Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 2 / 27K / iii

Table of contents
- Solution iii

Solution

Expand the payoff as

(S_{T} - 2 K + K^{2} / S_{T}) 1_{{S_{T} > K}}

. The same risk-neutral measure applies. For any real

p

, completing the square gives the truncated lognormal moment

E [S_{T}^{p} 1_{{S_{T} > K}}] = S_{0}^{p} e^{p r T + p (p - 1) σ^{2} T /2} Φ (d_{2} + p σ T) .

Using

p = 1, 0, - 1

and discounting by

e^{- r T}

gives

V_{0} = S_{0} Φ (d_{1}) - 2 K e^{- r T} Φ (d_{2}) + \frac{K ^{2}}{S _{0}} e^{- 2 r T + σ^{2} T} Φ (d_{2} - σ T) .

The inverse-price moment explains both the extra volatility factor and the shifted normal argument in the last term.

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