the greeks: derivatives of option prices - millersville...

The Greeks: Derivatives of Option PricesAn Undergraduate Introduction to Financial Mathematics

J. Robert Buchanan

2010

J. Robert Buchanan The Greeks: Derivatives of Option Prices

Sensitivity Analysis

In calculus a derivative gives you a measure of the rate ofchange of a dependent variable as an independentvariable is changed.




In the world a finance an option is an example of aderivative, any financial instrument whose value is derivedfrom that of an underlying security.




In the world a finance an option is an example of aderivative, any financial instrument whose value is derivedfrom that of an underlying security.

Here we will calculate the partial derivatives (in the senseof calculus) of option value formulas. These partialderivatives will allow us to determine how sensitive thevalues of options are to changes in independent variablesand parameters.


Black-Scholes Option Pricing Formulas

w =ln(S/K ) + (r + σ2/2)(T − t)

σ√

T − t

C(S, t) = Sφ(w) − Ke−r(T−t)φ(w − σ√

T − t)

P(S, t) = Ke−r(T−t)φ(σ√

T − t − w) − Sφ(−w)


Cumulative Distribution Function

The function φ(w) is the cumulative distribution function

φ(w) =1√2π

∫ w

−∞e−x2/2 dx


Cumulative Distribution Function

The function φ(w) is the cumulative distribution function

φ(w) =1√2π

∫ w

−∞e−x2/2 dx

which by the Fundamental Theorem of Calculus has derivative

φ′(w) =1√2π

e−w2/2.


Partial Derivatives of w

w =ln(S/K ) + (r + σ2/2)(T − t)

σ√

T − t



w =ln(S/K ) + (r + σ2/2)(T − t)

σ√

T − t∂w∂t

=1

2σ√

T − t

(

ln(S/K )

T − t− r − σ2

2

)



w =ln(S/K ) + (r + σ2/2)(T − t)

σ√

T − t∂w∂t

=1

2σ√

T − t

(

ln(S/K )

T − t− r − σ2

2

)

∂w∂S

=1

σS√

T − t



w =ln(S/K ) + (r + σ2/2)(T − t)

σ√

T − t∂w∂t

=1

2σ√

T − t

(

ln(S/K )

T − t− r − σ2

2

)

∂w∂S

=1

σS√

T − t∂w∂r

=

√T − tσ



w =ln(S/K ) + (r + σ2/2)(T − t)

σ√

T − t∂w∂t

=1

2σ√

T − t

(

ln(S/K )

T − t− r − σ2

2

)

∂w∂S

=1

σS√

T − t∂w∂r

=

√T − tσ

∂w∂σ

=√

T − t − wσ


Theta Θ (1 of 3)

Theta Θ is the partial derivative with respect to time t .

Time is the only independent variable we are certain willchange before expiry. It is also the only deterministicindependent variable.


Theta Θ (2 of 3)

C = Sφ(w) − Ke−r(T−t)φ(w − σ√

T − t)∂C∂t

= Sφ′(w)∂w∂t

− rKe−r(T−t)φ(w − σ√

T − t)

− Ke−r(T−t)φ′(w − σ√

T − t)[

∂w∂t

+σ

2√

T − t

]


Theta Θ (2 of 3)


T − t)∂C∂t

= Sφ′(w)∂w∂t

− rKe−r(T−t)φ(w − σ√

T − t)


T − t)[

∂w∂t

+σ

2√

T − t

]

=Se−w2/2 − Ke−r(T−t)−(w−σ

√T−t)2/2

2σ√

2π(T − t)

(

ln(S/K )

T − t− r − σ2/2

)

− Ke−r(T−t)

(

rφ(w − σ√

T − t) +σe−(w−σ

√T−t)2/2

2√

2π(T − t)

)


Theta Θ (3 of 3)

For a European Put

∂P∂t

=Se−w2/2

2σ√

2π(T − t)

(

ln(S/K )

T − t− r − σ2/2

)

+ Kre−r(T−t)φ(σ√

T − t − w)

− Ke−r(T−t)−(w−σ√

T−t)2/2

2σ√

2π(T − t)

(

ln(S/K )

T − t− r + σ2/2

)


Delta ∆ (1 of 2)

∆ was involved in the derivation of the Black-Scholes PDE andis defined to be the partial derivative with respect to the price ofthe security.


T − t)∂C∂S

= φ(w) +(

Sφ′(w) − Ke−r(T−t)φ′(w − σ√

T − t)) ∂w

∂S


Delta ∆ (1 of 2)



T − t)∂C∂S

= φ(w) +(


T − t)) ∂w

∂S= φ(w)


Delta ∆ (1 of 2)



T − t)∂C∂S

= φ(w) +(


T − t)) ∂w

∂S= φ(w)

Question: what is the range of Delta for a European calloption?


Delta ∆ (2 of 2)

Recall the Put-Call Parity formula:

P + S = C + Ke−r(T−t)


Delta ∆ (2 of 2)


P + S = C + Ke−r(T−t)

∂

∂S(P + S) =

∂

∂S(C + Ke−r(T−t))


Delta ∆ (2 of 2)


P + S = C + Ke−r(T−t)

∂

∂S(P + S) =

∂


∂P∂S

+ 1 =∂C∂S

∂P∂S

= φ(w) − 1


Delta ∆ (2 of 2)


P + S = C + Ke−r(T−t)

∂

∂S(P + S) =

∂


∂P∂S

+ 1 =∂C∂S

∂P∂S

= φ(w) − 1

Question: what is the range of Delta for a European putoption?


Example (1 of 2)

Example

The current price of a stock is $77 and its volatility is 35% peryear. The risk-free interest rate is 3.25% per year. A portfolio isconstructed consisting of one six-month European call optionwith a strike price of $80 and the cash obtained from shorting ∆shares of the stock. The portfolio’s value is non-random. Whatis ∆?


Example (2 of 2)

The assumption the portfolio’s value is non-random is theassumption

(∆)S − C =

(

∂C∂S

)

S − C = 0

made in deriving the Black-Scholes equation.


Example (2 of 2)


(∆)S − C =

(

∂C∂S

)

S − C = 0


S = 77 σ = 0.35 T =612

r = 0.0325 K = 80 t = 0

Using these values

w =ln(S/K ) + (r + σ2/2)(T − t)

σ√

T − t∂C∂S

= ∆ = φ(w)


Example (2 of 2)


(∆)S − C =

(

∂C∂S

)

S − C = 0


S = 77 σ = 0.35 T =612

r = 0.0325 K = 80 t = 0

Using these values

w =ln(S/K ) + (r + σ2/2)(T − t)

σ√

T − t≈ 0.0349666

∂C∂S

= ∆ = φ(w)


Example (2 of 2)


(∆)S − C =

(

∂C∂S

)

S − C = 0


S = 77 σ = 0.35 T =612

r = 0.0325 K = 80 t = 0

Using these values

w =ln(S/K ) + (r + σ2/2)(T − t)

σ√

T − t≈ 0.0349666

∂C∂S

= ∆ = φ(w) ≈ 0.513947


Gamma Γ

Gamma is the second partial derivative with respect to S, thus

Γ =∂

∂Sφ(w)

= φ′(w)∂w∂S

∂2P∂S2 =

∂2C∂S2 =

e−w2/2

σS√

2π(T − t).


Gamma vs. S(0)

50 100 150 200SH0L

0.005

0.010

0.015

G

K = 100, σ = 0.25 T = 1 r = 0.0325


Gamma vs. K and T

60 80 100 120 140

0.0

0.2

0.4

0.6

0.8

1.0

K

T

S(0) = 100, σ = 0.25 r = 0.0325


Relationships Between ∆, Θ, and Γ

Remember the Black-Scholes PDE:

rF = Ft + rSFS +12

σ2S2FSS


Relationships Between ∆, Θ, and Γ

Remember the Black-Scholes PDE:

rF = Ft + rSFS +12

σ2S2FSS

Since Ft = Θ, ∆ = FS , and FSS = Γ then the Black-Scholesequation can be thought of as

rF = Θ + rS∆ +12

σ2S2Γ.


Vega V (1 of 2)

Vega is the partial derivative with respect to volatility σ.


T − t)∂C∂σ

= Sφ′(w)∂w∂σ


T − t)(

∂w∂σ

−√

T − t)


Vega V (1 of 2)

Vega is the partial derivative with respect to volatility σ.


T − t)∂C∂σ

= Sφ′(w)∂w∂σ


T − t)(

∂w∂σ

−√

T − t)

=S√

T − t√2π

e−w2/2


Vega V (2 of 2)

According to the Put-Call Parity formula:

P + S = C + Ke−r(T−t)


Vega V (2 of 2)


P + S = C + Ke−r(T−t)

∂

∂σ(P + S) =

∂

∂σ(C + Ke−r(T−t))


Vega V (2 of 2)


P + S = C + Ke−r(T−t)

∂

∂σ(P + S) =

∂

∂σ(C + Ke−r(T−t))

∂P∂σ

=∂C∂σ

∂P∂σ

=S√

T − t√2π

e−w2/2

Remark: vega is identical for puts and calls.


Example (1 of 2)

Example

Consider a three-month European put option on a stock whosecurrent value is $200 and whose volatility is 30%. The optionhas a strike price of $195 and the risk-free interest rate is6.25%.

1 Find the vega of the option.2 If the volatility of the stock increases to 31%, approximate

the change in the value of the put.


Example (2 of 2)

S = 200 σ = 0.30 T = 3/12

K = 195 r = 0.0625 t = 0


Example (2 of 2)

S = 200 σ = 0.30 T = 3/12

K = 195 r = 0.0625 t = 0

Using these values

w =ln(S/K ) + (r + σ2/2)(T − t)

σ√

T − t

V =S√

T − t√2π

e−w2/2


Example (2 of 2)

S = 200 σ = 0.30 T = 3/12

K = 195 r = 0.0625 t = 0

Using these values

w =ln(S/K ) + (r + σ2/2)(T − t)

σ√

T − t≈ 0.347952

V =S√

T − t√2π

e−w2/2


Example (2 of 2)

S = 200 σ = 0.30 T = 3/12

K = 195 r = 0.0625 t = 0

Using these values

w =ln(S/K ) + (r + σ2/2)(T − t)

σ√

T − t≈ 0.347952

V =S√

T − t√2π

e−w2/2 ≈ 37.5509


Example (2 of 2)

S = 200 σ = 0.30 T = 3/12

K = 195 r = 0.0625 t = 0

Using these values

w =ln(S/K ) + (r + σ2/2)(T − t)

σ√

T − t≈ 0.347952

V =S√

T − t√2π

e−w2/2 ≈ 37.5509

Using the linear approximation,

dP = Vdσ = (37.5509)(0.01) = 0.375509.


Rho ρ (1 of 2)

Rho is the partial derivative with respect to the risk-free interestrate r .


T − t)∂C∂r

= Sφ′(w)∂w∂r

+ K (T − t)e−r(T−t)φ(w − σ√

T − t)


T − t)∂w∂r


Rho ρ (1 of 2)

Rho is the partial derivative with respect to the risk-free interestrate r .


T − t)∂C∂r

= Sφ′(w)∂w∂r

+ K (T − t)e−r(T−t)φ(w − σ√

T − t)


T − t)∂w∂r

= K (T − t)e−r(T−t)φ(w − σ√

T − t)


Rho ρ (2 of 2)

Starting with the Put-Call Parity formula:

P + S = C + Ke−r(T−t)


Rho ρ (2 of 2)


P + S = C + Ke−r(T−t)

∂

∂r(P + S) =

∂

∂r(C + Ke−r(T−t))


Rho ρ (2 of 2)


P + S = C + Ke−r(T−t)

∂

∂r(P + S) =

∂

∂r(C + Ke−r(T−t))

∂P∂r

=∂C∂r

− K (T − t)e−r(T−t)

∂P∂r

= −K (T − t)e−r(T−t)φ(σ√

T − t − w)


Example (1 of 2)

Example

Consider a three-month European put option on a stock whosecurrent value is $200 and whose volatility is 30%. The optionhas a strike price of $195 and the risk-free interest rate is6.25%.

1 Find the rho of the option.2 If the interest rate increases to 7.00%, approximate the

change in the value of the put.


Example (2 of 2)

S = 200 σ = 0.30 T = 3/12

K = 195 r = 0.0625 t = 0


Example (2 of 2)

S = 200 σ = 0.30 T = 3/12

K = 195 r = 0.0625 t = 0

Using these values

w =ln(S/K ) + (r + σ2/2)(T − t)

σ√

T − t

ρ = −K (T − t)e−r(T−t)φ(σ√

T − t − w)


Example (2 of 2)

S = 200 σ = 0.30 T = 3/12

K = 195 r = 0.0625 t = 0

Using these values

w =ln(S/K ) + (r + σ2/2)(T − t)

σ√

T − t≈ 0.347952

ρ = −K (T − t)e−r(T−t)φ(σ√

T − t − w)


Example (2 of 2)

S = 200 σ = 0.30 T = 3/12

K = 195 r = 0.0625 t = 0

Using these values

w =ln(S/K ) + (r + σ2/2)(T − t)

σ√

T − t≈ 0.347952

ρ = −K (T − t)e−r(T−t)φ(σ√

T − t − w) ≈ −20.2315


Example (2 of 2)

S = 200 σ = 0.30 T = 3/12

K = 195 r = 0.0625 t = 0

Using these values

w =ln(S/K ) + (r + σ2/2)(T − t)

σ√

T − t≈ 0.347952

ρ = −K (T − t)e−r(T−t)φ(σ√

T − t − w) ≈ −20.2315

Using the linear approximation,

dP = ρdr = (−20.2315)(0.0075) = −0.151737.


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