0 ⤠t < infty, ââ
A Black-Scholes Formula for a Market in a Random Environment by Richard J.Criego Nothern Arizona University and Anatoly V.Swishchuk National Academy of Sciences of Ukraine Key Words: random evolution process; Feymann-Kac formula; Black-Scholes formula; European call option; (B,S)-market.
0
Introduction.
A standard model for a securities market is given by the following system: ( dB(t) = rB(t)dt, B(0) > 0 (0.1) dS(t) = µS(t)dt + σS(t)dW (t), S(0) > 0 where B(t) is the (riskless) bond with interest rate r ≥ 0 and S(t) is the risky asset or stock with appreciation rate µ and volatility σ > 0 ; here W(t) is the standard one-dimensional Wiener or Brownian motion process. We call the process ( B(t),S(t)) a (B,S)-market. Suppose that, in addition, the parameters r=r(x),µ = µ(x) and σ = σ(x) depend on some (environmental) parameter x, so that for a fixed value of x, the system (0.1) will depend on x; futheremore, now assume that x varies at random in time according to a finite Markov chain: ( dB(t) = r(x(t))B(t)dt, B(0) > 0 (0.2) dS(t) = µ(x(t))S(t)dt + σ(x(t))S(t)dW (t), S(0) > 0 The Markov chain is thus an additional source of randomness (besides the Brownian motion W(t)) for the market and the resulting process (B(t),S(t)) is called a (B,S)-market in a random environment or (B,S,X)-incomplete market. We will use the theory of random evolutions via an Ito-type formula to obtain a Black-Scholes option price formula for the (B,S,X) -incomplete market assuming a European call option. We can think of the Markov chain as a ”driving process” for the (B,S)market. Random evolution processes driven by Markov chains are studied in [1] and those processes driven by semi-Markov processes were investigated in [4-8]. The hedging of options under a mean-square criterion and with semi-Markov volatility was treated in [6].
1
Random Processes Driven by a Markov Chain
Let X = {Ω, F, Ft , Pi , i ∈ E} be a Markov chain with finite state space E. For simplicity, and without loss of generality, the space E will be taken to be 1
E = {2, 22 , . . . , 2n }; this choice is made since we will be considering below differences of the values in E we wish for a difference 2j − 2i to be a unique number for i 6= j. We also suppose for simplicity that the infintesimal matrix Q = (q2i 2j ), i, j = 1, 2, ..., n, of the Markov chain satisfies: (1.1a)
F or all i ∈ E, q2i 2i ≤ 0, and q2i 2j ≥ 0 f or i 6= j
(1.1b)
F or all i ∈ E, q2i 2i > −∞ (i.e., each state 2i is stable);
(1.1c)
T here exists k so that q2k 2k < 0 (i.e., 2k is a nonabsorbing state); n X q2i 2j = 0 (i.e., Q is conservative).
(1.1d)
j=1
As was done in [1], we construct a Markov chain with infinitesimal matrix Q defined as a stochastic integral with respect to a Poisson random measure. Let U be the ”mark” space U = {aij = 2j − 2i : i 6= j; i, j = 1, 2, ..., n} and on the same common probality space (Ω, F, P ) we construct ”marks” U1 , U2 , . . ., which are independent random variables with a common distribution: P (U = aij ) =
q2 i 2j f or i 6= j; −(q22 + · · · + q2n 2n )
and let Π(B) = P (U ∈ B) be the distribution of each Ui . On the probability space (Ω, F, P ) let {N (t), t ≥ 0} be a standard Poisson process with intensity X λ= q2i 2j = −(q22 + · · · + q2n 2n ); i,j=1 i6=j
note that 0 < λ < ∞ by conditions (1.1b) and (1.1c). We assume that {N (t), t ≥ 0} is independent of U1 , U2 , . . ., which means that {N (t), U1 , U2 , . . . , t ≥ 0, i = 1, 2, ...} is a compound Poisson process. The Poisson random measure associated with the latter process is denoted M (dt×du) where if T ⊂ [0, ∞) and B ⊂ U,then (1.2)
M (T × B, ω) =
∞ X
IT ×B (τi (ω), Ui (ω)),
i=1
where the jump times τ1 , τ2 , ...,of N (t) are such that the increments τk −τk−1 , for k = 1, 2, ..., with τ0 = 0 are independent and have an exponential distribution with parameter λ. We note that X q2i 2j E[M ([0, t] × B)] = λ · t · Π(B) = λ · t · ≡ t · m(B) λ aij ∈B
2
P where we define m(B) ≡ aij ∈B q2i 2j . Let b(t, u, ω) be a real-valued function defined for 0 ≤ t < ∞, u ∈ U, ω ∈ Ω,which is left continuous in t for almost all ω,measurable in u for almost all ω, adapted in ω,that is b(t, u, ·) is Ft -measurable for fixed u. We can define a stochastic integral Z 0Z (1.3) b(s, u)M (ds × du) t
B
for 0 ≤ t < ∞, B ⊂ U, provided for example, Z tZ E[ (b(s, u))2 dsΠ(du)] < ∞. 0
B
The stochastic integral in (1.3) satisfies a useful transformation formula (Itotype formula) [2,p.269], namely, if Z t Z tZ x(t) = x(0) + a(s)ds + b(s, u)M (ds × du) 0
0
U
then, t ∂φ ∂φ [ (1.4) φ(t, x(t)) = φ(0, x(0)) + + a(s) ]M (ds × du)+ ∂s ∂x 0 Z tZ + [φ(s, x(s−) + b(s, u)) − φ(s, x(s−))]M (ds × du)
Z
0
U
where a(·, ω) is a measurable real-valued function for 0 ≤ t < ∞ and a(t, ·) is adapted, b(t, ω) is above and φ(t, x) is a real-valued function defined for 0 ≤ t < inf ty, −∞ < x < ∞, which is bounded for finite t and x, and so that ∂φ ∂φ ∂t and ∂x are contuous. The following result follows [1, Theorem 1.1], which we prove for compeleness. Theorem 1. Let Q = (q2i 2j ) be a matrix that satisfies (1.1) and let M (dt × du) be the associated Poisson random measure defined in (1.2). Define the random process {x(t), t ≥ 0} by the stochastic integral equation: Z tZ (1.5) x(t) = x(0) + b(x(s−), u)M (ds × du) 0
U
where (1.6)
b(x, u) ≡
n X
u · I{2i } (x) · I{aij ∈U:i6=j} (u),
i=1
for x ∈ E = {2, 22 , ..., 2n }, u ∈ U = {aij = 2i − 2j : i 6= j, i, j = 1, 2, ..., n}. Then, X = {x(t), t ≥ 0} is a Markov chain with infitesimal matrix Q. 3
Proof. Let f : E → R = (−∞, ∞) and substituting f (x) for φ(t, x) in (1.4) we obtain Z tZ f (x(t)) = f (x(0)) + [f (x(s−) + b(x(s−), u)) − f (x(s−))]M (ds × du). 0
U
Defining ν([0, t] × B) ≡ tm(B) = E[M ([0, t] × B)] we have that ηt (B) = M ([0, t] × B) − ν([0, t] × B) is a zero-mean martingale for each B, and thus each side of the following expression is a zero-mean martingale: Z tZ (1.7) f (x(t)) − f (x(0)) − [f (x(s−) + b(x(s−), u)) − f (x(s−))]m(du)ds 0 U Z tZ = [f (x(s−) + b(x(s−), u)) − f (x(s−))](M − ν)(du × ds) 0
U
Dididing each side of (1.7) by t, taking the expectation E2i of each side and taking the limit as t → 0 we obtain that if A is the infinitesimal generator of the Markov chain, then Z i A(f (2 )) = [f (2i + b(2i , u)) − f (2i )]m(du) U
=
Z
[f (2i +
U
=
=
=
=
n X j=1 j6=i n X j=1 j6=i n X
j=1 j6=i n X
n X
u · I2k (2i ) · I{akj ∈U:k6=j,j=1,...,n} (u)) − f (2i )]m(du)
k=1
[f (2i + aij ) − f (2i )]q2i 2j
[f (2j ) − f (2i )]q2i 2j
f (2j )q2i 2j − f (2i )
n X
q 2i 2j
j=1 j6=i
f (2j )q2i 2j .
j=1
Thus, the infinitesimal generator A is given by the matrix Q = (q2i 2j ). Our next step is to let Yx = {yx (t), t ≥ 0} be a diffusion processon R for each x ∈ E = {2, 22 , ..., 2n }, that is determined by the stochastic differential equation: dyx (t) = µ(x, yx (t))dt + σ(x, yx (t))dW (t), yx (0) = 0 for a fixed Brownian motion process W = {W (t), t ≥ 0} where for each x, µ(x, ·) and σ(x, ·) are real-valued cntinuous functions that staisfy a Lipschitz condition: (1.8)
|µ(x, y) − µ(x, y‘)| + |σ(x, y) − σ(x, y‘)| ≤ K · |y − y‘|. 4
Thus, Yx = {yx (t), t ≥ 0} is well-defined. Let X be a Markov chain given by (1.5) and assume that the process W and X are defined on the same probability space with respect to the same filtration and are independent. Now let the process Y = {y(t), t ≥ 0} be defined by (1.9)
dy(t) = µ(x(t), y(t))dt + σ(x(t), y(t))dW (t), y(0) = 0
The process Y describes the position y(t) of a random relay race among the processes Yx , whereby if x(0) = x0 , then the baton is handed initially to the process yx0 (t), and this process evoloves in time until the first jump of the Markov chain x(·) to a new state x1 ; the baton is then passed to the process yx1 (t) which begins at the point at which yx0 (t) left off; the process yx1 (t) evoloves until the next jump of the Markov chain; the process continues in this manner. The position of this process at time t is y(t). While {y(t), t ≥ 0} itself is not a Markov process, the joint process z(t) = (x(t), y(t)) is Markov process as shown in the following result [1, Theorem 2.1]. Theorem 2. Let {x(t), t ≥ 0} be the Markov chain defined and let {y(t), t ≥ 0} be defined by (1.9). Then Z ≡ {z(t) = (x(t), y(t)), t ≥ 0} is a uniqly defined Markov process with state space E × R and with infintesimal operator L given by (1.10)
Lf (x, y) = Ax f (x, y) + Qf (x, y),
where (1.11)
Ax f (x, y) = µ(x, y)
∂ 2 f (x, y) ∂f (x, y) 1 2 + σ (x, y) ∂y 2 ∂y 2
Qf (x, y) =
n X
q2i 2j f (2j , y).
j=1
Proof. The process Z can be written in vector form as follows: Z t Z t Z tZ ¯b(z(s), u)M ¯ (s) + ¯ (ds × du) µ ¯(z(s))ds + σ ¯ (z(s))dW z(t) = z(0) + 0
0
0
U
¯ (t) = (0, W (t)), where we let µ ¯(x, y) = (0, µ(x, y)), σ ¯ (x, y) = (0, σ(x, y)), W ¯b((x, y), u) = (b(x, u), 0), M ¯ (ds × du) = (M (ds × du), 0). Now, the existance and uniqueness of Z is given by [2,p.273] and, furthemore, Z is a Markov process with state space E×R [2, p.288]. Furthemore, an Ito-type formula for Z [2,p.269] yeilds, in the same manner as above, that the infintesimal generator L of Z is given by n
Lf (x, y) = µ(x, y)
∂f (x, y) 1 2 ∂ 2 f (x, y) X + σ (x, y) + q2i 2j f (2j , y). ∂y 2 ∂y 2 j=1
We call Z = {z(t) = (x(t), y(t)), t ≥ 0} the random evolution process obtained from the (driven)processes {yx (t), t ≥ 0} with process {x(t), t ≥ 0}. 5
2
A Feynmann-Kac Formula for the Random Evolution Process Z.
Let X = {xs (t), t ≥ s}be a Markov chain with state space E and infinitesimal matrix Q as before,but with xs (s) = x ∈ E, and let {ys (t), t ≥ s} be the analogue of the process given by (1.9), that is: (2.1)
ys (t) = y +
Z
t
µ(ν, xs (ν), ys (ν))dν +
s
Z
t
σ(ν, xs (ν), ys (ν))dW (ν)
s
where ys (s) = y. Also let Lt be the differential operator Lt = µ(t, x, y)
d 1 d2 + σ 2 (t, x, y) 2 dy 2 dy
where the functions µ and σ are real-valued continuous and satisfy a Lipschitz condition. The following theorem presents a Feynman-Kac formula (2.3) for the random evolution process. Theorem 3. Let r(t, x, y) be a bounded continuous function and consider a backward Cauchy problem for the function u(t, x, y): ( ∂u ∂t + Lt u + r(t, x, y)u + Qu = 0 (2.2) u(T, x, y) = ϕ(x, y) where ϕ is a bounded continuous function E × R. Then the Cauchy problem (2.2) has the solution (2.3)
Z u(t, x, y) = Et,x,y [ϕ(xt (T ), yt (T )) · exp(
T
r(ν, xt (ν), yt (ν))dν)].
t
Here, Et,x,y is the integral with respect to the measure Pt,x,y (·) = P (·|(xt (t), yt (t)) = (x, y)). Proof. Let 0 ≤ s < t ≤ T and let us consider the process Z t (2.4) ζ(t) ≡ u(t, xs (t), ys (t)) · exp( r(ν, xs (ν), ys (ν))dν). s
We now show that ζ(t) is an
Fst -martingale
6
where Fst ≡ σ(W (ν), xs (ν) : s ≤ ν ≤
t) with respect to Ps,x,y . We note that Ft := F0t . We have from (2.4) that: Es,x,y [ζ(t + h) − ζ(t)|Fst ] Z t = exp( r(ν, xs (ν), ys (ν))dν) s
Z t · Es,x,y [u(t + h, xs (t + h), ys (t + h)) · exp( r(ν, xs (ν), ys (ν))dν) − u(t, xs (t), ys (t))|Fst ] s Z t = exp( r(ν, xs (ν), ys (ν))dν) s
· Es,x,y [u(t + h, xs (t + h), ys (t + h)) + h · u(t + h, xs (t + h), ys (t + h)) · r(t, xs (t), y( t)) − u(t, xs (t), ys (t))|Fst ] + o(h) Z t ∂u = exp( r(ν, xs (ν), ys (ν))dν) · Es,x,y [( + Lt u + Qu + ru)h|Fst ] + o(h) = o(h) ∂t s where we have applied (2.1) and (2.2).Thus, we obtain Es,x,y [ζ(t+h)−ζ(t)|Fst ] = o(h). Hence, if 0 ≤ s < t < u ≤ T and h = u−t N , then summing the following equalities for k = 1, 2, ..., N , Es,x,y [ζ(t + k · h) − ζ(t + (k − 1) · h)|Fst+(k−1)h ] = o(h) and applying the conditional expectation with respect to Fst we obtain: Es,x,y [ζ(u) − ζ(t)|Fst ] = N · o(h) =
u−t · o(h). h
Thus, we have as h → 0, (2.5)
Es,x,y [ζ(u) − ζ(t)|Fst ] = 0,
that is, ζ(t) is an Fst -martingale with respect to Ps,x,y . ¿From (2.4) and (2.5) it follows that Es,x,y [ζ(T )] = Es,x,y [ζ(s)]. But, taking into account (2.2) we have Z t (2.6) Es,x,y [ζ(T )] = Es,x,y [u(T, xs (T ), ys (T )) · exp( r(ν, xs (ν), ys (ν))dν)] s Z t (2.7) = Es,x,y [ϕ(xs (T ), ys (T )) · exp( r(ν, xs (ν), ys (ν))dν)] s
and (2.8)
Es,x,y [ζ(s)] = Es,x,y [u(s, xs (s), ys (s))] = u(s, x, y).
Hence, from (2.6) and (2.8) we can conclude (2.9)
Z t u(s, x, y) = Es,x,y [ϕ(xs (T ), ys (T )) · exp( r(ν, xs (ν), ys (ν))dν)]. s
Representation (2.3) follows fom (2.9) with s = t. 7
3
Black-Scholes Equation for a (B, S, X)-incomplete Securities Market.
In the famous Black-Scholes model of a (B, S)-market [3] that is used for the evaluation of option prices by riskless Bt (bonds or bank account) and risky St (stocks or assets), it is supposed that the dynamics of Bt and St are determined by the following system ( dB(t) = rB(t)dt, B(0) > 0 (3.1) dS(t) = µS(t)dt + σS(t)dW (t), S(0) > 0 with interest rate r ≥ 0, appreciation rate µ ∈ R and volatility σ > 0. We suppose that, in addition, tha coefficients r,µ and σ depend on the Markov chain {x(t), t ≥ 0} which is assumed to be independent of the Brownian motion {W (t), t ≥ 0}; we thus obtain the following system for Bt and St : ( dB(t) = r(x(t))B(t)dt, B(0) > 0 (3.2) dS(t) = µ(x(t))S(t)dt + σ(x(t))S(t)dW (t), S(0) > 0 Due to the additional source of randomness induced by X = {x(t), t ≥ 0}, we denote the securities market as being (B, S, X)-incomplete. In this section we extend a Black-Scholes-type formula to the market (B, S, X). We note that we can construct a family of measures P µ−r such that, for µ−r Pt ≡ P µ−r /Ft (that is, Ptµ−r is P µ−r restricted to Ft ), we have [8]: Ptµ−r (dω) = Ztµ−r Pt (dω)
(3.3) where (3.4)
( s )−r(xs ) Ztµ−r = exp(−intt0 µ(xσ(x dW (s) − s) Pt ≡ P/Ft
1 2
Rt 0
s )−r(xs ) 2 ( µ(xσ(x ) ds) s)
We note that Ft := F0t . Let Z
T
(
µ(xs ) − r(xs ) 2 ) ds < +∞a.s. σ(xs )
1 2
Z
0
and Ex,y [exp(
0
t
(
µ(xs ) − r(xs ) 2 ) ds)] < +inf ty. σ(xs )
Ztµ−r
It can be shown that ≥ 0, E[Ztµ−r ] = 1, Ptµ−r is a probability measure and P 0 ≡ P is an initial measure. By a Girsanov-type result, it follows that the process Z t µ(xs ) − r(xs ) µ−r (3.5) Wt ≡ Wt + ds σ(xs ) 0 8
is a Brownian motion with respect to P µ−r . Hence, Law(W µ−r /P µ−r ) = Law(W/P )
(3.6)
where W 0 = W and P 0 = P . It follows from (3.3)-(3.6) that for any µ(·) ∈ C(E), if dStµ = Stµ (r(xt )dt + σ(xt )dWtµ−r )
(3.7) then
Law(S µ /P µ−r ) = Law(S r /P ).
(3.8)
Let X be a Markov chain on E with infinitesimal matrix Q, and let Bt and Str be defined by (3.2) and (3.7), respectively, and let L(x) be the fifferential operator d 1 d2 L(x) = r(x) · s · + σ 2 (x) · s2 · 2 . ds 2 ds Theorem 4. The backward Cauchy problem for C(t, x, y), ( ∂C ∂t + L(x)C − r(x)C + QC = 0 (3.9) C(T, x, S) = ϕ(x, S) where ϕ(x, S) is a bounded continuous function on E × S,has the solution (3.10)
C(t, x, S) = Et,x,S [ϕ(xt (T ), Str (T ))exp(−
Z
T
r(xt (ν))dν)].
t
Proof. The theorem follows directly from Theorem 3 with r(t, x, y) ≡ −r(x), for all t ≥ 0 and y ∈ R. We call (3.9) a Black-Scholes equation for the (B, S, X)-incomplete market.
4
European Call Option Under a (B, S, X)-market.
We now consider a European call option under a (B, S, X)-incomplete market with cost function fT = fT (Stµ (T )).It follows from (3.8) that (4.1)
Law(fT (Stµ (T ))/P µ−r ) = Law(fT (Str (T ))/P ).
For the backward Cauchy problem, ( ∂C ∂t + L(x)C − r(x)C + QC = 0 (4.2) C(T, x, S) = fT (S)
9
we have from (3.9) and (3.10) that C(t, x, S) =
Et,x,S [fT (Str (T ))exp(−
T
Z
r(xt (ν))dν)]
t µ−r = Et,x,S [fT (Stµ (T ))exp(−
Z
T
r(xt (ν))dν)].
t
The dynamics of the summary capital xrt (T ) under (B, S, X)-market are defined by the following formula [8]: Z T xrt (T ) = Et,x,S [fT (Str ))exp(− r(xt (ν))dν)|Ft ]. t
We note that xrT (T ) = fT (STr (T )) and STr (T ) = S0r (T ) (see (5.2) below).
5
Black-Scholes Formula for a (B, S, X)-market.
Let fT (S) = (ST − K)+ = max{ST − K, 0}, where T is the expiration date and K is the exercise price for the risk asset. For a standard European call option with the cost function fT (S) = (ST − + K) , the rational cost CTx,S is defined by the formula Z T (5.1) CTx,S = E0,x,S [(S0r (T ) − K)+ exp(− r(xν )dν)] 0
where (5.2)
( RT RT S0r (T ) = Sexp( 0 r(xs )ds)exp( 0 σ(xs )dW (s) − 12 σ 2 (xs )ds) xs = x0 (s), S = S0
The formula for CTx,S is obtained from Theorem 4 by lelling fT (S) = (ST −K)+ . The value CTx,S can be calculated in some cases more simply [6]; for example, letting r(x) = 0 for all x ∈ E, it follows from (5.1) (5.2) that CTx,S = E0,x,S [max(S0 (T ) − K, 0)], where Z S0 (T ) = Sexp( 0
T
1 σ(xs )dW (s) − σ 2 (xs )ds) 2
We note that the finction (5.3)
C(t, x, S) = Ex,S [f (ST −t )]
is the solution of the Cauchy problem ( 2 ∂C + 12 σ 2 (x)s2 ∂∂sC2 + QC = 0 ∂t (5.4) C(T, x, S) = f (S) 10
where dSt = σ(xt )St dW (t), S0 = S. RT Let FTx be the distribution of the random variable ZTx ≡ 0 σ 2 (xs )ds. Then, from (5.3) and (5.4) it follows that (5.5) Z Z y 1 x,S CT := C(T, x, S) = E[f (ST )] = ( f (y)y −1 ψ(z, ln + z)dy)FTx (dz) s 2 2
where ψ(z, ν) = (2πz)−1/2 exp( ν2z ). In particular, for f (s) = (s − K)+ , we have from (5.5) for all x ∈ E, Z z CTx,S = CTBS (( )1/2 , T )FTx (dz) T where CTBS (σ, T ) is a Black-Scholes value [3] for a European call option with volatility σ, expiration date and interest rate r = 0.
11
References. 1.Garcia,M.A. and Griego,R.J., Stochastic integral representation of random evolution processes, 1996, to appear. 2.Gihman,I.I. and Skorohod,A.V.,Stochastic Differential Equations, SpringVerlag, New York, 1972. 3.Black,F. and Scholes,M., The pricing of options and corporate liabilities, Journal of Political Economy, 1973, No.3, pp.637-659. 4.Swishchuk, A.V., Limit theorems for stochastic differential equations with semi-Markov switchings, Analytical Methods in Probabilitstic Problems, Kiev Institute of Mathematics, 1988,pp.82-90. 5.Swishchuk,A.V., Limit theorems for stochastic differential equations with semi-Markov switchings, Proceedings of the International School - Evolution of Stochastic Systems: Theory and Applications, Ukraine, Crimea, 3-14, May, 1992-TVP/VSP, 1995, pp.331-347. 6.Swishchuk, A.V., Hedging of options under mean-square criterion and with semi-Markov volatility, Ukrainian Math. J., 1995, V.47,No.7,pp.976-983. 7.Swishchuk, A.V. and Burdeainy, A.G., Stability of evolutionary stochastic systems and its application in financial mathematics,Ukrainian Math. J., 1996, V.48, No.10, pp.1386-1401 (in Ukrainian). 8.Swishchuk, A.V. and Zhuravitsky, D.G., Applications of discounting evolutionary stochastic systems in financial mathematics, Reports of the National Acad. Sci. of Ukraine, 1997, No.6 (to appear in Ukrainian).
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