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Replicating a Squared Call Payoff with a Continuum of Calls

Article Quant Q&A · Author: user34971

Summary

The document examines how to replicate the square of a European call’s value across time. It distinguishes a terminal payoff from a value process: the squared call price does not generally have discounted-martingale dynamics, so a terminal claim alone cannot reproduce that price at every time. Under a Black–Scholes framework, the proposed replication includes the squared intrinsic payoff at maturity and an intermediate cash-flow stream derived by substituting the squared call value into the pricing PDE.

A second approach uses static replication. Applying the vanilla option payoff decomposition and restricting it to outcomes above the strike gives an integral of calls across strikes at or above the threshold, weighted by two. The document then approximates this continuum with a weighted sum over listed strikes. This is a theoretical construction whose practical accuracy depends on available strikes and discretization; the PDE cash-flow expression also relies on its stated model assumptions.

Key ideas

  • A squared call price is generally not the discounted expectation of a single terminal payoff at every time.
  • A dynamic replication can combine a squared terminal payoff with intermediate cash flows obtained from the pricing PDE.
  • Static replication represents the squared call payoff as a strike integral of vanilla calls above the threshold strike.
  • A finite set of quoted strikes provides only an approximation to the continuous static replication.

Tags

Full text
# Replicating the square of an option $C^2 (S,K,t,T)$


# Replicating the square of an option $C^2 (S,K,t,T)$












Given a vanilla options market, i.e. $C(S,K,t, T)$ for all strikes $K$, is it possible to replicate $C^2 (S,K,t,T)$? So I am looking for a self-financing portfolio which has a price equal to $C^2(S,K,t,T)$ for a fixed strike $K$ and for all $t$.

Thanks.

## Answer by Antoine Conze (score 3, accepted)

https://quant.stackexchange.com/a/44666

There is no terminal $\mathcal{F}_T$ mesurable payoff $g$ such that $e^{-r(T-t)} E_t[g] = C(S_t, t, T, K)^2$, simply because $E_t[g]$ must be a martingale and $e^{r(T-t)} C(S_t, t, T, K)^2$ is not.

So any deal that has npv $C(S_t, t, T, K)^2$ must involve a stream of intermediary payoffs $ h(S_t,t) dt$, which you can solve for by plugging $V(S,t) = C(S, t, T, K)^2$ in the BS PDE $$ \frac{\partial V}{\partial t} + r S \frac{\partial V}{\partial S} + \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} -rV + h(S,t) = 0 $$ to obtain $$ h(S,t) = -\left(\frac{\partial V}{\partial t} + r S \frac{\partial V}{\partial S} + \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} - rV\right) $$ along with the terminal payoff $g(S) = \max(S-K,0)^2$

## Answer by Daneel Olivaw (score 5)

https://quant.stackexchange.com/a/44609

I assume your trade $V(S,K,t,T)$ is European. Its payoff is: $$\begin{align} V(S,K,T,T)&=C^2(S,K,T,T) \\[3pt] &=\max(S_T-K,0)^2 \\[3pt] &=\boldsymbol{1}_{\{S_T\geq K\}}(S_T-K)^2 \\[3pt] &=\boldsymbol{1}_{\{S_T\geq K\}}f(S_T) \end{align}$$ where $f(x)=(x-K)^2$. By Carr-Madan's static replication formula (see this question or this paper), we have(1): $$\begin{align} f(S_T)&=f(K)+f'(K)(S_T-K)+\int_0^{K}f''(k)(k-S_T)^+\text{d}k+\int_{K}^{\infty}f''(k)(S_T-k)^+\text{d}k \\[3pt] &=2\int_0^{K}(k-S_T)^+\text{d}k+2\int_{K}^{\infty}(S_T-k)^+\text{d}k \end{align}$$ where $(x)^+=\max(x,0)$. Multiplying by $\boldsymbol{1}_{\{S_T\geq K\}}$: $$\boldsymbol{1}_{\{S_T\geq K\}}f(S_T)=2\int_{K}^{\infty}(S_T-k)^+\text{d}k$$ Multiplying by the discount factor $D(t,T)$ and taking the conditional expectation under the risk-neutral measure $Q$, we get the following theoretical replicating strategy: $$V(S,K,t,T)=2\int_{K}^{\infty}C(S,k,t,T)\text{d}k$$ Given in practice there is no availability of a continuum of call options, the following approximation is made: $$V(S,K,t,T)\approx2\sum_{i=0}^nC(S,k_i,t,T)\delta_i$$ where $\{k_i:i=0,\dots,n\}$ are the quoted strikes with $k_0=K$ and $\delta_i=k_{i+1}-k_i$.

(1) We have chosen as threshold value the strike $K$.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.