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Really new to financial Maths. I am currently having problems with the Carr-Madan Formula.

$$f(S_T)=f(F_t) + f'(F_t) (S_T - F_t) + \int_0^{F_t} f''(K) (K-S_T)^+ \ d K + \int_{F_t}^{\infty} f''(K) (S_T-K)^+ \ d K$$

I am struggling to understand what it is used for and I can't seem to find any good articles to explain what is going on. I was wondering could someone recommend any readings, given that I quite new to this.

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  • $\begingroup$ It's Carr-Madan not Mahan, and it's used for replicating and hedging exotics $\endgroup$
    – Kiwiakos
    Jun 15, 2016 at 15:03
  • $\begingroup$ I appreciate your input, however I was just wondering do you know any articles where the carr-mahan formula is explained in laymans terms. Like I said I am really new to it. $\endgroup$
    – user3223190
    Jun 15, 2016 at 15:20
  • $\begingroup$ This is nothing special and more than the Taylor expansion formula. $\endgroup$
    – Hans
    Jan 7, 2020 at 2:30

3 Answers 3

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For a sufficiently smooth function $f$, positive constant $a$, and $x>0$, Note that, \begin{align*} f(x) -f(a) &= \int_a^{x} f'(v) dv \\ &= \int_a^{x} \big[f'(v) -f'(a) + f'(a) \big] dv \\ &= f'(a) (x-a) + \int_a^{x}\!\! \int_a^v f''(u)du dv\\ &= f'(a) (x-a) + \int_a^{x}\!\! \int_u^{x} f''(u)dv du\\ &= f'(a) (x-a) + \int_a^{x}f''(u)(x-u)du. \end{align*} Then, \begin{align*} f(x) &= f(a) + f'(a) (x-a) + \int_a^{x}(x-u)f''(u)du \\ &= f(a) + f'(a) (x-a) + \int_a^{x}\big(\pmb{1}_{a \leq x} + \pmb{1}_{a > x} \big)(x-u)f''(u)du \\ &= f(a) + f'(a) (x-a) + \int_a^{x} \pmb{1}_{a \leq x}\,(x-u)f''(u)du + \int_x^{a} \pmb{1}_{a > x}\,(u-x)f''(u)du \\ &= f(a) + f'(a) (x-a) + \int_a^{x} \pmb{1}_{a \leq x}\,(x-u)^+f''(u)du + \int_x^{a} \pmb{1}_{a > x}\,(u-x)^+f''(u)du \\ &= f(a) + f'(a) (x-a) + \int_a^{\infty} \pmb{1}_{a \leq x}\, (x-u)^+f''(u)du + \int_{0}^a \pmb{1}_{a \geq x}\, (u - x)^+f''(u)du \\ &=f(a) + f'(a) (x-a)\\ &\qquad + \int_a^{\infty}(1- \pmb{1}_{x < a})\, (x-u)^+f''(u)du + \int_{0}^a (1-\pmb{1}_{x>a})\, (u - x)^+f''(u)du \\ &= f(a) + f'(a) (x-a) + \int_a^{\infty}(x-u)^+f''(u)du + \int_{0}^a(u - x)^+f''(u)du. \end{align*} This formula is used in the valuation of a variance swap, and, as an approximation, the constructuion of the VIX index; see https://www.cboe.com/micro/vix/vixwhite.pdf.

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    $\begingroup$ Thanks @Siron. The delta function is nice, but I like to derive the formula using a rudimentary approach. $\endgroup$
    – Gordon
    Jun 16, 2016 at 18:39
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    $\begingroup$ Since vanilla option prices are readily observable, why doesn't this formula solve option pricing, period. Any European claim can be priced this way, doesn't it? Then why bother with modeling? $\endgroup$
    – starovoitovs
    Mar 28, 2019 at 19:23
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    $\begingroup$ It is the Taylor expansion. However, to reach the final form is not trivial. Moreover, the Taylor is not related to options. I will think it worth a new name. $\endgroup$
    – Gordon
    Jan 7, 2020 at 3:07
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    $\begingroup$ Does the this proof of Carr-Madan formula work also for every $x\in\mathbb{R}$ and every $a\in\mathbb{R}$? $\endgroup$
    – B_B
    Aug 30, 2020 at 9:58
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    $\begingroup$ @B_B: By going through the proof, you will notice that the proof works for all $x, a \in \mathbb{R}$. $\endgroup$
    – Gordon
    Aug 31, 2020 at 12:12
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The main interest of the formula is that it allows you, at least theoretically, to replicate any European option with payoff $f(\cdot)$ using only Call and Put options. As simple examples, consider $f(S)=S$ and $f(S)=(S-K)^+$.

The formula also implies that knowing all Puts and Calls for all strikes for a given maturity gives you the price of any European option with the same maturity.

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    $\begingroup$ we need $f$ to be smooth, however, $f(s)=(s-k)^+$ is not. $\endgroup$
    – Gordon
    Jun 15, 2016 at 17:20
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    $\begingroup$ One could understand the right hand side in the sense of distributions in that case $\endgroup$
    – user22171
    Jun 15, 2016 at 17:22
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    $\begingroup$ Non-smooth payoffs can often be written $f(S)1_{S\in \mathcal{A}}$ where $\mathcal{A}$ is some set thus the formula can be applied to $f(S)$ and the whole result can then be multiplied by $1_{S\in \mathcal{A}}$. $\endgroup$
    – Daneel Olivaw
    Mar 27, 2019 at 13:44
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If $f\colon\mathbf{R}\to\mathbf{R}$ has a piecewise continuous second derivatve, then \begin{align*} f(x) = f(a) + f'(a)(x-a) + \int_{-\infty}^a (k - x)^+ f''(k)\,dk + \int_a^\infty (x - k)^+ f''(k)\,dk. \end{align*} Note this formula holds for $x = a$. Taking a derivative with respect to $x$ yields \begin{align*} f'(x) &= f'(a) + \int_{-\infty}^a -1(x \le k) f''(k)\,dk + \int_a^\infty 1(x \ge k) f''(k)\,dk\\ &= f'(a) - \int_{\min\{x, a\}}^a f''(k)\,dk + \int_a^{\max\{x,a\}} f''(k)\,dk\\ \end{align*} Note this formula holds for $x = a$. Taking a derivative with respect to $x$ yields \begin{align*} f''(x) = f''(x)1(x < a) + f''(x)1(x > a)(k) \end{align*} for $x\not= a$. Note the left and right limits as $x\to a$ equal $f''(a)$. This proves the original formula is valid.

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