BIRKBECK (University of London) MATHEMATICAL AND NUMERICAL METHODS. Thursday, 29 May 2008, 10.00am-13.15pm (includes 15 minutes reading time).

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1 BIRKBECK (University of London) MSc EXAMINATION FOR INTERNAL STUDENTS MSc FINANCIAL ENGINEERING School of Economics, Mathematics, and Statistics MATHEMATICAL AND NUMERICAL METHODS ANSWERS EMMS011P Thursday, 29 May 2008, 10.00am-13.15pm (includes 15 minutes reading time). The paper is divided into two sections. There are four questions in each section. Answer TWO questions from Section A and TWO questions from Section B. All questions carry the same weight. page 1 of 12 Please turn over

2 SECTION A (Answer TWO questions from this section.) 1. (a) [Bookwk] Completing the square in the exponent, we have Ee cz = e ct (2π) 1/2 e t2 /2 dt R = (2π) 1/2 e (t c)2 /2+c 2 /2 dt R = e c2 /2 (2π) 1/2 e s2 /2 ds = e c2 /2, using the substitution s = t c in the penultimate line. R 4 pts (b) [Bookwk] Brownian motion describes the stochastic process satisfying the following conditions: i. B 0 = 0; ii. B t+h B t N(0,h), f any t 0 and h > 0; iii. Choose any positive integer m and points 0 t 1 < t 2 < < t m. Then the increments B tk B tk 1, k = 2,...,m, are independent random variables. To prove that it s a martingale, we note that, f s > t, EB(s) F t = E(B(s) B(t)+B(t)) F t = E(B(s) B(t)) F t +EB(t) F t = 0+B t. 4 pts (c) [Standard problem.] Since the Brownian motions are independent, we have Eexp(c T Z t ) = e c iz i,t, and since Z i,t tu i, where U 1,...,U n are independent N(0,1) random variables, we infer from part (a) that i=1 Eexp(c T Z t ) = e tc2 i /2 = e t c 2 /2. i=1 4 pts (d) [Fairly standard problem.] If Z 1,t, Z 2,t and Z 3,t denote independent (univariate) Brownian motions, set Z t = (Z 1,t,Z 2,t,Z 3,t ) T, and define W t = M 1/2 Z t. Then EW t W T t = M 1/2 EZ t Z T t M 1/2 = M 1/2 (ti)m 1/2 = tm, page 2 of 12

3 where I R 3 3 is the identity matrix. Thus the key is to calculate M 1/2 f the given matrix M. The attentive student should observe that [ ] a b X = b a satisfies X 2 = [ a 2 +b 2 2ab 2ab a 2 +b 2 ]. Hence the required matrix square-root is given by a 0 b M 1/2 = b 0 a [It s perfectly acceptable, albeit slightly longer, if the student first calculates the eigenvalues and eigenvects of M.] 8 pts page 3 of 12 Please turn over

4 2. (a) [Standard problem] We have, using the Itô rules, X t+dt = e iλwt = X t (1+iλdW t 1 ) 2 λ2 dt, dx t = X t (iλdw t 1 ) 2 λ2 dt. Then, taking expectations and using EdW t = 0, together with the independence of W t and dw t, we deduce dm/dt = (λ 2 /2)m(t), which has solution m(t) = exp( λ 2 t/2), since m(0) = Eexp(iλW 0 ) = 1. (b) [Fairly standard problem] Restating part (a), since W t N(0,t), we have just proved that Writing c = t 1/2, this becomes Ee i tz = e t/2. Ee icz = e c2 /2. (1) Now ˆp(z) = p(x)e ixtz dx R n = (2πσ 2 ) 1/2 e iz kx k e x2 k /(2σ2) dx k = k=1 Ee iσz kw k, k=1 where W 1,W 2,...,W n N(0,1) are independent nmalized Gaussian random variables. Hence, using (1), we deduce ˆp(z) = e (1/2)σ2 zk 2 = e (1/2)σ 2 z 2. k=1 7 pts (c) [Unseen problem] We use the spectral factization A = QDQ T, where Q R n n is an thogonal matrix of eigenvects of A and D R n n is the cresponding diagonal matrix of eigenvalues of A; the latter are positive numbers because A is positive definite. Thus we can define the unique symmetric positive definite square-root A 1/2 = QD 1/2 Q T, and we let y = A 1/2 x in the Fourier transfm integral, page 4 of 12

5 noting that dy = det(a 1/2 ) dx = (deta) 1/2 dx. Thus, setting σ = 1 in part (b) and using the symmetry of A 1/2, we obtain ˆq(z) = (2π) n/2 2 yty exp( iz T (A 1/2 y))dy = (2π) n/2 = e 1 2 A1/2 z 2 = e 1 2 zt Az. e 1 R n e 1 R n 2 yty exp( i(a 1/2 z) T y)dy [It would, of course, be equally acceptable to use the substitution y = Q T x, so avoiding the explicit use of the matrix square-root; students should be familiar with both methods.] 8 pts page 5 of 12 Please turn over

6 3. (a) [Standard problem] Using the usual Itô calculus rules, we obtain Y t+dt = e ct W t+dt = e ct W t e ct dw t ( = Y t 1+c T dw t + 1 ) 2 (ct dw t ) 2 = Y t (1+c T dw t + 1 ) 2 ct dw t dwt T c = Y t (1+c T dw t + 1 ) 2 ct Mc dt. and subtracting Y t from both sides then provides the answer. (b) [Fairly standard problem] Taking the expectation of the SDE derived in the first part of the question, and recalling that EdW i,t = 0 and EY t dw i,t = 0, we obtain the differential equation with solution dm dt = 1 ( c T Mc ) m(t), 2 m(t) = m(0)e (ct Mc)t/2. (c) [Variant on standard problem] Applying part (b) with c = e i, we obtain ES i (t) = S i (0)e a it e 1 2 tet i Me i = S i (0)e a it e M iit/2. Thus risk-neutrality (i.e. ES i (t) = S i (0)e rt requires that a i +M ii /2 = r, a i = r M ii /2. (d) Setting a i = r M ii /2 and applying part (b) with c = 2e i, we have ES i (t) 2 = S i (0) 2 e (2r M ii)t e 2eT i Me i = S i (0) 2 e (2r M ii)t e 2M iit = S i (0) 2 e (2r+M ii)t. page 6 of 12

7 4. (a) Writing y(t) X t f brevity, we have i.e. the differential equation Integrating, we obtain dy = α(θ y)dt, dy θ y = αdt. [ ln(θ y)] y(τ) X 0 = ατ, Thus θ y(τ) θ X 0 = e ατ. X t = θ+(θ X 0 )e αt. Hence lim t Xt = θ. (b) If Y t = X t θ, then dy t = dx t and Ornstein Uhlenbeck becomes dy t = αy t dt+σdw t, dy t +αy t dt = σdw t. Multiplying both sides by the integrating fact e αt, we find (c) Integrating we obtain d ( Y t e αt) = σe αt dw t. d(y s e αs ) = σe αs dw s, T Y t e αt Y 0 = σ e αs dw s, 0 T X t = θ +e αt (X 0 θ)+σe αt e αs dw s, which is the stated solution. 0 (d) Weseethatvar X t = e 2αt var I(t)ff(s) = σe αs. Thusthestochastic integration theem [ ( t ) 2 ] t E g(s)dw s = g(s) 2 ds 0 0 page 7 of 12 Please turn over

8 implies var X t = e 2αt t 0 ( ) e σ 2 e 2αs ds = e 2αt σ 2 2αt 1, 2α var X t = σ 2 ( 1 e 2αt 2α ). page 8 of 12

9 SECTION B (Answer TWO questions from this section.) 5. (a) [Bookwk] The random variable X X n has the binomial distribution with probability p: exactly l of X 1,X 2,...,X n are equal to unity with probability ( n l) p l (1 p) n l, those remaining being equal to 1, f 0 l n. Hence Ef(S T,T) = n l=0 ( ) n p l (1 p) n l f(s 0 e lα (n l)α,t), l and discounting provides the stated equation. (b) [Unseen problem] We have f DP (S 0,0) = e rt l<n/2 ( ) n 2 n, l because S 0 exp(lα (n l)α) < S 0 if and only if 2l < n. Now the binomial coefficients are symmetric about their central value and, since n is odd, there are an even number of binomial coefficients. Further, the student should be aware that n ( ) n 2 n = (1+1) n =, k which implies k=0 f DP (S 0,0) = e rt 2 n 2 n 1 = e rt /2. (c) [Unseen problem] This time, we have f DC (S 0,0) = e ( ) n rt 2 n = e rt 2 n 2 n 1 = e rt /2. l l n/2 Alternatively, the student could simply observe the obvious put-call parity relation f DP (S t,t) + f DC (S t,t) = exp( rt), since they sum to unity at expiry. (d) [Slight variant on standard bookwk.] We have ) f DP (S 0,0) = e P( rt 2rWT < 0 = e rt /2, and similarly f f DC (S 0,0), with < being replaced by. page 9 of 12 Please turn over

10 6. (a) [Bookwk] The Fourier transfm of the Black Scholes equation is 0 = [ r +iz(r σ 2 /2) 1 2 σ2 z 2 ] f + t f. Solving this first-der dinary differential equation f fixed z R, we obtain [ ] f(z,t) = f(z,0)e r+iz(r σ 2 /2) 1 2 σ2 z 2, setting t = T and rearranging, [ r+iz(r σ 2 /2) 12 σ2 z 2 ] t, f(z,0) = f(z,t)e T. 10 pts (b) [Variant on standard problem]iff(x,t) = G ω 2(x), thenweobtain f(z,t) = e iµz Ĝ ω 2(z) = e iµz e ω2 z 2 /2, using part(b) above. Using the final displayed equation of the previous part, we obtain Thus f(z,0) = e rt e iz([r σ2 /2]T µ) e (σ2 T+ω 2 )z 2 /2 = e rt T [r σ 2 /2]T µg σ 2 T+ω 2(z). f(x,0) = e rt G σ 2 T+ω 2(x+[r σ2 /2]T µ). To obtain the required option price, we now take a linear combination, obtaining M f(x,0) = e rt a l G σ 2 T+σl 2(x+[r σ2 /2]T µ l ). l=1 [Some students might recognize this as the Green s function f the Black Scholes PDE.] 10 pts page 10 of 12

11 7. [Standard bookwk.] (a) In matrix fm, we obtain T(1+µ, µ/2)u n+1 = T(1 µ,µ/2)u n, U n+1 = T(1+µ, µ/2) 1 T(1 µ,µ/2)u n. Now every tridiagonal symmetric Toeplitz (TST) matrix has the same eigenvects. Thus the eigenvalues of this product of TST matrices are given by λ k = 1 µ+µcos(πk M ) 1+µ µcos πk M = 1 2µsin2 ( πk 2M ) 1+2µsin 2 ( πk 2M ). We deduce λ k ( 1,1) f all µ 0. Hence Crank Nicolson is spectrally stable f all µ > pts (b) The associated functional equation is u n+1 (x) µ 2 (u n+1(x+h) 2u n+1 (x)+u n+1 (x h)) = with Fourier transfm where u n (x)+ µ 2 (u n(x+h) 2u n (x)+u n (x h)), û n+1 (z) = û n (z)r(hz), r(w) = 1 2µsin2 w/2 1+2µsin 2 w/2. Thus 1 r(hz) 1, f all h > 0 and z R, because of the elementary inequality 1 (1 x)/(1 + x) 1, f x 0. Hence Crank Nicolson is von Neumann stable f all µ pts page 11 of 12 Please turn over

12 8. [Unseen problems] (a) We must satisfy the relations ET k = exp(rh) and ET 2 k = exp((2r + σ 2 )h). In other wds, we have and e rh = ET k = u+d 2 e (2r+σ2)h = E(Tk) 2 = u2 +d 2, 2 as required. The sketch of these equations will, of course, be the circle, centre (0,0), radius 2e (r+σ2 /2)h, together with the line passing through the points ((2e rh,0) and (0,2e rh ). It is easily checked that the line always intersects the circle in exactly two points when h > 0. However, one of the intersection codinates becomes negative when exp(σ 2 h) > 2. Thus we obtain positive solutions f h σ 2 ln2. 7 pts (b) If we eliminate d from the equations in part (a), we find the new quadratic equation i.e. Completing the square, we obtain i.e. Thus u 2 +(u 2e rh ) 2 = 2e (2r+σ2 )h, 2u 2 4e rh u+4e 2rh = 2e (2r+σ2 )h, u 2 2e rh u+2e 2rh = e (2r+σ2 )h. (u e rh ) 2 e 2rh +2e 2rh = e (2r+σ2 )h, ) (u e rh ) 2 = e (2r+σ2)h e 2rh = e (e 2rh σ2h 1. u = e rh ±e rh e σ2 h 1. (c) We have, by independence of the {T i }, ESn m = E = = i=1 i=1 T m i E(T m i ) (u m /2+d m /2) i=1 ( ) u m +d m n =. 2 7 pts 8 pts page 12 of 12