Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 39 / 2 / a

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- Solution a

Solution

For a deterministic

s > 0

, the positive part

(s - K)_{+}

vanishes for

K \geq s

, so direct integration gives

\int_{0}^{\infty} K^{ε - 1} (s - K)_{+} d K = s \frac{s ^{ε}}{ε} - \frac{s ^{1 + ε}}{1 + ε} = \frac{s ^{1 + ε}}{ε ( 1 + ε )} .

At

s = 0

both sides vanish. Applying this pointwise to the nonnegative random variable proves the power payoff static call representation

S^{1 + ε} = ε (1 + ε) \int_{0}^{\infty} K^{ε - 1} (S - K)_{+} d K .

By Tonelli theorem, the corresponding moment identity is

M (1 + ε) = ε (1 + ε) \int_{0}^{\infty} K^{ε - 1} C (K) d K,

including the possibility that both sides are infinite. No higher-moment assumption is needed to interchange these nonnegative integrals.

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