Chapter 19

The Black-Scholes-Vasicek Model

The Black-Scholes-Vasicek model is given by a standard time-dependent Black-Scholes

model for the stock price process St, with time-dependent but deterministic volatility σS,t,

and with interest rates r = rt which are assumed to be not constant, but stochastic. That is, under the risk neutral measure, (we ignore dividends)

dSt S = rt dt + σS,t dBt (19.1) St with interest rates given by a mean reverting Ornstein-Uhlenbeck or Vasicek process,

r drt = κ(¯r − rt)dt + σrdBt (19.2)

with a constant interest rate volatility σr. This model is in particular relevant for pricing longer dated options with maturities of 5, 10 or 20 years. The Brownian motions dBS and dBr may have a non vanishing correlation,

S r dBt · dBt = ρ dt (19.3)

The solutions to (19.1) and (19.2) are given by (t > s)

R t 2 R t S (ru−σ /2)du+ σS,udBu St = Sse s S,u s (19.4)

and

−κ(t−s) −κ(t−s) −κt R t κu r rt = e rs +r ¯(1 − e ) + σre s e dBu (19.5)

For the model (19.1,19.2), there is a closed form solution for plain vanilla calls and

puts. For a constant (time independent) stock volatility σS,t = σS, a derivation can be found in [1]. For time dependent volatility the derivation is similar. There is the following

153 154 Chapter 19

Theorem 19.1: Let, under the risk neutral measure, the stock and interest rate processes

(St, rt) be given by (19.1,19.2) and let gκ(t) be given by

1 − e−κt gκ(t) := κ (19.6) Then the following statements hold:

a) Let H = H(ST , rT ) be some non exotic payoff with maturity T . Then:

h R T i − rsds V (St, rt, t, T ) := E e t H(ST , rT ) | St, rt

Z v √ 2 2  S t,T  φ +ξ dφ dξ t − 2 + vt,T φ − 2 = P (t, T ) H P (t,T ) e , rt,T (φ, ξ) e 2π (19.7) R2 where

2 −κ(T −t) σr 2 rt,T (φ, ξ) :=r ¯ + e (rt − r¯) − 2 gκ(T − t)

p  2 1/2  + σr g2κ(T − t) × ρt,T φ + (1 − ρt,T ) ξ (19.8)

R T −κ(T −u) σr 2 ρ t σS,u e du + 2 gκ(T − t) ρt,T := √ p (19.9) vt,T g2κ(T − t) vt,T is given in (b) below and P (t, T ) is the price of a zero bond in the Vasicek model,

R T  − rsds  P (t, T ) := E e t | rt

= A(t, T ) e−gκ(T −t) rt (19.10)

2 2 σr
σr 2 r¯− {gκ(T −t) − [T −t]} − gκ(T −t) A(t, T ) = e 2κ2 4κ (19.11)

b) The price of a plain vanilla european call/put is given by ( = +1 for call,  = −1 for put)

V (St, t, T ) =  {St N(d+) − P (t, T ) KN(d−)} (19.12) where

h St/K i log P (t,T ) ± vt,T /2 d± = √ (19.13) vt,T Z T n 2 2 2o vt,T = σS,u + 2 ρ σr σS,u gκ(T − u) + σr gκ(T − u) du (19.14) t

Remarks: (i) Since interest rates have dimension 1/time, the IR volatility σr has dimen- 3 √ sion 1/time 2 . The stock volatility has dimension 1/ time and κ has dimension 1/time.

Thus vt,T in (19.14) above is indeed a dimensionless quantity as it should be. Chapter 19 155

(ii) For constant stock volatility σS,u = σS =: σ,

−κ(T −u) 2 −κ(T −t) −2κ(T −t) 2 R T 1−e σr 1−e 1−e
vt,T = σ (T − t) + 2ρ σr σ t κ du + κ2 T − t − 2 κ + 2κ

−κ(T −t) 2 −κ(T −t) −2κ(T −t) 2 1 1−e
σr 1−e 1−e
= σ (T − t) + 2ρ σ σr κ T − t − κ + κ2 T − t − 2 κ + 2κ

or, with τ := T − t,

2 τ−gκ(τ) 2 τ−2gκ(τ)+g2κ(τ) v = v(τ) = σ × τ + 2ρ σ σr × κ + σr × κ2 (19.15)

and

σr 2 ρ σ gκ(τ) + 2 gκ(τ) ρt,T = ρ(τ) = p p (19.16) v(τ) g2κ(τ)

Proof of Theorem: We have to compute the expectation

h R T i − rsds V (St, rt, t, T ) = E e t H(ST , rT ) | St, rt (19.17)

h R T  R T 2 R T S  i − rsds (ru−σ /2)du σS,udBu = E e t H St e t S,u e t , rT | St, rt with

R T 2 R T S (ru−σ /2)du+ σS,udBu ST = Ste t S,u t (19.18) −κ(T −t) −κ(T −t) −κT R T κu r rT = e rt +r ¯(1 − e ) + σr e t e dBu (19.19) satisfy the SDE’s

dSt = r dt + σ dBS (19.20) St t S,t t r drt = κ(¯r − rt)dt + σrdBt (19.21)

with correlated Brownian motions

S r dBt · dBt = ρ dt (19.22)

We have

R T 1−e−κ(T −t) 1−e−κ(T −t)
R T 1−e−κ(T −v) r t rudu = κ rt +r ¯ T − t − κ + σr t κ dBv

R T r =: µt,T + σr t gκ(T − v) dBv (19.23) 156 Chapter 19

where we introduced the notation

1−e−κ(T −t)  1−e−κ(T −t)  µt,T = κ rt +r ¯ T − t − κ (19.24)

1−e−κt gκ(t) = κ (19.25)

Introducing further

1 R T 2 ηt,T = 2 t σS,u du (19.26)

and

r dBt ≡ dBt (19.27) S p 2 dBt = ρ dBt + 1 − ρ dZt (19.28)

with two independent Brownian motions dBt and dZt, the expectation (19.17) reads

Z R T −µt,T −σr gκ(T −u) dBu V (St, rt, t, T ) = e e t × (19.29) √  R T R T 2  µt,T −ηt,T +σr gκ(T −u)dBu + σS,u(ρdBu+ 1−ρ dZu) H St e t t , rT dW (B)dW (Z) where dW (B) = dW ({Bu}t

and leave Z unchanged. Then, by Girsanov’s Theorem

−σ R T g (T −u) dB − 1 σ2 R T g (T −u)2du e r t κ u 2 r t κ dW (B) = dW (B˜) (19.31)

such that

1 2 R T 2 −µt,T σr gκ(T −u) du V (St, rt, t, T ) = e e 2 t × (19.32) Z  R T µt,T −ηt,T +σr gκ(T −u)[dB˜u−σrgκ(T −u) du] H St e t ×

n √ o R T σ ρ[dB˜ −σ g (T −u) du]+ 1−ρ2dZ  t S,u u r κ u ˜ e , rT dW (B)dW (Z)

−µ + 1 σ2 R T g (T −u)2du = e t,T 2 r t κ × Z  2 R T 2 R T µt,T −ηt,T −σr gκ(T −u) du−ρσr σS,ugκ(T −u) du H St e t t × √ R T ˜ 2 R T  t {ρσS,u+σrgκ(T −u)}dBu+ 1−ρ t σS,udZu ˜ e , rT dW (B)dW (Z) Chapter 19 157

with rT given by

−κ(T −t) −κ(T −t) R T −κ(T −u) r rT = e rt +r ¯(1 − e ) + σr t e dBu −κ(T −t) −κ(T −t) R T −κ(T −u) ˜
= e rt +r ¯(1 − e ) + σr t e dBu − σrgκ(T − u)du 2 T −κ(T −t) −κ(T −t) σr 2 R −κ(T −u) ˜ = e rt +r ¯(1 − e ) − 2 gκ(T − t) + σr t e dBu

=: qt,T + Yt,T (19.33) where we abbreviated

2 −κ(T −t) −κ(T −t) σr 2 qt,T = e rt +r ¯(1 − e ) − 2 gκ(T − t) (19.34) R T −κ(T −u) ˜ Yt,T = σr t e dBu (19.35) and we used the relation

R T −κ(T −u) R T e−κ(T −u)−e−2κ(T −u) t e gκ(T − u)du = t κ du 1 −κ(T −t) 1−e−2κ(T −t)
= κ2 1 − e − 2 1 −κ(T −t) −2κ(T −t)
= 2κ2 1 − 2e + e 1 2 = 2 gκ(T − t) (19.36)

The random variables

R T ˜ p 2 R T Xt,T := t {ρσS,u + σrgκ(T − u)} dBu + 1 − ρ t σS,udZu (19.37) R T −κ(T −u) ˜ Yt,T = σr t e dBu (19.38) are Gaussian with mean 0 and variances

R T n 2 2 2 o vt,T := V[Xt,T ] = t ρσS,u + σrgκ(T − u) + (1 − ρ )σS,u du (19.39) 2 R T −2κ(T −u) wt,T := V[Yt,T ] = σr t e du 2 = σr g2κ(T − t) (19.40) R T  −κ(T −u) 2 −κ(T −u) Cov(Xt,T ,Yt,T ) = t ρσrσS,u e + σr gκ(T − u)e du 2 R T −κ(T −u) σr 2 = ρσr t σS,u e du + 2 gκ(T − t) (19.41) where we used (19.36) again in the last line. Thus, introducing

2 R T 2 R T Ft,T := µt,T − ηt,T − σr t gκ(T − u) du − ρσr t σS,ugκ(T − u) du (19.42) 158 Chapter 19

and recalling rT = qt,T + Yt,T , we obtain

1 2 R T 2 −µt,T + σr gκ(T −u) du V (St, rt, t, T ) = e 2 t ×   Z 1 −1 Xt,T − (Xt,T ,Yt,T )Ct,T Ft,T +Xt,T
2 Yt,T dXt,T dYt,T H St e , qt,T + Yt,T e det Ct,T 2π R2 −µ + 1 σ2 R T g (T −u)2du = e t,T 2 r t κ × (19.43) Z √ √ 1 −1 x Ft,T + vt,T x
− 2 (x,y)Rt,T ( y ) dx dy H St e , qt,T + wt,T y e det Rt,T 2π R2 (19.44) where   V[Xt,T ] Cov(Xt,T ,Yt,T ) Ct,T = (19.45) Cov(Xt,T ,Yt,T ) V[Yt,T ] denotes the covariance matrix of Xt,T and Yt,T and Rt,T the corresponding correlation matrix,   1 ρt,T Rt,T = (19.46) ρt,T 1 with

Cov(Xt,T ,Yt,T ) ρt,T = p V[Xt,T ]V[Yt,T ] T 2 ρσ R σ e−κ(T −u)du + σr g (T − t)2 = r t S,u 2 κ √ p 2 vt,T σr g2κ(T − t)

R T −κ(T −u) σr 2 ρ t σS,u e du + 2 gκ(T − t) = √ p (19.47) vt,T g2κ(T − t)

Transforming to independent Gaussians,

−µ + 1 σ2 R T g (T −u)2du V = e t,T 2 r t κ × (19.48) Z 2 2  √ √  x +y dx dy Ft,T + vt,T x 2 1/2
− 2 H St e , qt,T + wt,T ρt,T x + (1 − ρt,T ) y e 2π R2

For the special case H ≡ 1, (19.17) and the last equation reduce to

R T  − rudu  V = P (t, T ) = E e t | rt −µ + 1 σ2 R T g (T −u)2du = e t,T 2 r t κ (19.49) Chapter 19 159

With that, we obtain

F µ −σ2 R T g (T −u)2du−ρσ R T σ g (T −u) du−η e t,T = e t,T r t κ r t S,u κ t,T 1 2 R T 2 R T − σ gκ(T −u) du−ρσr σ gκ(T −u) du−η  = e 2 r t t S,u t,T P (t, T ) =: emt,T /P (t, T ) (19.50) where we introduced

1 2 R T 2 R T mt,T := − 2 σr t gκ(T − u) du − ρσr t σS,ugκ(T − u) du − ηt,T (19.51)

From the definition of vt,T in (19.39)

R T n 2 2 2 o vt,T = t ρσS,u + σrgκ(T − u) + (1 − ρ )σS,u du R T  2 2 2 = t σS,u + 2ρσrσS,ugκ(T − u) + σr gκ(T − u) du (19.52) such that, recalling (19.26),

vt,T mt,T = − 2 (19.53)

Thus, now for a general H again, continuing with (19.48),

Z √ 2 2  S  x +y dx dy t mt,T + vt,T x − 2 V (St, t, T ) = P (t, T ) H P (t,T ) e , xt,T (x, y) e 2π R2 Z v √ 2 2  S t,T  x +y dx dy t − 2 + vt,T x − 2 = P (t, T ) H P (t,T ) e , xt,T (x, y) e 2π (19.54) R2 where

√ 2 1/2
xt,T (x, y) = qt,T + wt,T ρt,T x + (1 − ρt,T ) y 2 −κ(T −t) σr 2 =r ¯ + e (rt − r¯) − 2 gκ(T − t) p  2 1/2  +σr g2κ(T − t) × ρt,T x + (1 − ρt,T ) y (19.55)

In particular, for a plain vanilla call H(ST ) = (ST − K)+, the same computation as in the Black-Scholes case (see chapter 6) gives n o St VCall(St, rt, t, T ) = P (t, T ) P (t,T ) N(d+) − KN(d−)

= St N(d+) − P (t, T ) KN(d−) (19.56) where

v h St/K i t,T log P (t,T ) ± 2 d± = √ (19.57) vt,T 160 Chapter 19 and this proves the theorem. 

Calibration of the Model to European Call Options

We assume that the interest rate process has already been calibrated to the initial R T R T yield curve ( by using a suitable shift function 0 rtdt → 0 {rt + ϕ(t)}dt with a piecewise constant ϕ(t) ) as well as to swaptions and caplets. Now we want to fit the model to ATM call options on S with the maturities T1,T2, ..., Tn. Let σimp(Ti) be the implied volatilities of those call options. We assume that the instantaneous stock volatility

σS,t =: σ(t) is piecewise constant on the intervalls [0,T1), [T1,T2), ... ,[Tn−1,Tn]. From the Theorem 19.1 above we find that the implied volatility σBSVas,imp (squared, times maturity) of a plain vanilla call in the BS-Vasicek model with maturity T is given by

2 v0,T ≡ vT = σBSVas,imp(T ) × T (19.58)

R T  2 2 2 = 0 σ(u) + 2ρσrσ(u)gκ(T − u) + σr gκ(T − u) du For piecewise constant

σ(u)|u∈[Ti−1,Ti) =: σi

r the above equation becomes (we write temporarily σ instead of σr for the interest rate vol, to avoid confusion with the equity σi’s)

i X R Tj 2 r r 2 2 vT = {σ(u) + 2ρσ σ(u)gκ(Ti − u) + (σ ) gκ(Ti − u) } du (19.59) i Tj−1 j=1 i X n 2 r R Tj r 2 R Tj 2 o = σ (Tj − Tj−1) + 2ρσ σj gκ(Ti − u)du + (σ ) gκ(Ti − u) du j Tj−1 Tj−1 j=1 with T0 := 0. Now the calibration is done by requiring

model ! market σBSVas,imp(Ti) = σimp (Ti) = σimp (Ti) ≡ σimp(Ti) (19.60) for all i = 1, 2, ..., n. The right hand side of (19.60) is taken from market data, and the left hand side is given by the expression (19.59). Thus we are led to the following system of quadratic equations for σi: For all i = 1, 2, ..., n

i n r r 2 o 2 X 2 ρσ R Tj (σ ) R Tj 2 ∆Tj σ (Ti) = σ + 2σj gκ(Ti − u)du + gκ(Ti − u) du (19.61) imp j ∆Tj Tj−1 ∆Tj Tj−1 Ti j=1 Chapter 19 161

where we put ∆Tj = Tj − Tj−1. From this set, σ1, σ2, ..., σn can be determined inductively starting with σ1: For i = 1, the above equation simplifies to

r 2 2 n 2 ρσr R T1 (σ ) R T1 2 o σ (T1) = σ + 2σ1 gκ(T1 − u)du + gκ(T1 − u) du (19.62) imp 1 ∆T1 T0 ∆T1 T0 which gives

σr R T1 σ1 = −ρ gκ(T1 − u)du ∆T1 T0 q 2 2 σr R T1
2 (σr)2 R T1 2 ± σ (T1) + ρ gκ(T1 − u)du − gκ(T1 − u) du (19.63) imp ∆T1 T0 ∆T1 T0

We have

1 R t 1 1−e−κ(t−s)
t−s s gκ(t − u)du = κ 1 − κ(t−s) (19.64) 1 R t 2 1 1−e−κ(t−s) 1−e−2κ(t−s)
t−s s gκ(t − u) du = κ2 1 − 2 κ(t−s) + 2κ(t−s) (19.65)

For small x := κ(t − s) < 1 we may Taylor expand to obtain

1−e−x 1−(1−x+x2/2−x3/6+O(x4)) x = x x x2 3 = 1 − 2 + 6 + O(x ) (19.66)

Thus,

1 R t 1  x x2 3
 t−s s gκ(t − u)du = κ 1 − 1 − 2 + 6 + O(x ) 1 x x2 3
= κ 2 + 6 + O(x ) t−s 2
= 2 + O κ(t − s) (19.67) 1 R t 2 1   x x2  2x2 3  t−s s gκ(t − u) du = κ2 1 − 2 1 − 2 + 6 + 1 − x + 3 + O(x ) 1 x2 3
= κ2 3 + O(x ) (t−s)2 3
= 3 + O κ(t − s) (19.68)

Realistic data from IR calibration are κ around 10 percent and σr around 0.5 to 1 percent, (at least in pre-financial-crisis times)

κ ≈ 10% , σr ≈ 1% (19.69)

Then, since σimp(T ) is around 20 to 40 percent, for maturities less than, say, 5 years such that κTi ≤ 0.5 to justify a Taylor expansion,

r 2 2 (σ ) R T1 2 2 r 2 2 σ (T1) − gκ(T1 − u) du ∼ σ (T1) − (σ ) T /3 imp ∆T1 T0 imp 1 2 2 (T1/100) ∼ σimp(T1) − 3 (19.70) 162 Chapter 19

should be positive. If this is not the case σ1 = 0 may be a reasonable choice. Now suppose

that σ1, ..., σi−1 have already been determined. Then σi is obtained from

n r r 2 o 2 2 ρσ R Ti (σ ) R Ti 2 ∆Ti σ (Ti) = σ + 2σi gκ(Ti − u)du + gκ(Ti − u) du imp i ∆Ti Ti−1 ∆Ti Ti−1 Ti

+ si−1(σ1, ..., σi−1) (19.71)

where we put

i−1 n r r 2 o X 2 ρσ R Tj (σ ) R Tj 2 ∆Tj si−1 = σ + 2σj gκ(Ti − u)du + gκ(Ti − u) du (19.72) j ∆Tj Tj−1 ∆Tj Tj−1 Ti j=1

If one would substitute in the above expression Ti by Ti−1, one would obtain the expression 2 for σimp(Ti−1). Therefore we abreviate

2 2 ∆σimp,i := σimp(Ti) − si−1(σ1, ..., σi−1) (19.73)

Then,

σr R Ti σi = −ρ gκ(Ti − u)du (19.74) ∆Ti Ti−1 q ∆σ2 r 2 imp,i 2 σr R Ti
2 (σ ) R Ti 2 + Ti + ρ gκ(Ti − u)du − gκ(Ti − u) du ∆Ti ∆Ti Ti−1 ∆Ti Ti−1

if the above expression is a positive real number and σi = 0 otherwise. Finally,

1 R t 1 e−κ(T −t)−e−κ(T −s)
t−s s gκ(T − u)du = κ 1 − κ(t−s) 1 R t 2 1 e−κ(T −t)−e−κ(T −s) e−2κ(T −t)−e−2κ(T −s)
t−s s gκ(T − u) du = κ2 1 − 2 κ(t−s) + 2κ(t−s)