Lecture 12 Sandor Zoltan N´emeth - web.mat.bham.ac.ukweb.mat.bham.ac.uk/S.Z.Nemeth/3g11/presentation12-3g11.pdf · PerpetualAmericanoptions 1 Same as regular American Options but

Mathematical Finance

Lecture 12

Sandor Zoltan Nemeth

University of Birmingham

Autumn

S Z Nemeth (University of Birmingham) Mathematical Finance Autumn 1 / 12

Outline

MATERIAL COVERED: Lecture Notes, page 58

Subection 5.5.3

Exercise Sheet 5, Question 2

1 Perpetual Options


Outline


Subection 5.5.3


1 Perpetual Options

2 American Vanilla Call Option with a Constant Dividend Yield:Equations (reminder)


Outline


Subection 5.5.3


1 Perpetual Options

2 American Vanilla Call Option with a Constant Dividend Yield:Equations (reminder)

3 Solving the American Perpetual Vanilla Call


Perpetual American options

1 Same as regular American Options but T = ∞




2 Not traded because:




2 Not traded because:1 Impossible to predict change of r , σ, D0 in time




2 Not traded because:1 Impossible to predict change of r , σ, D0 in time2 Risk: Writer or underlying company go out of business before option

exercised





exercised

3 Analytical Solutions tested against the corresponding Americanoption with a finite expiry date





exercised


4 Independent of time: If underlying same value today as yesterday,then it should have same value today as yesterday ←− infinite time

available for excercise





exercised




5 Steady state solutions for corresponding American option withfinite time to expiry





exercised




5 Steady state solutions for corresponding American option withfinite time to expiry

6 All ∂

∂tterms are zero and Sf (t) is replaced by Sf = constant


American Vanilla Call Option with a Constant Dividend

Yield: Equations

1 For S > Sf (t):C = S − E

and



Yield: Equations

1 For S > Sf (t):C = S − E

and

1

2σ2S2 ∂2C

∂S2+

∂C

∂t− rC + (r −D0)S

∂C

∂S< 0



Yield: Equations

1 For S > Sf (t):C = S − E

and

1

2σ2S2 ∂2C

∂S2+

∂C

∂t− rC + (r −D0)S

∂C

∂S< 0

2 For S < Sf (t):C ≥ max(S − E , 0)

and



Yield: Equations

1 For S > Sf (t):C = S − E

and

1

2σ2S2 ∂2C

∂S2+

∂C

∂t− rC + (r −D0)S

∂C

∂S< 0

2 For S < Sf (t):C ≥ max(S − E , 0)

and

1

2σ2S2 ∂2C

∂S2+

∂C

∂t− rC + (r −D0)S

∂C

∂S= 0



Yield: Equations 2

3 At S = Sf (t):C = Sf (t)− E

and



Yield: Equations 2

3 At S = Sf (t):C = Sf (t)− E

and

∂C

∂S= 1



Yield: Equations 2

3 At S = Sf (t):C = Sf (t)− E

and

∂C

∂S= 1

4 At S = 0:C = 0



Yield: Equations 2

3 At S = Sf (t):C = Sf (t)− E

and

∂C

∂S= 1

4 At S = 0:C = 0

5 At t = T :C = max(S − E , 0)



Yield: Equations 2

3 At S = Sf (t):C = Sf (t)− E

and

∂C

∂S= 1

4 At S = 0:C = 0

5 At t = T :C = max(S − E , 0)

6 Here D0, r , σ,E ,T = constants



Yield: Equations 2

3 At S = Sf (t):C = Sf (t)− E

and

∂C

∂S= 1

4 At S = 0:C = 0

5 At t = T :C = max(S − E , 0)


7 It is necessary to find C (S , t) for 0 ≤ S ≤ Sf (t) and for−∞ < t ≤ T



Yield: Equations 2

3 At S = Sf (t):C = Sf (t)− E

and

∂C

∂S= 1

4 At S = 0:C = 0

5 At t = T :C = max(S − E , 0)


7 It is necessary to find C (S , t) for 0 ≤ S ≤ Sf (t) and for−∞ < t ≤ T

8 It is also necessary to find Sf (t) for −∞ < t ≤ T


Solving the American Perpetual Vanilla Call with a

Constant Dividend Yield: Exercise Sheet 5, Question 2

1 For S > Sf :C = S − E

and




1 For S > Sf :C = S − E

and

1

2σ2S2d

2C

dS2− rC + (r −D0)S

dC

dS< 0




1 For S > Sf :C = S − E

and

1

2σ2S2d

2C

dS2− rC + (r −D0)S

dC

dS< 0

2 For S < Sf :C ≥ max(S − E , 0)

and




1 For S > Sf :C = S − E

and

1

2σ2S2d

2C

dS2− rC + (r −D0)S

dC

dS< 0

2 For S < Sf :C ≥ max(S − E , 0)

and

1

2σ2S2d

2C

dS2− rC + (r −D0)S

dC

dS= 0


Solving the Perpetual American Vanilla Call with a


3 At S = Sf :C = Sf − E

and




3 At S = Sf :C = Sf − E

and

dC

dS= 1




3 At S = Sf :C = Sf − E

and

dC

dS= 1

4 At S = 0:C = 0




3 At S = Sf :C = Sf − E

and

dC

dS= 1

4 At S = 0:C = 0

5 Here D0, r , σ,E = constants




3 At S = Sf :C = Sf − E

and

dC

dS= 1

4 At S = 0:C = 0


6 It is necessary to find C (S) for 0 ≤ S ≤ Sf




3 At S = Sf :C = Sf − E

and

dC

dS= 1

4 At S = 0:C = 0


6 It is necessary to find C (S) for 0 ≤ S ≤ Sf

8 It is also necessary to find Sf




Insert C ∼ Sλ into

1

2σ2S2 d

2C

dS2+ (r −D0)S

dC

dS− rC = 0





1

2σ2S2 d

2C

dS2+ (r −D0)S

dC

dS− rC = 0

=⇒ 1

2σ2S2

d2CdS2

︷︸︸︷

λ(λ− 1)Sλ−2+(r −D0)S

dC

dS︷︸︸︷

λSλ−1−r

S︷︸︸︷

Sλ = 0





1

2σ2S2 d

2C

dS2+ (r −D0)S

dC

dS− rC = 0

=⇒ 1

2σ2S2

d2CdS2

︷︸︸︷

λ(λ− 1)Sλ−2+(r −D0)S

dC

dS︷︸︸︷

λSλ−1−r

S︷︸︸︷

Sλ = 0

=⇒ 1

2σ2λ(λ− 1)Sλ + (r −D0)λS

λ − rSλ = 0





1

2σ2S2 d

2C

dS2+ (r −D0)S

dC

dS− rC = 0

=⇒ 1

2σ2S2

d2CdS2

︷︸︸︷

λ(λ− 1)Sλ−2+(r −D0)S

dC

dS︷︸︸︷

λSλ−1−r

S︷︸︸︷

Sλ = 0

=⇒ 1

2σ2λ(λ− 1)Sλ + (r −D0)λS

λ − rSλ = 0

=⇒ 1

2σ2λ(λ− 1) + (r −D0)λ− r = 0

Divide by (1/2)σ2:

=⇒ λ(λ− 1) + (k − kD)λ− k = 0,

k = 2rσ2 and kD = 2D0

σ2




λ(λ− 1) + (k − kD)λ− k = 0




λ(λ− 1) + (k − kD)λ− k = 0

=⇒ λ2 + (k − kD − 1)λ− k = 0




λ(λ− 1) + (k − kD)λ− k = 0

=⇒ λ2 + (k − kD − 1)λ− k = 0

∆ = (k − kD − 1)2 + 4k > 0 =⇒ distinct real roots




λ(λ− 1) + (k − kD)λ− k = 0

=⇒ λ2 + (k − kD − 1)λ− k = 0


λ1 > λ2 roots




λ(λ− 1) + (k − kD)λ− k = 0

=⇒ λ2 + (k − kD − 1)λ− k = 0


λ1 > λ2 roots

=⇒ (λ− λ1)(λ− λ2) = λ2 + (k − kD − 1)λ− k




λ(λ− 1) + (k − kD)λ− k = 0

=⇒ λ2 + (k − kD − 1)λ− k = 0


λ1 > λ2 roots

=⇒ (λ− λ1)(λ− λ2) = λ2 + (k − kD − 1)λ− k

=⇒ λ2 − (λ1 + λ2)λ + λ1λ2 = λ2 + (k − kD − 1)λ− k




λ(λ− 1) + (k − kD)λ− k = 0

=⇒ λ2 + (k − kD − 1)λ− k = 0


λ1 > λ2 roots

=⇒ (λ− λ1)(λ− λ2) = λ2 + (k − kD − 1)λ− k

=⇒ λ2 − (λ1 + λ2)λ + λ1λ2 = λ2 + (k − kD − 1)λ− k

=⇒ λ1λ2 = −k < 0 =⇒ λ1 > 0 > λ2




λ(λ− 1) + (k − kD)λ− k = 0

=⇒ λ2 + (k − kD − 1)λ− k = 0


λ1 > λ2 roots

=⇒ (λ− λ1)(λ− λ2) = λ2 + (k − kD − 1)λ− k

=⇒ λ2 − (λ1 + λ2)λ + λ1λ2 = λ2 + (k − kD − 1)λ− k

=⇒ λ1λ2 = −k < 0 =⇒ λ1 > 0 > λ2

=⇒

λ1 = 12

(

1− k + kD +√

(k − kD − 1)2 + 4k)

> 0

λ2 = 12

(

1− k + kD −√

(k − kD − 1)2 + 4k)

< 0




General Solution: C = a1Sλ1 + a2S

λ2





λ2

C = 0 at S = 0





λ2

C = 0 at S = 0

=⇒ 0 = limS→0 a1Sλ1 + a2S

λ2 = a1 × 0+ a2 ×∞





λ2

C = 0 at S = 0

=⇒ 0 = limS→0 a1Sλ1 + a2S

λ2 = a1 × 0+ a2 ×∞

=⇒ 0 =

0︷︸︸︷

a1 × 0+a2 ×∞ = a2 ×∞ =⇒ a2 = 0 =⇒ C = a1Sλ1C = a1Sλ1C = a1Sλ1





λ2

C = 0 at S = 0

=⇒ 0 = limS→0 a1Sλ1 + a2S

λ2 = a1 × 0+ a2 ×∞

=⇒ 0 =

0︷︸︸︷


C = Sf − E at S = Sf =⇒ a1Sλ1f

a1Sλ1f

a1Sλ1f

= C (Sf ) = Sf − E= Sf − E= Sf − E





λ2

C = 0 at S = 0

=⇒ 0 = limS→0 a1Sλ1 + a2S

λ2 = a1 × 0+ a2 ×∞

=⇒ 0 =

0︷︸︸︷



a1Sλ1f

a1Sλ1f

= C (Sf ) = Sf − E= Sf − E= Sf − E

dC

dS= 1 at S = Sf =⇒ λ1a1S

λ1−1|S=Sf= 1 =⇒ λ1a1S

λ1−1f

= 1λ1a1Sλ1−1f

= 1λ1a1Sλ1−1f

= 1





λ2

C = 0 at S = 0

=⇒ 0 = limS→0 a1Sλ1 + a2S

λ2 = a1 × 0+ a2 ×∞

=⇒ 0 =

0︷︸︸︷



a1Sλ1f

a1Sλ1f

= C (Sf ) = Sf − E= Sf − E= Sf − E

dC

dS= 1 at S = Sf =⇒ λ1a1S

λ1−1|S=Sf= 1 =⇒ λ1a1S

λ1−1f

= 1λ1a1Sλ1−1f

= 1λ1a1Sλ1−1f

= 1

{

a1Sλ1f

= Sf − E

λ1a1Sλ1−1f

= 1




{

a1Sλ1f

= Sf − E

λ1a1Sλ1−1f

= 1




{

a1Sλ1f

= Sf − E

λ1a1Sλ1−1f

= 1

Multiply the first equation by λ1 and the second equation by Sf




{

a1Sλ1f

= Sf − E

λ1a1Sλ1−1f

= 1


=⇒{

λ1a1Sλ1f

= λ1(Sf − E )

λ1a1Sλ1f

= Sf




{

a1Sλ1f

= Sf − E

λ1a1Sλ1−1f

= 1


=⇒{

λ1a1Sλ1f

= λ1(Sf − E )

λ1a1Sλ1f

= Sf

λ1(Sf − E ) = Sf =⇒ λ1Sf − λ1E = Sf =⇒ λ1Sf − Sf = λ1E




{

a1Sλ1f

= Sf − E

λ1a1Sλ1−1f

= 1


=⇒{

λ1a1Sλ1f

= λ1(Sf − E )

λ1a1Sλ1f

= Sf

λ1(Sf − E ) = Sf =⇒ λ1Sf − λ1E = Sf =⇒ λ1Sf − Sf = λ1E

=⇒ (λ1 − 1)Sf = λ1E =⇒ Sf =λ1E

λ1 − 1, a1 =

Sf − E

Sλ1f




Thus if S < Sf , then C =

a1︷︸︸︷

Sf − E

Sλ1f

Sλ1





a1︷︸︸︷

Sf − E

Sλ1f

Sλ1

If S ≥ Sf , then C = S − E





a1︷︸︸︷

Sf − E

Sλ1f

Sλ1


Note: If D0 = 0, then kD = 2D0σ2 = 0 which implies

λ1 = 12

(

1− k + kD +√

(k − kD − 1)2 + 4k)

=

12

(

1− k +√

(k − 1)2 + 4k)

= 12

(

1− k +√

(k2− 2k + 1) + 4k)

=

12

(

1− k +√k2 + 2k + 1

)

= 12

(

1− k +√

(k + 1)2)

=

12 (1− k + k + 1) = 1





a1︷︸︸︷

Sf − E

Sλ1f

Sλ1


Note: If D0 = 0, then kD = 2D0σ2 = 0 which implies

λ1 = 12

(

1− k + kD +√

(k − kD − 1)2 + 4k)

=

12

(

1− k +√

(k − 1)2 + 4k)

= 12

(

1− k +√

(k2− 2k + 1) + 4k)

=

12

(

1− k +√k2 + 2k + 1

)

= 12

(

1− k +√

(k + 1)2)

=

12 (1− k + k + 1) = 1

Thus Sf = λ1E

λ1−1 = λ1E

0 = ∞

American Perpetual Vanilla Call Without Dividends −→ NO early exercise


Documents

Lecture 12 Sandor Zoltan N´emeth - web.mat.bham.ac.ukweb.mat.bham.ac.uk/S.Z.Nemeth/3g11/presentation12-3g11.pdf · PerpetualAmericanoptions 1 Same as regular American Options but