 Solution

ID: past-exam-of-the-mathematical-models-of-financial-derivatives-course-of-the-university-of-oxford/2017/2/c/solution

The Black-Scholes reflection symmetry becomes a weighted reflection after taking a logarithm. Put

x = lo g (S / B)

and

v (x, t) = V (B e^{x}, t)

v_{t} + \frac{1}{2} σ^{2} v_{xx} + b v_{x} - r v = 0, b = r - σ^{2} /2.

For

w (x, t) = e^{γ x} v (- x, t)

, its left-hand side, after removing the factor

e^{γ x}

, is

v_{t} + \frac{1}{2} σ^{2} v_{xx} - (σ^{2} γ + b) v_{x} + (\frac{1}{2} σ^{2} γ^{2} + bγ - r) v,

evaluated at

(- x, t)

. Taking

γ = - 2 b / σ^{2}

b

and the additional zeroth-order term zero. Thus

W (S, t) = (S / B)^{1 - 2 r / σ^{2}} V (B^{2} / S, t)

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