

rX X tt t

t tt t

r T t X X

T t

Tt T

t

dS dSt T dt dX rdt dXS S

S S e

S Y N r T t T tS

;

2

;

2

2

T

t

t

TY r T t

T tY

0, ,rt

tA er T tT

t t t t t TT

Vt T V A V e VA

3.1/ EMM & Probability Calculations

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27. From the definition of expectation of Yt;T, where g is the payoff function:

28.The SNRV Z~N(0,1) is such that:

29.Hence:

30. In the case of a Euro-Call:

28 March 2016 83

0, , r T t r T t yt t T tt T V e V e g S e p y dy

2

222

Y r T tZ Y r T t Z T t

T t

22

22

=pdf of SNRV=T-payoff RV @t

2

zr T t z T t

r T tt t

eV e g S e dz

22

22

2

zr T t z T t

r T tT T t t

eg S S K V e S e K dz

3.1/ EMM & Probability Calculations

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31.Notice that:

32. Therefore:

28 March 2016 84

22

2

2

0

0 ln 02

ln2

r T t Z T tt

t

t

SS e K r T t T tZK

S r T tK

Z zT t

22

220

0

22

... 0 22

2

2

z

zr T t z T t

r T tt t

zr T t z T t

r T ttz

eV e S e K dz

ee S e K dz

3.1/ EMM & Probability Calculations

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33.Then:

28 March 2016 85

2 22

0 0

2 2 2

0 0

22

0 0

2

0

2 22

2 2 2

12 2

12

2 2

2 2

2 2

2

z zr T t z T t

r T tt tz z

z zT t z T tr T t

t z z

zz T tr T t

t z z

z T tz

t

e eV e S e dz K dz

e eS dz e K dz

e eS dz e K dz

eS dz

2

0

0,1,100

2

2NN T t

z

zr T t

P Z zP Z z

ee K dz

3.1/ EMM & Probability Calculations

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34. From there:

28 March 2016 86

,1 0,10 0

2

2

1

12

0,10 2

ln2

ln2

ln2

N T t N

t

t

t t

t

N

P Z z P Z z T t

S r T tK

N Z T tT t

S r T tK

N Z N dT

Z

t

V S N d eS r T tK

P Z z NT t

N d

2r T t N dK

BSMF

3.1/ EMM & Probability Calculations

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3.2/ Risk Premium,Equivalent Martingale Measures

with Change of Numeraire

28 March 2016 87

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35.Start from Feynman-Kac in (23); to lighten the notation, look at the Euro-Call (for Euro-Put: opposite payoff or C-P parity):

28 March 2016 88

3.2/ EMM & Change of Numeraire

2

2

0

A deterministic2

0 20 0

0, , Feynman-Kac

Euro Call case

V

tt T

T

Tt t t

T

Tt t

T T T

r T t X Xt

t tT

r T XtrT

Vt T V AA

S KA

A S K

A S e KA

V e S e K

2

20

2

1

@ 0

T

T

T

r T XrT

S K S KrT

t

e S e e K

1 1

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36. For the second term, quite simple (the expectation of the indicator function is the probability):

28 March 2016 89

2

probability of the expectation of indicator conditionthe indicator

20

20

2

ln2

T

T

T TS K S KrT rT rT

r T XrT

rT

S K

K

K

e K e K e K

e K S e

Se K r

1 1

2

20

2

ln2

0

2

,1

T

rT TT

T

d

rT

T X

S r TX Xe K N

T TK

TX

e

T

KN d

3.2/ EMM & Change of Numeraire

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37. Form the first tem:

38.Notice on the RHS the Doléans Exponential, hence a new RND and its corresponding new measure :

39.Change to new measure , with θs=σ:

28 March 2016 90

2 2

0 02 20

T TtTT X dt dX T

tde e dXd

22

expression of X with

220 0

0,

= tt

t t

ABM

r t Xt X

t t

X X t

t T

S A S e S e

2 2

2 20 01

T

T T

TS

r T X T XTS

rK Ke S e S e

1 1

3.2/ EMM & Change of Numeraire

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40.Go back to the first term calculation:

28 March 2016 91

2 2r deterministic(constant)

2 20 0 0

, aka RNDexpectation

0

to change of mea

1T T T

T

T Tr T X T XrT

dS K S K S K

S K

d

de S e S e Sd

dS dd

1 1 1

1

2

0 0 0

-probability of the -probability expectation of the indicator conditionindicator functionsure with RND

20 0

T T TS K

dd

r

S KS d S S

S S e

S K

1 1

1

20

0

20

0 1

0 0

ln2

ln2

1

,1

T

T

T X

T

T TT

d

S

KK

K

SS r T X

S r TX XS T

S

N XT

N

T

d

T

3.2/ EMM & Change of Numeraire

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41.Conclusion:

42.Corollaries:– N is the normal distribution, which is pervasive throughout

the demonstration– N(d2) is the -probability of exercising the option at

maturity, under the risk neutral measure with the currency numeraire

– N(d1) is the -probability of exercising the option at maturity, under the risk neutral measure with the asset numeraire.

28 March 2016 92

10 20rTV S N e Nd K d

3.2/ EMM & Change of Numeraire

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4/ Dual Variables Parameters with Forward Kolmogorov Equation (aka Fokker-Planck Equation) and with

Transition Probability Density Function

28 March 2016 93

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4/ Dual Variables Parameters with FKE1. Fundamental Asset Pricing Formulae of Feynman-Kac for

Euro-Style:

2. Switch around T (expiration) and K (Strike), for given Stand t. For the Call Option case (i.e. with ε=+1):

28 March 2016 94

Feynman-Kac, aka Fondamental Asset Pricing Formulae

risk free risk neutral processrisk neutral discout payoff @Texpectation@t for PV@t

, t (T)T

trdt

t tV S e Payoff

with derterministic

any measure works with FK, provided measure identical for expectation and process

= 1 for Euro-Call or Euro-Put Options

T

trdt

t Te S K

Euro-Call option payoff case

r, t r T t

t t TV S e S K

risk freeparameters variablesvariable parameters transition probabilitycalloptionpresent densityfunctionpayoffvaluation

S , t ; K,T , ; , , = , ; ,r T tt t T T tV V K T S t V K T e S K p S T S t

Risk NeutralExpectation@t

0

(tpdf)

TdS

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4/ Dual Variables Parameters with FKE3. For the Call case write, at given St and t:

4. Derive V w.r.t. K, once & twice:

28 March 2016 95

0

0

...dS 0

, ,

, ,

KT

r T tT T T

r T tT T TK

V K T e S K p S T dS

V K T e S K p S T dS

( ) ( ) 0 ( ) (.) 0

Leibniz integral

ru

0

2

le

2

,= , ,

= ,

,,

,

T

r T tT T T T KK

S K p S and K K p

r T tT TK

r T tT K

r T t

V K Te p S T dS S K p S T

K

e p S T dS

V K Te p S T

Ke p K T

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5. Derive V wrt to T, once:

6. Forward Kolmogorov Equation (forward wrt derivation vav STand T) is such that:

7. Replace the time derivation in (5) RHS, by expression in (6):

2 2 2,

2

, ,, 12

K T T T T TT

T T

S p S T rS p S Tp S TFKE

T S S

, ,, r T t T

T TK

V K T p S TrV K T e S K dS

T T

2 2 2,

2

,1, 2

= ,,

T K T T

r T t TT TK

T T

T

S p S TV K T S

rV K T e S K dST rS p S T

S

4/ Dual Variables Parameters with FKE

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4/ Dual Variables Parameters with FKE8. Simplified notations, with the understanding of the above

variables, and write:

9. In , split the RHS sum under the integral into A and B:

10.And calculate A and B in integration by parts.

28 March 2016 97

2 2 2

2

2

2

1=2

r T t

K

r T t

K

r T t r T t

K

V e S K pdS

S p rSpV rV e S K dST S S

V Ve pdS e pK K

2 2 2

2=2

r T tr T t

K K

A B

S p rSpV erV S K dS e S K dST S S

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4/ Dual Variables Parameters with FKE11.Calculation of A:

28 March 2016 98

2 2 2

integration2 by parts

2 2 2 2 2 2 2

2

2 2 2 2

2 22 2

2 20 0 0 assuming

K

K K

KK

S pA S K dS

SS p S p S p

S K S KS S S S

S p S pA S K dS dS

S S S

S pS K S p

S

K p

2 2

2 2

2+

3+

2 2

22 2

2

0

' 0

r T t r T t

S

S

V Ve p p eK K r T t

S p

S p

K p

VA e KK

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4/ Dual Variables Parameters with FKE12.Calculation of B:

28 March 2016 99

integrationby parts

20 0 assuming S 0

r T t r T t

K K

K

K K

K K

SK

K

V e S K pdS SpdS Ve

rSpB S K dS

SrSp

S K rSp S K rSpS S

B S K rSp dS rSpdSS

S K rSp r SpdS

r SpdS p

r SpdS

K

r T t r T t

K K

K pdSr T t

KV Ve pdS pdS eK K r T t r T t

B rVe rK pdS

VB rVe rKeK

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4/ Dual Variables Parameters with FKE13.As a result:

14.And finally, the terminal BSM equation (notice the opposite “Γ” sign):

28 March 2016 100

22 2

2

2 2 2

2

2 2 2

2

=2

=2

2

2

r T tr T t

r T tr T t r T t r T t r T t

V erV A e BT

e V VrV e K e rVe rKeK K

K V VrV rV rKK K

K V VrKK K

2 2 2

2risk nullfree discountstrikeexpiration strike strikestrike termdiffusiontheta gamma sensitivitydrift

02

V K V VrKT K K

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• Remarks:– The arrival variables satisfy the FKE– The departure variables satisfy the BKE (see the original

BSME)

28 March 2016 101

4/ Dual Variables Parameters with FKE


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