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Replicating a Squared Payoff with Out-of-the-Money Calls

Article Quant Q&A · Author: chinsoon12
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Summary

The note explains how a squared terminal asset price can be represented using a continuum of call payoffs across strikes. It starts from the general replication formula for a smooth payoff, which uses calls above a reference level and puts below it. For the square function, the second derivative is constant, so both option strips receive uniform weight.

The derivation applies put-call parity inside the lower-strike integral. The added put-minus-call terms reduce to a deterministic integral that cancels the reference-level and linear asset terms, leaving only the call strip over all positive strikes. This establishes equality between the mixed call-and-put representation and the calls-only expression. The result is a theoretical payoff identity; the note does not address discounting, discrete strike availability, transaction costs, or practical hedging error.

Key ideas

  • A twice differentiable payoff can be represented with options on either side of a chosen positive reference level.
  • For a squared payoff, the second derivative gives constant weights across the option strips.
  • Put-call parity converts the lower-strike put terms into calls plus terms that cancel the reference-level components.
  • The resulting squared payoff is represented by a weighted continuum of calls across positive strikes.

Tags

OptionsDerivatives pricingVolatility
Full text
# Pricing squared derivative: Equating S^2 + a strip of otm calls + a strip of otm puts = only calls


# Pricing squared derivative: Equating S^2 + a strip of otm calls + a strip of otm puts = only calls












In Peter Carr, Dilip Madan, Towards a Theory of Volatility Trading, 1998, (also derived here by Gordon), both calls and puts are used to replicate any twice differentiable payoff. I suppose one would choose the atm fwd to be the kappa and then use otm options to price the derivative.

In the answer to Replicating a square derivative with calls and puts, the payoff of the squared derivative is replicated with only call options.

How do we show that the original formula in Carr and Madan paper and the replication using only call options are equal?

This feels like its related to put-call parity and integrating the spot to get the first term on the original formula. But I am lost with how the strike term in put-call parity are handled.

## Answer by Gordon (score 2, accepted)

https://quant.stackexchange.com/a/69894

For a sufficiently smooth function $f$ and positive constant $a$, \begin{align*} f(x) &= f(a) + f'(a) (x-a) + \int_a^{\infty}(x-u)^+f''(u)du + \int_{0}^a(u - x)^+f''(u)du. \end{align*} Then \begin{align*} S_T^2 &= a^2 + 2a(S_T-a) + 2\int_a^{\infty}(S_T-u)^+du + 2\int_{0}^a(u - S_T)^+du\\ &=a^2 + 2a(S_T-a) + 2\int_0^{\infty}(S_T-u)^+du + 2\int_{0}^a\Big[(u - S_T)^+-(S_T-u)^+\Big]du\\ &=a^2 + 2a(S_T-a) + 2\int_0^{\infty}(S_T-u)^+du +2\int_{0}^a (u - S_T)du\\ &=a^2 + 2a(S_T-a) + 2\int_0^{\infty}(S_T-u)^+du + a^2-2aS_T\\ &=2\int_0^{\infty}(S_T-u)^+du. \end{align*}

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