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The Numerical Solution of the American Option Pricing by Carl Chiarella

By Carl Chiarella

The early workout chance of an American alternative makes it demanding to cost. The Numerical resolution of the yank alternative Pricing challenge makes a speciality of 3 numerical equipment that experience proved invaluable for the numerical answer of the partial differential equations with unfastened boundary challenge coming up in American choice pricing, specifically the strategy of traces, the sparse grid procedure and the quintessential rework method. It truly explains and demonstrates the benefits and barriers of every of them utilizing numerous examples.

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Additional info for The Numerical Solution of the American Option Pricing Problem: Finite Difference and Transform Approaches

Example text

21). 17) to recover V (x, τ ), the American call price in the x-τ plane. 23) [CE (S, τ ) + CP (S, τ )] = C A (S, τ ), K K where CE (S, τ ) = KVE (x, τ ) is the value of the corresponding European call written on S and CP (S, τ ) = KVP (x, τ ) is the early exercise premium for C A (S, τ ). 4 below. 3. 23) is given by ∞ CE (S, τ ) = e−λτ (λτ )n (n) EQ {CBS [SXn e−λkτ , K, K, r, q, τ, σ2 ]}, n! 25) page 22 September 5, 2014 8:1 The Numerical Solution of the American Pricing Problem. . 9in x 6in American Call Options under Jump-Diffusion Processes b1863-ch03 23 N [·] is the cumulative normal density function, and we define Xn ≡ Y1 Y2 .

An evenly spaced grid is used, and the free boundary is estimated at each time step using cubic spline interpolation of the price profile, combined with the bisection method. To improve the accuracy of the Crank-Nicolson solution, we use a twostep procedure at each time step. After determining an initial solution at (1) time step i, denoted here as Ci (S), using the estimate of Ci−1 (S) in the integral term, we then find an updated estimate by repeating the solution (1) process, now using the Ci (S) values in the integral estimate.

By inspection lim f (x) = K(r + λ). x→0 Also, lim f (x) = lim(Kr − qx) x→∞ because 0 0 then f is strictly decreasing from positive to negative values on (0, ∞). Hence there exists a unique x∗ ∈ (0, ∞). We observe that f (K) = K (r − q) + 1 0 (1 − Y )G(Y )dY . If f (K) ≤ 0 then x∗ ∈ (0, K], otherwise x∗ ∈ (K, ∞). If q = 0 then f (x) → Kr as x → ∞. Since f ≤ 0 there is no x∗ such that f (x∗ ) = 0. 54) we see that for q > 0 a(0+) = Kmax (1, x∗ ).

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