Advanced Financial Models Example shee 4 - Michaelmas 8 Michael Tehranchi Problem. (Hull Whie exension of Black Scholes) Consider a marke wih consan ineres rae r and wih a sock price modelled as d = (µ d + σ()dw ) for a given non-random coninuous funcion σ : [, ) (, ). (a) Compue he replicaion cos of a European call opion. (b) A call opion is said o be a-he-money if Ke rt = S. Explain how you could use your answer o par (a) and quoed ime- prices of a-he-money calls of differen mauriies T o esimae he funcion σ. Soluion. (a) The replicaion cos is given by C(T, K) = E Q [e rt (S T K) + ] for each mauriy T where Q is he unique risk-neural measure. Bu, noe ha d log = (r σ() )d + σ()dŵ by Iô s formula, where dŵ = dw + (µ r)/σ()d defines a Q- Brownian moion. Since σ is no random, we can conclude from Example shee 3, Problem, ha ( log N log(s e rt ) ) σ(s) ds, σ(s) ds Hence, he replicaion cos of he call of he a-he-money call of mauriy T is ( ) C(T, K) = S F σ(s) ds, Ke rt /S where F is he funcion from Example shee 3, Problem. (b) Noe ha for K = S we have [ ( T ) / ] } C(T, S ) = S {Φ σ(s) ds. Assuming ha he quoed call prices are heir minimal replicaion coss, we need only choose σ in such a way ha [ ( C(T, σ(s) ds = Φ S ) + )] S by differeniaing boh sides wih respec o T. Problem. (variance swap) Consider a marke a sock wih price S, where S be a posiive Iô process, and ineres rae r =. A variance swap is a European coningen claim wih payou N ( log S ) n. n n= where < < N = T are fixed non-random daes. We know from sochasic calculus ha he payou of he variance swap, for large n, is approximaely given by ξ T = log S T.
The goal of his exercise is o show ha ξ T can be replicaed in an asympoic sense. (a) Confirm he ideniy (b) Confirm he ideniy log(s T /S ) = log x = x d log S T. (k x) + dk (x k) + dk. (c) Explain how o approximaely replicae ξ T by rading in he sock, cash, and a family of call and pu opions of differen srikes bu all wih mauriy dae T. Show ha he number of shares of he sock varies dynamically bu he porfolio of calls and pus is saic. To wha exen is his replicaion sraegy dependen of he deails of he model for S? Soluion. (a) Noe ha by Iô s formula: and hence d log = d The conclusion follows. (b) Since (k x) + only if k x we have Similarly, (k x) + dk = d S. S d log S = d S. S (x k) + dk = x ( k x ) dk = log(x ) ( x) + x ( x k ) dk k = (x ) + log(x ) (c) By he consrucion of quadraic variaion in lecure, we have ξ T log S T = = d log(s T /S ) d S T + (log S + ) + P T (T, K) dk + K C T (T, K) dk K where P T (T, K) = (K S T ) + is he payou of a pu opion and C T (T, K) = (S T K) + is he payou of a call. Therefore, a replicaion sraegy is o hold he saic posiion dk K pus of srike K and dk calls of srike K >, and he dynamic posiion of K shares of he sock a ime. This replicaion sraegy only requires ha S is posiive and ha Iô s formula applies, bu is oherwise model-independen. (Bu don forge abou noions of admissibiliy, and even
more fundamenally, ha we need o be in a siuaion where we can safely ignore bid-ask spread, price impac, ransacion coss, ec.) Problem 3. Consider a marke wih zero ineres rae r = and a sock wih price dynamics d = σ dw where σ is independen of he Q-Brownian moion W. Le C(T, K) = E Q [(S T K) + ]. (a) Show ha here is a family of measures µ T on [, ) such ha C(T, K) = S F (v, K/S )µ T (dv) where F (v, m) = Φ( log m/ v + v/) mφ( log m/ v v/). (b) If here are consans a b such ha a σ b a.s., show ha he implied volailiy saisfies a Σ(T, K) b. where he implied volailiy Σ(T, K) is defined implicily as he unique non-negaive soluion σ of he equaion F (T σ, K/S ) = C(T, K)/S. (c) Show he equaliy Σ(T, K) = Σ(T, S /K), i.e. he funcion x Σ(T, S e x ) is even. Hin: Firs prove he ideniy F (v, m) = m + mf (v, /m). Soluion 3. (a) Noice ha condiional on he process σ, he disribuion of log S T = log S σ sds + is normal, since σ and W are independen. Hence ( E[(S T K) + σ] = S F σ s dw s σ sds, K/S ). The conclusion follows from he ower propery of condiional expecaions, where he measure µ T is he law of he non-negaive random variable σ s ds. (b) Since v F (v, m) is increasing, we have F (T a, K/S ) F (v, K/S )µ T (dv) F (T b, K/S ). [a,b] Bu since he middle erm is jus F (T Σ(T, K), K/S ), we can conclude a Σ(T, K) b. 3
(c) Noice he Black Scholes call price funcion saisfies he following ideniy: F (v, m) = Φ( log m/ v + v/) mφ( log m/ v v/) = Φ(log m/ v v/) m [ Φ(log m/ v + v/) ] = m + m [ Φ(log m/ v + v/) (/m)φ(log m/ v v/) ] = m + m F (v, /m) Now use he above calculaion: S F (T Σ(T, K), K/S ) = = S F (v, K/S )µ T (dv) [S K + K F (v, S /K)] µ T (dv) = S K + K F (T Σ(T, S /K), S /K) = S F (T Σ(T, S /K), K/S ). In paricular, he implied volailiy smile in his model (where he volailiy is uncorrelaed wih he driving Brownian moion) is symmeric as a funcion of log-moneyness log(k/s ). This observaion is due o Renaul and Touzi in 996. Problem 4. * (Hull Whie exension of Cox Ingersoll Ross) Consider he shor rae model given by dr = λ( r() r ) d + γ r dw for posiive consans λ and γ and a deerminisic funcion r : R + R. Find he iniial forward rae curve T f T for his model. Soluion 4. Consider he PDE V (, T, r) + λ( r() r) V r (, T, r) + γ r V (, T, r) = rv (, T, r) r V (T, T, r) =. As usual we can make he ansaz ra(,t ) B(,T ) V (, T, r) = e for some funcions A and B which saisfy he boundary condiions A(T, T ) = B(T, T ) =. Subsiuing his ino he PDE yields which yields he coupled sysem A r B γ λ( r() r)a + ra = r A γ (, T ) = λa() + A() B (, T ) = λ r()a(, T ). 4
The equaion for A is a Riccai equaion, whose soluion is A(, T ) = B(, T ) = (e β(t ) ) (β + λ)e β(t ) + (β λ) λ r(s)a(s, T )ds where β = λ + γ. Hence, he ime forward raes are given by f T = T log P T = Problem 5. (Vasicek model) Suppose 4β e βt [(β + λ)e βt + (β λ)] r + dr = λ( r r )d + γdw 4β e β(t s) λ r(s)ds [(β + λ)e β(t s) + (β λ)]. where W is a Q-Brownian moion. Fix a ime horizon T > and le Q T be he T -forward measure. Using he fac ha he forward rae (f T ) [,T ] is a Q T -maringale, show ha he process W T defined by is a Q T -Brownian moion. dw T = dw + γ λ ( e λ(t ) )d Soluion 5. Recall he formula for he forward rae in he Vasicek model f(, T ) = e λ(t ) r + ( e λ(t ) ) r γ λ ( e λ(t ) ) for T. Use Iô s formula o ge df(, T ) = λe [r λ(t ) r + γ ] λ ( e λ(t ) ) d + e λ(t ) dr [ = e λ(t ) γ dw + σ ] λ ( e λ(t ) )d = e λ(t ) γdw S Since his is a maringale, we mus conclude ha W T is a Q S -Brownian moion. (Acually, we firs conclude ha W T is a local maringale. Bu since W T = W =, we know ha W T is acually a Brownian moion by Lévy s characerisaion heorem.) Problem 6. Le W,..., W m be m independen Brownian moions, and le X,..., X m evolve as dx i = ax i d + b dw i given iniial condiions X,..., X m and fixed consans a, b. Le m Z = (X) i = X. i= Show ha here exiss consans α, β, γ and a scalar Brownian moion Ŵ such ha dz = (αz + β)d + γ Z dŵ. 5
Soluion 6. Le f(x) = x so ha f = x i and f = if i = j and oherwise. By x i x i x j Iô s formula n n dz = XdX i i + d X i i= = (az + nb )d + i= n i= bx i dw i The conclusion now follows wih α = a, β = nb and γ = b, by defining Ŵ = X s dw s X s and noing ha Ŵ is a Brownian moion by Lévy s characerisaion heorem, since W =. Problem 7. * Given posiive consans λ, r, γ such ha r < γ and λ r > γ /, and an iniial condiion < r < γ and a Brownian moion W, i is possible o show ha here exiss a process (r ) saisfying dr = λ( r r )d + (γ r ) r dw such ha < r < γ for all almos surely. Define he funcion H : R + (, γ) (, ) by E[e rs ds r = r] = H(, r) Show ha H(, ) is a quadraic funcion for each. Soluion 7. We will show ha here is a funcion G : R + (, γ) (, ) such ha G(, ) is quadraic and furhermore saisfies he PDE G = λ( r r) G r + (γ r) r G r rg wih boundary condiion G(, r) = for all r. Assuming for he momen ha we have such a funcion G, noe ha for fixed >, he process (M s ) s T defined by M s = e s ru du G( s, r s ) is a local maringale by Iô s formula. Furhermore, since M is bounded (because he process (r s ) s is bounded), he local maringale M is acually rue maringale. Hence Tha is o say, G = H. Now o find G, we make he ansaz ha G(, r ) = M = E(M ) = E[e rs ds ]. G(, r) = g () + g ()r + g ()r wih g () = and g () = g () =. Plugging his in yields g + g r + g r = λ( r r)(g + g r) + (γ r) rg r(g + g r + g r ) = λ rg r[g + λg (λ r + γ )g ] r [g + (λ + γ)g ] 6
(he coefficiens are no really imporan wha one should noice is ha he coefficien of r 3 vanishes on he righ-hand side). Leing g = (g, g, g 3 ), by comparing coefficiens, we need only solve he sysem of linear ODEs where S = g = Sg, g() = (,, ) λ λ r + γ λ r (λ + γ) Problem 8. Le X be exponenially disribued wih E(X) =, and se Y = (+)e X for. Show ha here exiss a posiive maringale ( ) such ha he random variables and Y have he same disribuions for all. Hin: Compue E[(Y K) + ] and use Dupire s formula. Soluion 8. Here is one consrucion. Le W be a wo-dimensional Brownian moion. Le Z = + s dw s Firs noe ha ( ) Z N, + I so + Z exp() by sandard properies of normal random variables. Hence Y ( + )e (+) Z So consider = ( + )e (+) Z. By Iô s formula we have (afer some calculaion) ha d = Z dw so S is a local maringale. Bu since E( ) = for all, and since S is a supermaringale by Faou s lemma, we have ha S is a rue maringale by example shee, problem 8. Now, how on earh can we come up wih somehing like his? Before answering his quesion, here happens o be anoher very shor soluion o his problem. There is a heorem of Kellerer ha says he following: suppose ha he funcion C : R + R + [, ] is such ha C(, K) is increasing wih C(, K) = ( K) + for all K, and ha C(, ) is convex and C(, ) = for all. Then here is a non-negaive maringale S such ha C(, K) = E[( K) + ] In our case, noe U = e X is uniform on [, ] so le C(, K) = E[(Y K) + ] = (( + )u K) + du { = K + ( K +/ +) if K + if K + We re now done since he funcion C saisfies he hypoheses of Kellerer s heorem. Indeed, by he Breeden Lizenberger formula, we can now derive ha Y for all. 7.
Bu if you don know Kellerer s heorem (i wasn menioned in he lecures), we can ry o find he maringale explicily. Here is one mehod. Define he local volailiy by Dupire s formula: σ local (, K) = C K C K = ( ) + + log for K +. K Suppose for he momen ha here exiss a soluion o he SDE (*) d = σ local (, )d W, S =. where W is a scalar Brownian moion, such ha S is a maringale and le C(, K) = E[( K) + ]. Assuming for he momen ha Dupire s PDE is saisfied, we have C = σ local(, K) C K wih boundary condiion C(, K) = ( K) +. Assuming for he momen ha he PDE has a unique soluion, we have C = C (since we know ha C saisfies he PDE by consrucion). Then, afer all hese assumpions are checked, Breeden Lizenberger formula says ha Y for all. To implemen his approach, we firs need o show ha ha he SDE (*) has a soluion. Unforunaely, he coefficien is no Lipschiz, so he sandard heorems (from he sochasic calculus, no lecured in AFM) don apply. Bu a useful echnique is o suppose momenarily ha a soluion exiss, and hen apply a one-o-one ransformaion o derive anoher SDE. If he ransformed SDE has a soluion, so does he original one. Given he form of σ local we ry leing X = + log ( ) + in he SDE (*). Iô s formula yields he ransformed SDE (**) dx = ( + ) d X d + W Now, inspired by problem 6 above, we can guess he form of he soluion inroduced above by leing a = and b() = (i.e. we le he coefficien be ime- varying), where he scalar + Brownian moion W is relaed o he wo-dimensional Brownian moion W by Z s dw w W = Z s. Forunaely, in his case we know he disribuion of X so i is no necessary o verify he remaining assumpions. (Oherwise, we would have o pull ou a book on PDEs o find condiions ha a linear parabolic equaion has a unique soluion.) 8