Carr-Madan Decomposition of an Arbitrary Payoff
You are given a random variable $S$ (think of it as a stock price at expiry), its expectation $E[S]$, and the prices of European calls and puts at every strike:
$C(K) = E[\max(S - K,\, 0)], \qquad P(K) = E[\max(K - S,\, 0)]$
Let $f$ be an arbitrary twice-differentiable function.
- Find a closed-form expression for $E[f(S)]$ in terms of $f$, $E[S]$, $C(K)$, and $P(K)$.
- Explain the financial interpretation: what does each term in the decomposition correspond to as a traded instrument?
- Show how this result lets you extract the risk-neutral density of $S$ from option prices (Breeden-Litzenberger identity).
Hints
- Think about how you would decompose an arbitrary smooth function into pieces that match traded instruments -- what are the building blocks?
- Write a second-order Taylor expansion of $f(S)$ around a reference point $K^{*}$, but express the remainder using call and put payoffs $(S-K)^+$ and $(K-S)^+$ instead of the usual $(S-K^{*})^2$ form.
- Split the remainder integral at $K^{*}$: for $K > K^{*}$ the term $(S-K)^+$ is a call payoff, and for $K < K^{*}$ the term $(K-S)^+$ is a put payoff. Take expectations to get the formula in terms of $C(K)$ and $P(K)$.
Worked Solution
How to Think About It: The core idea is almost embarrassingly simple once you see it. Any smooth payoff $f(S)$ can be broken into three pieces: a constant, a linear piece, and curvature. Constants are bonds. Linear pieces are forwards. Curvature is what options give you. So if you know all call and put prices, you can replicate -- and therefore price -- any smooth payoff. This is the Carr-Madan formula, and it is one of the most useful identities in derivatives.
Before diving into the math, think about what must be true qualitatively. If $f$ is linear ($f'' = 0$), then $E[f(S)]$ should only depend on $E[S]$, not on option prices at all. That is exactly what the formula gives you -- the option integrals vanish. If $f$ is convex ($f'' > 0$ everywhere), the formula tells you $E[f(S)]$ is bigger than the linear approximation, which is just Jensen's inequality.
Quick Sanity Checks: - If $f(S) = S$: we should get $E[f(S)] = E[S]$. Check: $f'' = 0$, so the integrals vanish and we get $f(K^{*}) + f'(K^{*})(E[S] - K^{*}) = K^{*} + (E[S] - K^{*}) = E[S]$. - If $f(S) = (S - K_0)^+$ (a single call with strike $K_0$): the formula should reproduce $C(K_0)$. Check: $f''(K) = \delta(K - K_0)$, so the integral picks out $C(K_0)$. - If $f$ is convex, the curvature terms are non-negative, so $E[f(S)] \geq f(E[S])$ -- Jensen's inequality falls out for free.
Derivation:
*Part 1 -- The Carr-Madan formula.* Fix a reference point $K^{*}$ (often taken as $E[S]$ or the forward price). Write a second-order Taylor expansion with integral remainder:
$f(S) = f(K^{*}) + f'(K^{*})(S - K^{*}) + \int_0^{\infty} f''(K)\,(S - K)^+ \mathbf{1}_{K > K^{*}}\, dK + \int_0^{\infty} f''(K)\,(K - S)^+ \mathbf{1}_{K \leq K^{*}}\, dK$
More cleanly:
$f(S) = f(K^{*}) + f'(K^{*})(S - K^{*}) + \int_{K^{*}}^{\infty} f''(K)(S - K)^+ \, dK + \int_0^{K^{*}} f''(K)(K - S)^+ \, dK$
This is an exact identity (not an approximation) for any twice-differentiable $f$ with absolutely integrable second derivative. To verify it, note that the two integrals together reconstruct the second-order remainder $\int_0^{\infty} f''(K) \cdot (\text{payoff at strike } K) \, dK$.
Now take expectations:
$E[f(S)] = f(K^{*}) + f'(K^{*})\bigl(E[S] - K^{*}\bigr) + \int_{K^{*}}^{\infty} f''(K)\, C(K)\, dK + \int_0^{K^{*}} f''(K)\, P(K)\, dK$
This is the Carr-Madan decomposition.
*Part 2 -- Financial interpretation.* Each term maps to a traded instrument: - $f(K^{*})$: a position in zero-coupon bonds (deterministic cash) - $f'(K^{*})(S - K^{*})$: a position in forward contracts (linear exposure) - $\int_{K^{*}}^{\infty} f''(K)\, C(K)\, dK$: a continuum of OTM calls weighted by $f''(K)$ (captures upside curvature) - $\int_0^{K^{*}} f''(K)\, P(K)\, dK$: a continuum of OTM puts weighted by $f''(K)$ (captures downside curvature)
The second derivative $f''(K)$ tells you how many options at strike $K$ you need. Convex payoffs require buying options; concave payoffs require selling them.
*Part 3 -- Breeden-Litzenberger.* Apply the Carr-Madan formula with $f(S) = \mathbf{1}_{S \leq x}$ -- a digital put payoff. Actually, more directly: note that $C(K) = E[(S-K)^+] = \int_K^{\infty}(s - K)q(s)\,ds$, where $q$ is the risk-neutral density. Differentiate once:
$\frac{\partial C}{\partial K} = -\int_K^{\infty} q(s)\,ds = -(1 - F(K))$
Differentiate again:
$\frac{\partial^2 C}{\partial K^2} = q(K)$
So the risk-neutral density is the second derivative of the call price with respect to strike (undiscounted, or multiply by $e^{rT}$ if prices are discounted). In practice, you estimate this numerically from the volatility surface using a butterfly spread:
$q(K) \approx \frac{C(K + \Delta K) - 2C(K) + C(K - \Delta K)}{(\Delta K)^2}$
which is exactly the price of a butterfly centered at $K$, normalized by the wing width squared.
Practical Interpretation: This result is the foundation of variance swap pricing, volatility surface interpolation, and model-free hedging. When a desk prices an exotic payoff, they often start by computing $f''(K)$ and integrating against the vanilla option surface -- that gives the model-free price. Any deviation from that price is a bet on the dynamics (path-dependence, jumps, stochastic vol) not captured by the terminal distribution.
The Breeden-Litzenberger identity is how every options desk extracts implied distributions from the vol surface. If you see a kink in call prices at some strike, that means probability mass is concentrated there. Butterfly spreads are literally bets on the local density.
Answer: The Carr-Madan decomposition is:
$E[f(S)] = f(K^{*}) + f'(K^{*})(E[S] - K^{*}) + \int_{K^{*}}^{\infty} f''(K)\,C(K)\,dK + \int_0^{K^{*}} f''(K)\,P(K)\,dK$
Any smooth payoff decomposes into bonds (constant), forwards (linear), and a continuum of calls and puts (curvature). The risk-neutral density is $q(K) = \partial^2 C / \partial K^2$.
Intuition
The Carr-Madan formula says something profound in simple language: if you know the price of every vanilla option, you know the price of every smooth payoff. The constant piece is a bond, the linear piece is a forward, and all the interesting stuff -- the curvature -- lives in the options. The weight you put on the option at strike $K$ is just $f''(K)$, the curvature of your payoff at that point. Convex payoffs cost money (you are buying options); concave payoffs generate premium (you are selling them). Jensen's inequality is baked right in.
In practice, this is the backbone of model-free pricing. Variance swaps, for instance, correspond to $f(S) = \log(S)$, and the Carr-Madan formula turns their price into an integral over OTM option prices -- no model needed. The Breeden-Litzenberger identity ($q(K) = \partial^2 C / \partial K^2$) is the other side of the same coin: it lets you read the market's implied distribution directly from option prices. Every skew trade, every tail risk hedge, every exotic pricing sanity check starts here. If you remember one formula from derivatives theory, make it this one.