 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/iii/paper-48/3/b/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Write

h = Δ x

k = Δ t

and march backward from the terminal payoff. At the interior grid point

(x_{i}, t_{j})

use

D_{x} u_{i}^{j} = \frac{u _{i + 1}^{j} - u _{i - 1}^{j}}{2 h}, D_{xx} u_{i}^{j} = \frac{u _{i + 1}^{j} - 2 u _{i}^{j} + u _{i - 1}^{j}}{h ^{2}} .

u_{i}^{j - 1} = u_{i}^{j} + k [a_{i}^{j} \frac{u _{i + 1}^{j} - 2 u _{i}^{j} + u _{i - 1}^{j}}{h ^{2}} + b_{i}^{j} \frac{u _{i + 1}^{j} - u _{i - 1}^{j}}{2 h}] .

Here

a_{i}^{j} = σ (e^{x_{i}}, t_{j})^{2} /2

b_{i}^{j} = r - a_{i}^{j}

, and

u_{i}^{m} = (e^{x_{i}} - K)^{+}

. At a far-left boundary use

u ≃ 0

; at a far-right boundary the call asymptotics are

u ≃ e^{r (T - t)} e^{x} - K

. These boundary values are inserted into the adjacent interior updates. Smooth-solution Taylor expansion gives consistency order

O (k + h^{2})

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