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Replicating a Squared Stock Price Payoff with Calls

Article Quant Q&A · Author: Calculon

Summary

The document derives a static option replication for a payoff equal to the square of a non-dividend-paying stock’s terminal price. It rewrites the square as twice the integral, over strikes from zero to infinity, of the call payoff at each strike. This identity shows why a continuum of calls can represent the convex squared payoff, even though any individual call is only piecewise linear in the stock price.

For a finite partition of strikes, the integral becomes a weighted sum of calls, with each option’s weight set by the width of its strike interval. This gives an approximation to the payoff, with closer strike spacing improving the representation under suitable conditions. The answer says put-based replication is analogous but does not derive it. It also leaves out practical constraints such as available strike ranges, option prices, discounting, and hedging or discretization error.

Key ideas

  • The squared terminal stock price can be expressed as twice the integral of call payoffs across strikes.
  • A continuum of calls provides a static representation of the convex squared payoff.
  • A finite portfolio approximates the integral using strike intervals as call weights.
  • The finite-strike construction is an approximation, and practical replication depends on market availability and costs.
  • A put-based construction is possible by an analogous argument, though it is not shown.

Tags

Full text
# Replicating a square derivative with calls and puts


# Replicating a square derivative with calls and puts












I have a derivative that pays off $S_T^2$ at time $T > 0$ with $S_T$ denoting the price of a non dividend-paying stock at $T$. I came across a question about how one can statically replicate this derivative with vanilla calls and puts.

My guess is that it is impossible to do that on the entire support of $S_T$. Since the square function dominates a linear function eventually and the call option is linear in $S_T$ for $S_T$ large enough, there cannot be a sequence of linear combinations of calls and puts that converges to the payoff of this derivative pointwise. I was also given a hint that I should consider integration. I am aware that $S_T^2$ can be written as $S_T^2 = 2\int_0^{S_T}x\,dx$ but I am not sure if that is what the hint hints at. Any tips/solutions appreciated.

## Answer by Gordon (score 6, accepted)

https://quant.stackexchange.com/a/35633

Note that \begin{align*} S_T^2 = 2\int_0^{S_T} k dk. \end{align*} Then \begin{align*} S_T^2 &= 2S_T^2-2\int_0^{S_T} k dk\\ &=2S_T\int_0^{S_T}dk-2\int_0^{S_T} k dk\\ &=2\int_0^{S_T} (S_T-k)dk\\ &=2\int_0^{\infty} (S_T-k)^+dk. \end{align*} For the partition $0=k_0 < k_1 < \cdots < k_n < \infty$, \begin{align*} S_T^2 &=2\int_0^{\infty} (S_T-k)^+dk\\ &\approx 2\sum_{i=1}^n (k_i-k_{i-1})(S_T-k_i)^+. \end{align*} That is, it can be replicated by a portfolio of call options. The replication by put options is similar.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

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