Replicating Arbitrary Payoffs with Calls and Puts

Worked Solution

How to Think About It: This is about one of the most fundamental ideas in derivatives pricing: any smooth payoff can be decomposed into a static portfolio of vanilla instruments. Before writing any math, think about what tools you have. A bond gives you a constant payoff. A forward gives you a linear payoff. Calls give you convex kinks above a strike. Puts give you convex kinks below a strike. If $f$ has curvature (non-zero second derivative), you need options to replicate it -- and the amount of curvature at each strike tells you how many options at that strike you need. This is the Breeden-Litzenberger result, and it is the theoretical backbone of volatility surface extraction from option prices.

Quick Sanity Checks: If $f$ is linear ($f''= 0$ everywhere), the formula should collapse to just bonds and forwards -- no options needed. If $f(x) = (x - K)^+$ for some $K > 0$, the decomposition should recover a single call at strike $K$ (since $(x-K)^+$ has $f''(x) = \delta(x - K)$). If $f(x) = x^2$, we need options across all strikes because the curvature is constant.

Derivation:

Start with a second-order Taylor expansion with integral remainder. For any twice continuously differentiable $f$, we can write:

$$f(x) = f(0) + f'(0)\,x + \int_0^{\infty} f''(k)\,(x - k)^+ \, dk + \int_{-\infty}^{0} f''(k)\,(k - x)^+ \, dk$$

This identity can be verified by splitting into cases ($x > 0$ and $x < 0$) and applying integration by parts. For $x > 0$:

$$\int_0^{\infty} f''(k)(x-k)^+ \, dk = \int_0^{x} f''(k)(x-k) \, dk$$

Integrating by parts with $u = (x - k)$, $dv = f''(k)\,dk$:

$$= \big[(x-k)f'(k)\big]_0^x + \int_0^x f'(k)\,dk = -xf'(0) + f(x) - f(0)$$

The put integral vanishes when $x > 0$, confirming $f(x) = f(0) + f'(0)x + \int_0^x f''(k)(x-k)\,dk$. The case $x < 0$ works symmetrically via the put integral.

Taking expectations of both sides:

$$E[f(X)] = f(0) + f'(0)\,E[X] + \int_0^{\infty} f''(k)\,E[(X - k)^+] \, dk + \int_{-\infty}^{0} f''(k)\,E[(k - X)^+] \, dk$$

This is the Breeden-Litzenberger representation.

Practical Interpretation:

Each term corresponds to a standard financial instrument:

• $f(0)$: a zero-coupon bond (risk-free position paying a constant).
• $f'(0)\,E[X]$: a position in the forward/futures contract, with size $f'(0)$.
• $\int_0^{\infty} f''(k)\,E[(X-k)^+]\,dk$: a continuum of call options with strikes $k > 0$, weighted by $f''(k)$. Where $f$ is convex ($f'' > 0$), you buy calls; where concave ($f'' < 0$), you sell them.
• $\int_{-\infty}^{0} f''(k)\,E[(k-X)^+]\,dk$: a continuum of put options with strikes $k < 0$, weighted by $f''(k)$.

The given quantities map as follows:

• $E[X]$ is the forward price of $X$.
• $E[\max(0, X)] = E[(X - 0)^+]$ is the price of an at-the-money call (strike $K = 0$).
• $E[\min(0, X)] = -E[(0 - X)^+]$ is the negative of the at-the-money put price.

Note the consistency check: $E[X] = E[\max(0,X)] + E[\min(0,X)]$, which is just the decomposition $X = \max(0,X) + \min(0,X)$. In financial terms, this is a version of put-call parity at strike zero.

The key practical takeaway: if you can observe call and put prices at every strike from the market, you can price any smooth derivative by reading off $f''(k)$ and integrating against the option price surface. This is exactly how traders extract the implied distribution from the volatility smile and price exotic payoffs from vanilla options.

Answer: Any twice differentiable payoff $f(X)$ can be replicated as:

$$E[f(X)] = f(0) + f'(0)\,E[X] + \int_0^{\infty} f''(k)\,C(k)\,dk + \int_{-\infty}^{0} f''(k)\,P(k)\,dk$$

where $C(k) = E[(X-k)^+]$ and $P(k) = E[(k-X)^+]$ are call and put prices. The decomposition is: bonds ($f(0)$) + forwards ($f'(0)$) + a continuum of calls and puts weighted by curvature ($f''$).