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

For a deterministic $s>0$, the positive part $(s-K{)}_{+}$ vanishes for $K\ge s$, so direct integration givesAt $s=0$ both sides vanish. Applying this pointwise to the nonnegative random variable proves the power payoff static call representationBy Tonelli theorem, the corresponding moment identity isincluding the possibility that both sides are infinite. No higher-moment assumption is needed to interchange these nonnegative integrals.

Let $A_{\delta }={\sup }_{K\ge 0}{K}^{\delta }C(K)<\infty$. The call-price decay and moment threshold follows by splitting the preceding integral at one. Since $(S-K{)}_{+}\le S$, for $0<\epsilon <\delta$,For $K\ge 1$, the decay bound givesThe power payoff static call representation therefore yieldsThe $\epsilon =0$ case is the given finite first moment. The strict endpoint matters: a Pareto distribution with $\mathbb{P}(S>x)=x^{-(1+\delta )}$ for $x\ge 1$ has $C(K)=K^{-\delta }/\delta$ for $K\ge 1$, but its moment of order $1+\delta$ is infinite. Thus the stated decay condition does not generally imply the endpoint moment.

If $K=0$ or $S\le K$, the proposed sharp power-call inequality is immediate. Otherwise put $x=S/K>1$ and consider the ratioIts logarithmic derivative is $(1+\epsilon )/x-1/(x-1)$, whose sign is that of $\epsilon x-(1+\epsilon )$. The ratio decreases and then increases, with minimum at $x_{\ast }=(1+\epsilon )/\epsilon$. That minimum is $(1+\epsilon {)}^{1+\epsilon }/{\epsilon }^{\epsilon }$. Rescaling proves the sharp power-call inequalityThe constant is sharp because equality holds for $S=(1+\epsilon )K/\epsilon$ when $K>0$. This also explains the hinted minimization: with $a$ equal to that constant times $K^{\epsilon }$, the minimum of $s^{1+\epsilon }-as$ is $-aK$. Its denominator is ${\epsilon }^{\epsilon }$, including when reading the original PDF.

Take expected values in the sharp power-call inequality. If $M(1+\epsilon )$ is finite, then for every $K\ge 0$,The bound is independent of the strike, soCombined with the previous part, this relates finite moments to polynomial decay of expected European call option payoffs, while retaining the distinction at the moment threshold.