I know that a way of computing the price of a derivative paying $S^2$ at time $T$ is by making use of the following strategy:

$V=\int_{0}^{\infty} s^2 \frac{\partial^2 C}{\partial K^2}(K=s)ds$

Where $\frac{\partial^2 C}{\partial K^2}(K=s)$ is simply the risk neutral distribution of $S$.

Now, one should apply integration by parts twice in order to get an integral of the call prices. My question is, what is the name of this strategy/formula? Where can I find a derivation of it?

EDIT: By integrating by parts once, I get:

$V=\left[ s^2 \frac{\partial C}{\partial K}(K=s) \right]^{\infty}_{0} -\int_{0}^{\infty} 2s \frac{\partial C}{\partial K}(K=s)ds$

The first term is zero, but I don't know why. Can you please provide an explanation? Now by integrating by parts a second time the first derivative should become simply the price of options. But as I don't know why the first term is zero I hesitate to continue the derivation. Can you please help with next steps? EDIT: By taking into account the answer provided, the first term is zero,

$V= -\int_{0}^{\infty} 2s \frac{\partial C}{\partial K}(K=s)ds$

then we integrate by parts a second time:

$V= -\left(\left[2kC(k)\right]_{0}^{\infty}-\int_{0}^{\infty} 2 C(s)ds \right)$

Now, as



$V= 2\int_{0}^{\infty}C(K=s)ds $

Is that correct?

  • 2
    $\begingroup$ I think that one way to get a static replication formula is to use the Carr-Madan-formula and set $f(S_T)=S_T^2$. $\endgroup$
    – Count
    Sep 7, 2021 at 8:18
  • $\begingroup$ I cannot do the link between my question and Carr-adan formula. Maybe it is simpler if I continue the derivation. See the edit, can you help me with the next step? $\endgroup$
    – Joanna
    Sep 7, 2021 at 19:41

2 Answers 2


Proof sketch for

$$ \lim_{S\rightarrow \infty} S^2 \frac{\partial C}{\partial K}(K=S) = 0. $$

We have:

$$C(K) = E\left[(S_T-K)^+\right] = \int_K^\infty (u-K)f_{S_T}(u) du$$ $$ \frac{\partial C}{\partial K} = \int_K^\infty \frac{\partial }{\partial K}(u-K)f_{S_T}(u) du = -\int_K^\infty f_{S_T}(u) du =-(1-F_{S_T}(K))$$

$$ \frac{\partial C}{\partial K}(K=\infty) = 0 $$

Next, I think one might need to assume that the complementary cdf has a thin tail:

$$ \lim_{x\rightarrow \infty} x^2 (1-F_{S_T}(x)) = 0$$

In the case of a normal distribution this is true as, according to WolframAlpha, we have: $$ \lim_{x\rightarrow \infty} x^2 {\rm erfc(x)} = 0.$$

Note: See also this Quant SE direct solution.


Since your payoff only depends on $S_T$, you could use the Carr-Madan-formula

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

to get a static replication formula. In your example you have $f(S_T)=S_T^2$ and therefore $f'(F_t)=2F_t$ and $f''(F_t)=2$. Then choose $F_t=0$ to get:

$$ S_T^2=\int_0^\infty2(S_T-K)^+\mathrm{d}K=2\int_0^\infty(S_T-K)^+\mathrm{d}K $$

So you can replicate the payoff by using a portfolio of European-calls. Maybe this solution is easier than this integration by parts stuff.


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