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Analytic Approximation

To find an approximation to for small τ we follow the method outlined by Howison. We let

τ = �t′ 0 < � � 1 and assume the solution can be expanded as a series.

Option Pricing Under CEV Models

Call Option

0.1 find the call inner??

0.2 find the call outer ??

dS = Sµdv + σSγdX γ > 1

Vt + σ2S2γ

2 VSS + rSVS − rV

subject to

V (S,T) = max(S −K,0)

V (0, t) = 0

V (∞, t) ∼ S ( S −Ke−r(T−t)

) τ = T − t

Vτ − σ2S2γ

2 VSS − rSVS + rV = 0

Vτ − σ2S2γ

2 VSS−rSVS + rV = 0

V (S,0) = max(S −K,0)

V (0,τ) = 0

S →∞ V (S,τ) ∼ S

(→ S −Ke−rτ)

To find an approximation for small τ

let τ = �t′ 0 < � � 1 and assume a solution of the form

C(S,τ) = ∞∑ i=0

�iCi(S,t ′)

Show

∂C

∂S = ∞∑ i=0

�i ∂Ci ∂S

, ∂2C

∂S2 = ∞∑ i=0

�i ∂2Ci ∂S2

∂C

∂t′ = ∞∑ i=0

�i ∂Ci ∂t′

So DE is

Ct′ − �σ2S2γ

2 VSS − rS�CS + r�C = 0 (1)

Sub into (1)

[ ∂C0 ∂t′

+ � ∂C1 ∂t′

+ · · · ] − �2S2γ

[ ∂2C0 ∂S2

+ � ∂2C1 ∂S2

+ · · · ] −rS�

[ ∂C0 ∂S

+ � ∂C1 ∂S

+ · · · ] +r� [C0 + �C1 + · · · ]

(2)

equating coefficients of ε0 we get

∂C0 ∂t′

= 0 ⇒ C0 = C0(S)

At τ = 0

C0(S,0) = max(S −K,0)

⇒ C0(S,t′) = max(S −K,0)

Equating each coefficient of �′ in 2 we get

∂C1 ∂t1 − ∂2S2γ

∂2C0 ∂S2

− rS ∂C0 ∂S

+ rC0 = 0

C0 = S −k S > K

0 S < K

we get for S > K

∂C1 ∂t1

= rS − r(S −K) = rK

⇒ C1 = rKt′ S > K

and for S < K C0 = 0 so ∂C1 ∂t1

= 0 C1 = 0

So we have

C0(S,t ′) + �C1(S,t

  = S −K + rKt′� K −S � �K

= 0 K −S � �K

This is the outer solution to O(�). However, this solution is not differentiable at S = K. So

we introduce the inner variable

x = K −S � 1 2 K

and scale C to � 1 2 KQ.

Find put use formal put and call parity??

put option

C(t)−P(t) = S(t)−K ·B(t,T)

again, this is not differentiable at S = K. So we introduce the inner variable

x = S −K � 1 2 K

S = x� 1 2 K + K

= K ( 1 + x�

1 2

) and uscale P to �

1 2 KQ

Show (by taking partial derivative) that the DE becomes

K √ �

∂Q

∂t1 = σ2

[ K ( 1 + x�

1 2

)]2γ ×

1 √ �K

∂2Q

∂x2 + rK

( 1 + x�

1 2

) ∂P ∂S − r�

1 2 KQ

⇒ ∂Q

∂t1 = σ2

2 K2γ−2

( 1 + x�

1 2

)2γ ∂2Q ∂x2

+ r √ �(1 + x

√ �) ∂Q

∂x2 + r √ � ( 1 + x

√ � ) ∂Q ∂x − r�Q

which needs to be solved subject to

Q(x,0) = max(−x,0)

lim x→∞

Q(x,t1) = 0

lim x→−∞

Q(x,t1) = −x− rt1� 1 2 + O(�

3 2 )

for a put

p(s,0) = max(k −s,0)

p(∞,τ) = 0

p(0,τ) = ke−rτ

s → 0

x → −1 � 1 2

→−∞�issmall

For a x →−∞ (⇒ S → 0 x →−1/√�

andP → Ke−rτ = K(1− r τ+)

⇒ Q = P √ �V → Ke−rCt

Q ⇒ K (1− r�t1 + · · ·)

√ �K

Q → 1 √ � − rt1

√ � + O(�

3 2 )

Q →−x− rt1 √ � + O(�

3 2 )

now expand

Q(γ,τ) = ∞∑ i=0

� i/2Qi(x,t

so into DE get

∂

∂t1

[ Q0 + �

1 2 Q1 + · · ·

] = σ2

2 K2γ−2

( 1 + x�

1/2 )2γ ∂2

∂x2 [ Q0 + �

1/2Q1 + · · · ]

+ r √ � [ 1 + x

√ � ] ∂ ∂x

[ Q0 + �

1/2Q1 + · · · ]

− r� [ Q0 + �

1/2Q1 + · · · ]

(1 + α) p = 1 + p ·α +

p(p−1) 2!

α2 + · · ·

( 1 + x�

1/2 )2γ

= 1 + 2γx� 1/2 +

(2γ)(2γ −1) 2

x2� + · · ·

equating term of O(1) i.e �0term:

∂Q0 ∂t1

= σ2

2 K2γ−2

∂2Q0 ∂x2

(3)

to be solved subject to

Q0(x,0) = max(−x,0)

lim x→∞

Q0(x,t 1) = 0

lim x→−∞

Q0(x,t 1) = −λ

(4)

To solve this, we use a similarity reduction transformation

Let z = eαx

s = eβt1

ψ0 = e cQ0

( e−αz,e−βt1

) Find α,β,C so that PDE is unchanged

∂Q0 ∂t1

= σ2

2 k2γ−2

∂2Q0 ∂X2

→ Q0t1 = σ2

2 k2γ−2Q0xx

•

Z = eαX → x = e−αZ

S = eβt′ → t′ = eβS

ψ0 = e cQ0 → Q0 = e−cψ0

• Q0t′ =

∂Q0 ∂t′

= ∂

∂t′ ( e−cψ0

) = ec

∂ψ0 ∂t′

= e−ceβ ∂ψ0 ∂S′

= e(β−c) ∂ψ0 ∂S

• Q0x =

∂Q0 ∂x

= ∂

∂x

( e−cψ0

) = e−c

∂ψ0 ∂x

= e−ceα ∂ψ0 ∂Z

= e(α−c) ∂ψ0 ∂Z

• Q0xx =

∂2Q0 ∂x2

= ∂

∂x

( e(α−c)

∂ψ0 ∂Z

) = e(α−c)

∂

∂x

∂ψ0 ∂Z

= e(α−c)eα ∂2ψ0 ∂Z2

= e(2α−c) ∂2ψ0 ∂Z2

e(β−c) ∂ψ0 ∂S

= σ2

2 K2γ−2e(2α−c)

∂2ψ0 ∂Z2

so for the equation to be invariant, we need β = 2α

c is arbitrary, so let c = 0

so our transformation is

X = e−αZ

t′ = e−2αS

Q0 = ψ0

• Q0 ( e−αZ,e−βt′

) = f(η) → η =

e−α

e− β 2

Z √ t′

or lets say η = e(−α+β/2) Z √ t′

= e(−α+ β/2) ·eαX ·eβ/2S−1/2

= eβXS− 1 2 = e

β/2 x√ t′

= f

( e β/2 x√

t′

) Qt′ = f

′(η) ∂η

∂t′

= f ′(η) −eβ/2X 2t′

β/2

Q0x = f ′(η)

∂η

∂x = f ′(η)×

e β/2

√ t′

Q0xx = ∂

∂x

[ f ′(η)×

1 √ t

] = e β/2

√ t f ′′(η)×

∂η

∂x = eβ

t f ′(η)

sub into Q0t′ = σ2

2 K2γ−2Q0xx

−eβ/2X 2t′

3/2 f ′(η) =

σ2

2 K2γ−2

eβ

t f ′′(η)

∴ −eβ/2X 2t′

1/2 f ′(η) =

σ2

2 K2γ−2f ′′(η)

∴ −eβ/2η

2 f ′(η) =

σ2

2 K2γ−2f ′′(η)

S = x� 1 2 K + K

= K ( 1 + x�

1 2

) and uscale P to �

1 2 KQ

Show (by taking partial derivative) that the DE becomes

K √ �

∂Q

∂t1 = σ2

[ K ( 1 + x�

1 2

)]2γ ×

1 √ �K

∂2Q

∂x2 + rK

( 1 + x�

1 2

) ∂P ∂S − r�

1 2 KQ

⇒ ∂Q

∂t1 = σ2

2 K2γ−2

( 1 + x�

1 2

)2γ ∂2Q ∂x2

+ r √ �(1 + x

√ �) ∂Q

∂x2 + r √ � ( 1 + x

√ � ) ∂Q ∂x − r�Q

which needs to be solved subject to

Q(x,0) = max(−x,0)

lim x→∞

Q(x,t1) = 0

lim x→−∞

Q(γ,t1) = −x− rt1� 1 2 + O(�

3 2 )

For a x →−∞ (⇒ S → 0 x →−1/√�

andP → Ke−rτ = K(1− r τ+)

⇒ Q = P √ �V → Ke−rCt

Q ⇒ K (1− r�t1 + · · ·)

√ �K

Q → 1 √ � − rt1

√ � + O(�

3 2 )

Q →−x− rt1 √ � + O(�

3 2 )

now expand

Q(x,τ) = ∞∑ i=0

� i/2Qi(x,t

so into DE get

∂

∂t1

[ Q0 + �

1 2 Q1 + · · ·

] = σ2

2 K2γ−2

( 1 + x�

1/2 )2γ ∂2

∂x2 [ Q0 + �

1/2Q1 + · · · ]

+ r √ � [ 1 + x

√ � ] ∂ ∂x

[ Q0 + �

1/2Q1 + · · · ]

− r� [ Q0 + �

1/2Q1 + · · · ]

(1 + α) p = 1 + p ·α +

p(p−1) 2!

α2 + · · ·

( 1 + x�

1/2 )2γ

= 1 + 2γx� 1/2 +

(2γ)(2γ −1) 2

x2� + · · ·

equating term of O(1) i.e �0term:

∂Q0 ∂t1

= σ2

2 K2γ−2

∂2Q0 ∂x2

(5)

to be solved subject to

Q0(x,0) = max(−x,0)

lim x→∞

Q0(x,t 1) = 0

lim x→−∞

Q0(x,t 1) = −λ

(6)

To solve this, we use a similarity reduction transformation

x = eax?

t′ = ebt′?

Q0 = e cQ?0

sub into

∂Q0 ∂t

( σ2K2γ−2

) ∂2Q0 ∂x2

ec−b ∂Q?0 ∂t′?

= α ∂2Q?o ∂x?

e(c−2a)

so need b = 2a so far c is arbitrary.

As x →−∞→ x? →−∞

need Q0 →−x ecQ?0 →−eax?

To be univariant need a = c

so if a = 1/2, c = 1/2, b = 1

x = e 1/2x?

t′ = e(t′)?

Q0 = e 1/2Q?0

so Q0 = √ t′f(ξ) ξ =

x √ t′

This leads to

σ2K2γ−2 d2f

dξ2 + ξ

dξ −f = 0

Solving subject to BCS gives

Q0 =

√ t′ ·σKα−1 √ 2π

e − x

2t′σ2K2γ−2 − x

2 erfc

( x

√ 2t′ ·σKγ−1

) Then put x =

S −K √ �K

τ = �t′

P = � 1/2KQ0(S,τ)

To go to Q1

Equate terms of O(�1/2)

∂Q1 ∂t′

= σ2K2γ−2

[ ∂2Q1 ∂x2

+ 2γx ∂2Q0 ∂x2

] + r

∂Q0 ∂x

∂Q1 ∂t′

= σ2K2γ−2

∂2Q1 ∂x2

+ σ2K2γ−2γx ∂2Q0 ∂x2

+ r ∂Q0 ∂x

Subject to Q1(x,0) = 0

lim x→∞

Q1(x,t ′) = 0

lim x→∞

Q1(x,t ′) = −rt′

PDE is linear, so Q1 = Q11 + Q12 where

∂Q11 ∂t′

= σ2K2γ−2

∂2Q11 ∂x2

+ r ∂Q0 ∂x

and Q12 satisfies

∂Q12 ∂t′

= σ2K2γ−2

∂2Q1 ∂x2

+ xσ2K2γ−2γ ∂2Q0 ∂x2

For Q11

∂Q1 ∂t′

= σ2K2γ−2

[ ∂2Q1 ∂x2

+ r ∂Q0 ∂x

] (7)

the hit vt = 12vxx + u

let t′′ = αt′ what is α?

∂Q11 ∂t′

= ∂Q1 ∂t′′ × ∂t′′

∂t′ = α

∂Q1 ∂t′′

So in (7)

α ∂Q11 ∂t′′

× 1

σ2K2γ−2 =

[ ∂2Q1 ∂x2

] +

σ2K2γ−2 ∂Q0 ∂x

pick α = σ2K2γ−2

so coeff of ∂Q1 ∂t′′

= 1

Then

∂Q11 ∂t′′

= 1

∂2Q1 ∂x2

+ r

σ2K2γ−2 ∂Q0 ∂x

now using hint, solution is

Q11 = t ′′ ×

σ2K2γ−2 ∂Q0 ∂x

gn terms of original variable

Q11 = σ 2K2γ−2t′ ×

∂Q0 ∂x

then a particular solution is ⇒ v = 1 2 tuxx

For Q12

Q12 = r 2K2γ−2t′ ×γ

∂Q0 ∂x2

the hit vt = 12vxx + xu

t′′ = αt′ α = σ2K2γ−2

∂Q12 ∂t′

= ∂Q1 ∂t′′ · ∂t′′

∂′ = α

∂Q1 ∂t′′

α ∂Q12 ∂t′′

× 1

α =

∂2Q1 ∂x2

+ α

α γK

∂Q0 ∂x

∂Q12 ∂t′′

= 1

∂2Q1 ∂x2

+ Xγ ∂2Q0 ∂x

vt = 1

2 uxx + Xu

∂Q12 ∂t′′

= 1

∂2Q1 ∂x2

+ Xγ ∂Q0 ∂x

then a particular solution is ⇒ v = Xt′u + 1 2 t′′ux

Q0 definition

Q0 =

√ t′kγ−1σ √ 2π

e − x2k2−2γ

2t′σ2 − x

2 erfc

( x ·k1−r

σ √ 2t′

)

erfc(z) = 2 √ π

∫ −∞ z

e−t 2

dt = 1−erf(z)

= 1− 2 √ π

∫ z −∞

e−t 2

erfc

( xk1−γ

σ √ 2t′

) = 1−

2 √ π

∫ xk1−γ σ √ 2t′

−∞ e−t

let t = xk1−γ

σ √ 2t′

dt = k1−γ

σ √ 2t′ dx

erfc(t) = 1− 2 √ π

∫ e−t

2 k1−γ

σ √ 2t′

[ erfc

[ xk1−γ

σ √ 2t′

]]′ = −2 √ π

k1−γ

σ √ 2t′ e−

x2k2−2 2σt′

Find derivative of Q0

dQ0 dx

= −2x k2−2γ

2t′σ2

√ t′kγ−1σ √ 2π

e − x2k2−2γ

2t′σ2 − 1

2 erfc

( xk1−γ

σ √ 2t′

) + x

k1−γ

σ √ 2t′ e − x2k2−2γ

2σ2t′

= −xk1−γ √ 2πt′σ

e − x2k2−2γ

2t′σ2 − 1

2 erfc

( xk1−γ

σ √ 2t′

) +

xk1−γ

σ √ 2πt′

e − x2k2−2γ

2σ2t′

dQ0 dx

= − 1

2 erfc

( xk1−γ

σ √ 2t′

) d2Q0 dx2

= − 1

 −2√

k1−γ

σ √ 2t′ e − x2k2−2γ

2σ2t′

 

d2Q0 dx2

= 1 √ π

k1−γ

σ √ 2t′ e − x2k2−2γ

2σ2t′

d3Q0 dx3

= 1 √ π

k1−γ

σ √ 2t′

( −2xk2−2γ

2σ2t′

) e − x2k2−2γ

2σ2t′

d3Q0 dx3

= −2x √ π

k3−3γ

σ2 (√

2t′ )3e−

x2k2−2γ

2σ2t′

Then solved Q11

Q11 = t ′dq0 dx

Q11 = − 1

2 t′r erfc

( xk1−γ

σ √ 2t′

)

Then solved Q12

Q12 = xt ′r ·

d2Q0 dx2

+ 1

2 t′′ d3Q0 dx3

Q12 = xt ′r ·

1 √ π

k1−γ

σ √ 2t′ e − x2k2−2γ

2σ2t′ + 1

2 t′′ −2x √ π

( k1−γ

σ √ 2t′

)3 e − x2k2−2γ

2σ2t′

Q12 = xγ

2 √ 2π

√ t′σkγ−1e

− x2

2t′σ2k2γ−2

Then find Q1

Q1 = Q11 + Q12

Q1 = − t′r

2 erfc

( xk1−γ

σ √ 2t′

) + xt′r √ π

k1−γ

σ √ 2t′ e − x2k2−2γ

2σ2t′ + 1

2 t′′ ( −2x √ π

)( k1−γ

σ √ 2t′

)3 e − x2k2−2γ

2σ2t′

Q1 = xγ

2 √ 2π

√ t′σkγ−1e

− x2

2t′σ2k2γ−2

Q = Q1 + √ �Q0

Q0 =

√ t′kγ−1 √ 2π

σe − x2k2−2γ

2σ2t′ − x

2 erfc

( xk1−γ

σ √ 2t′

)

Q1 = xγ

2 √ 2π

√ t′σkγ−1e

− x2

2t′σ2k2γ−2

Assumption

x = s−k k √ �

τ = t′�

t′′ = γ2k2γ−2

x = s−k(special case) if � = 1

Q = Q0 + √ �Q1

P = √ �kQ

= √ �k [ Q0 +

√ �Q1 ]

P = √ � ·kQ0 + k�Q1

P = √ � ·kQ0 +

[ − kγτ

2 erfc

( s−k

σ2 √ ξτkγ

) +

(s−k)γσ 2 √ 2π

√ τkγ−1e

−x2 2t′σ2k2γ−2

] let Q0 =

√ t′

√ 2π σkγ−1e

−x2 2t′σ2k2γ−2 −

2 erfc

( x

√ 2t′σkγ−1

) Q0 =

√ τ

� · σ √ 2π kγ−1e

−x2 2t′σ2k2γ−2 −

2 erfc

( x

√ 2t′σkγ−1

)

k √ �Q0 =

√ τkγσ √ 2π

−(s−k)2

k2�2τ � σ2k2γ−2 −

(s−k)2

2k √ � erfc

  s−k k √ � √

2τ � ·σkγ−1

 

√ τkγσ √ 2π

−(s−k)2

2τσ2k2γ − (s−k)2

2k √ � erfc

( s−k σ √ 2τkγ

) sub from (2) into (1)

P =

√ τkγσ √ 2π

−(s−k)2

2τσ2k2γ − (s−k)2

2k √ � erfc

( s−k σ √ 2πkγ

) +

[ kγτ

2 erfc

( s−k

σ2 √ 2τkγ

) +

(s−k)γσ √ τkγ−1

e − (s−k)

2τσ2k2γ

] P =

√ τkγσ √ 2π

[ 1 +

(s−k) k

γ2

] e − (s−k)

2τσ2k2γ − [ (s−k)2

2k √ ξ

+ kγτ

] erfc

( s−k σ √ 2πkγ

) P =

√ τkγ−1σ √ 2π

[ k +

(s−k)γ 2

] e − (s−k)

2τσ2k2γ − [ (s−k)2

2k √ ξ

+ kγτ

] erfc

( s−k σ √ 2πkγ

)