Bergische Universität Wuppertal Applied Mathematics and Numerical Analysis Univ.-Prof. Dr. M. Günther Dr. C. Heuer Dipl.-Math. L. Teng Winter Term 2014/15 Exercise Sheet 4 in Computational Finance II Question 1:(13 points) The Black-Scholes equation is given by ∂V σ2 S 2 ∂ 2 V ∂V + + (r − δ)S − rV = 0 2 ∂t 2 ∂S ∂S (1) with S ∈ [0, Smax ] and t ∈ [0, T [. The final condition for a European Put with strike K > 0 is given by V (S, T ) = max(K − S, 0) for S ∈ [0, Smax ]. An equidistant grid for this problem setting is given by Si = i∆S, ∆S = Smax m tν = ν∆t, ∆t = T νmax for m, νmax > 1. Apply the finite difference approximations Vi+1 − Vi−1 ∂V (Si ) = + O (∆S)2 ∂S 2∆S and ∂ 2 V (Si ) Vi+1 − 2Vi + Vi−1 = + O (∆S)2 2 2 ∂S (∆S) and the Crank-Nicolson time-discretisation to (1). Use ωk,ν as the approximation of V (Sk , tν ) for k = 0, . . . , m and ν = 0, . . . , νmax as well as ω (ν) = (ω1,ν , . . . , ωm−1,ν )tr . Define A, B and d(ν) (which contains the boundary conditions) so that the resulting numerical scheme for a European Put is of the form Aω (ν+1) = Bω (ν) + d(ν) . Question 2:(7 points) For the Black-Scholes equation we can use V (S, t) ≈ S (2) as a boundary condition for a European Call at the boundary S → ∞. Apply the transformations ¯ −δ(T −t) , S = Se and S¯ = Kex , t=T− 1 V = Kexp − (qδ − 1)x − 2 2τ , σ2 q= 2r , σ2 qδ = 2(r − δ) σ2 1 2 (qδ − 1) + q τ y 4 to (2) and thus show that we have 1 1 2 (qδ + 1)x + (qδ + 1) τ y = exp 2 4 as a boundary condition for the transformed European Call y at the boundary x → ∞. Return the solutions until Monday, November 10, up to 4pm. You may submit in groups (up to two persons). 2