Carr–Madan Formula

The following equation is commonly referenced in the options-pricing literature. Let $g$ be a real-valued, twice-differentiable payoff function; let $S_t$ denote the value (unknown or random) of an underlying asset at time $t$; let $k$ denote the strike price of an option; and finally let $F$ be a constant. Then the following identity holds:

$g(S_t) = g(F) + g^{\prime}(F)(S_t - F) + \int_0^{F} g^{\prime\prime}(k) p(S_t, k) dk + \int_{F}^{\infty} g^{\prime\prime}(k) c(S_t, k) dk. \tag{1}$

Here $p$ and $c$ are the payoff functions of European options.

Let’s think about this equation. First, note that $g(S_t)$ is the value of a derivative, since it is a payoff that is a function of an underlying asset with value $S_t$. Second, note that this is a “European-style” payoff in the sense that the payoff occurs at time $t$ and not before. And finally, note that the integrals can be viewed as portfolios of options (weighted sums of calls and puts). So at a high level, this formula states that any European-style twice-differentiable derivative payoff can be replicated using a portfolio of European options.

This is particularly powerful for two reasons. First, the relationship makes very few assumptions. And second, the portfolio weights, the terms $g(k)$ inside each integral, do not depend on time or the underlying asset’s price. Thus, one can use this formula to construct a static hedge or a hedge that does not change with time.

I’ve sometimes seen this referred to as the Carr–Madan formula—it is Equation $1$ in (Carr & Madan, 2001)—, but even that paper states that the formula is “widely recognized”, and they cite several other earlier papers which reference this formula.

The primary goal of this post is to prove Equation $1$. I have seen it stated many times in the options-pricing literature and wanted to convince myself that it holds.

Taylor expanding with an integral remainder

The Carr–Madan formula is really just a special case of a Taylor expansion. For completeness, let’s rederive the Taylor expansion with an integral remainder.

We start with the second fundamental theorem of calculus. Let $f^{\prime}$ be a real-valued function on the closed interval $[a, x]$, and let $f$ be the antiderivative of $f^{\prime}$, a continuous function on $[a, x]$. The second fundamental theorem of calculus states

$f(x) - f(a) = \int_a^x f^{\prime}(t) dt. \tag{2}$

Now consider the following derivation (commentary to follow):

$\begin{aligned} f(x) - f(a) &= \int_a^x f^{\prime}(t) dt \\ &= \int_a^x \left[ f^{\prime}(t) + f^{\prime}(a) - f^{\prime}(a) \right] dt \\ &= \int_a^x f^{\prime}(a) dt + \int_a^x \left[ f^{\prime}(t) - f^{\prime}(a) \right] dt \\ &\stackrel{1}{=} f^{\prime}(a) (x - a) + \int_a^x \left[ \int_a^{u=t} f^{\prime\prime}(u) du \right] dt \\ &\stackrel{2}{=} f^{\prime}(a) (x - a) + \int_a^x \left[ \int_{t=u}^x f^{\prime\prime}(u) dt \right] du \\ &\stackrel{3}{=} f^{\prime}(a) (x - a) + \int_a^x f^{\prime\prime}(u) (x - u) du. \end{aligned} \tag{3}$

In step $1$, we re-applied the fundamental theorem of calculus. In step $2$, we interchanged the order of integration. If you forget how to change the order of integration, see these notes on the topic. The basic idea is that when integrating a 3D volume, we can switch which axis (on the $xy$-plane) to integrate over first. Finally, in step $3$, we eliminate the inner integral; doing this is the reason we changed the order of integration.

This allows us to rewrite the last line of Equation $3$ as

$f(x) = f(a) + f^{\prime}(a)(x-a) + \int_a^x f^{\prime\prime}(u)(x-u) du, \tag{4}$

which the reader may recognize as the Taylor expansion with an integral form of the remainder. (If we were trying to prove properties of the Taylor expansion, we might use integration by parts to further expand the function. See these notes if you’re curious about that.)

Here, we’re going to assume that $a$ is a constant, while $x$ is the independent variable of the function $f$. So $f(a)$ could be written $f(x=a)$, or the function $f$ with independent variable $x$ evaluated at the point $a$. Thus, we’re not assuming that $a \lt x$. Of course, this is implied by the conventional interpretation of the notation $[a, x]$ and $\int_a^x$ but nothing requires it. So here, we’re integrating over all points in a set defined by $x$ and $a$. This will be important in the next section.

Decomposing integrals with the Heaviside function

We want to convert Equation $4$ into Equation $1$. This suggests that we need to split the integral remainder in the Taylor expansion in Equation $4$ into two integrals. To do this, we’ll use the Heaviside function

$H(x) := \begin{cases} 1 & \text{$x \gt 0$}, \\ 0 & \text{$x \leq 0$}, \end{cases} \tag{5}$

and the ramp function

$[x]^{+} := \max(0, x). \tag{6}$

Using the fact that

$H(x) + H(-x) = 1, \tag{7}$

we can break up the integral in Equation $4$ as

$\begin{aligned} &\int_a^x f^{\prime\prime}(u)(x-u)du \\ &= \int_a^x \left[ H(x-a) + H(a-x) \right] (x-u) f^{\prime\prime} du \\ &= \int_a^x H(x-a) (x-u) f^{\prime\prime}(u)du + \int_a^x H(a-x) (x-u) f^{\prime\prime}(u)du \\ &\stackrel{1}{=} \int_a^x H(x-a) (x-u) f^{\prime\prime}(u)du + \int_x^a H(a-x) (u-x) f^{\prime\prime}(u)du \\ &= I_1 + I_2. \end{aligned} \tag{8}$

In step $1$, we used the fact from calculus that if we switch the limits of integration, we flip the sign. (See here if needed.) We then “pushed” the negative sign into the term $(x-u)$.

Now let’s rewrite each integral in the last line of Equation $8$ into a form that matches their respective integrals in Equation $1$. To do this, consider the following derivation of the left integral $I_1$ (commentary to follow):

$\begin{aligned} I_1 &= \int_a^x H(x-a)(x-u)f^{\prime\prime}(u)du \\ &\stackrel{1}{=} \int_a^x H(x-a)[x-u]^{+} f^{\prime\prime}(u)du \\ &\stackrel{2}{=} \int_a^{\infty} H(x-a)[x-u]^{+} f^{\prime\prime}(u)du \\ &\stackrel{3}{=} \int_a^{\infty} \left(1 - H(a-x) \right) [x-u]^{+} f^{\prime\prime}(u)du \\ &= \int_a^{\infty} [x-u]^{+} f^{\prime\prime}(u)du - \int_a^{\infty} H(a-x) [x-u]^{+} f^{\prime\prime}(u)du \\ &\stackrel{4}{=} \int_a^{\infty} [x-u]^{+} f^{\prime\prime}(u)du. \end{aligned} \tag{9}$

Now let’s explain each step.

Step $1$ holds because the Heaviside function filters for $x \gt a$:

$H(x-a) = \begin{cases} 1 & x \gt a, \\ 0 & x \leq a. \end{cases} \tag{10}$

Thus, the dummy variable $u$ is less than or equal to $x$, so $[x-u]^{+} = (x-u)$.

Step $2$ holds because the ramp function filters for $u \lt x$. When $u \geq x$, then $[x-u]^{+} = 0$. So we can extend the limits of integration without changing the values being integrated.

Step $3$ holds by Equation $7$, which is easy to convince yourself is true.

Step $4$ is a bit tricky. Let’s focus only on the right integral, the one that equals zero. Recall that we haven’t assumed anything about the ordering of $a$ and $x$. However, step $4$ holds because in either case, the integral is zero since

$\begin{aligned} a \leq x &\quad\implies\quad H(a-x) = 0, \\ a \gt x &\quad\implies\quad u \gt x \quad\implies\quad [x-u]^{+} = 0. \end{aligned} \tag{11}$

Finally, we can rewrite the right integral $I_2$ as:

$\begin{aligned} I_2 &= \int_x^a H(a-x) (u-x) f^{\prime\prime}(u)du \\ &= \int_x^a H(a-x) [u-x]^{+} f^{\prime\prime}(u)du \\ &= \int_{-\infty}^a H(a-x) [u-x]^{+} f^{\prime\prime}(u)du \\ &= \int_{-\infty}^a \left[1 - H(x-a) \right] [u-x]^{+} f^{\prime\prime}(u)du \\ &= \int_{-\infty}^a [u-x]^{+} f^{\prime\prime}(u)du - \int_{-\infty}^a H(x-a) [u-x]^{+} f^{\prime\prime}(u)du \\ &= \int_{-\infty}^a [u-x]^{+} f^{\prime\prime}(u)du. \end{aligned} \tag{12}$

I won’t add commentary here, since the logic is completely analogous to the logic used in Equation $9$.

A financial interpretation

Putting the results from the last two sections together, we have shown a slight generalization of the desired result:

$f(x) = f(a) + f^{\prime}(a) (x - a) + \int_{-\infty}^a [u-x]^{+} f^{\prime\prime}(u)du + \int_a^{\infty} [x-u]^{+} f^{\prime\prime}(u) du. \tag{13}$

Now let’s make some financial assumptions. Clearly, $g = f$, $S_t = x$, $F = a$, and $K = u$. Since $S_t \geq 0$, then clearly the lower limit for the left integral above can be written as $u=0$. And of course, the ramp functions represent the values of calls and puts or

$\begin{aligned} p(S_t, k) &:= [k - S_t]^{+}, \\ c(S_t, k) &:= [S_t - k]^{+}. \end{aligned} \tag{14}$

A common framing is to set $F$ in Equation $1$ to the forward price (hence the notation), so

$\mathbb{E}[S_t] = F. \tag{15}$

Here, the expectation is with respect to the risk-neutral measure. Then the second term in the Taylor expansion in Equation $1$ disappears, and we’re left with

$\mathbb{E}[g(S_t)] = g(F) + \int_{0}^{F} \mathbb{E}\left[p(S_t, k)\right] g^{\prime\prime}(k)dk + \int_{F}^{\infty} \mathbb{E}\left[c(S_t, k)\right] g^{\prime\prime}(k) dk. \tag{16}$

which looks quite promising. In words, the expected value of the payoff of a derivative of $S_t$ can be expressed as its intrinsic value plus a portfolio of options on the underlying. This relationship is the conceptual basis for static hedging or static replication, since we can hedge a possibly complex position with a basket of standard European options.