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I would like to understand the derivation that $$\phi(S)=\frac{\partial^2C}{\partial K^2}$$ from $$C(K) = \int_{0}^{\infty} max(S-K,0)\phi(S)dS$$ where $\phi$ is a probability density function.

(This equality is relevant for example to estimate transition densities for a European call from the prices of call options.)

Edit: the following does not work since the derivative of $max(S-K,0)$ is not continuous:

I see that I get this result by differentiating two times with respect to K, but only if I can exchange the differential operator with the integral operator ds:

$$\frac{\partial C}{\partial K}(\int_{0}^{\infty} max(S-K,0)\phi(S)dS$$ $$=\int_{0}^{\infty}\phi(S)\frac{\partial}{\partial K}( max(S-K,0))dS$$ $$=\int_{0}^{\infty}\phi(S)(-1)dS\quad\text{if } K<S, \quad 0 \text{ otherwise}$$ If I can do this, the desired result appears, since the dirac function has derivative 1 exactly at $K=S$.

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Note that $$C(S,K)=\int_{K}^{\infty}(S-K)\phi(S)\;dS$$ and so assuming $\phi(S)$ and $S\phi(S)$ is integrable we're fine. However, since $K$ appears in both the limits and the integrand, we have boundary terms to deal with and therefore we need Leibniz's rule; incidentally however, the boundary terms are all zero in this case. We have then $$\frac{\partial C}{\partial K}=\int_{K}^{\infty}-\phi(S)\;dS.$$ Differentiating a second time we get (this time just using the fundamental theorem of calculus, which is a special case of Leibniz's rule) $$\frac{\partial^{2}C}{\partial K^{2}}=-(-\phi(K)) = \phi(K)$$ as desired.

Daniel Fischer
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Sargera
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  • very nice. i learnt something today. The first observation is exceptionally crucial. – Lost1 Feb 17 '14 at 20:24
  • What you have not done is justifying the limit swap. do you know how to do that? curiously, i asked http://math.stackexchange.com/questions/673146/what-is-the-requirement-on-f-for-the-fracddx-intx-0fx-xdx-to-exis question. I think the link to Leibniz rule answer this. – Lost1 Feb 17 '14 at 20:28
  • Removing the max by changing the boundary of integration, didn't think of this. Thanks! –  Feb 17 '14 at 20:43
  • @Taylor I think before applying the fundamental theorem of calculus, it is neccesary to exchange the integral boundaries, so that the minus sign vanishes, right? –  Feb 19 '14 at 15:41
  • @Raphael No, the fundamental theorem works fine with the lower integral limit variable. You have - for continuous $f$ - $$\frac{d}{dx} \int_x^b f(t),dt = - f(x).$$ – Daniel Fischer Feb 19 '14 at 16:10
  • @DanielFischer But the minus sign in the last step above should be removed, right? –  Feb 19 '14 at 16:21
  • @Raphael It need not. Set $f(t) = -\phi(t)$ in the above (and replace $x$ with $K$), the differentiation leads to $-(-\phi(K))$ then. Of course, you can also write it $\int_\infty^K \phi(S),dS$ to get rid of the minus sign, but that's optional. – Daniel Fischer Feb 19 '14 at 16:27
  • @DanielFischer Thanks for your answers so far! I'm somehow stuck, so one more question: The second derivative, should it be $\phi$ or $-\phi$? –  Feb 19 '14 at 16:30
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    @Raphael $\frac{\partial^2 C}{\partial K^2}$? That's $\phi$. Taylor made a mistake at the end, since we differentiate the integral with respect to the lower bound, that differentiation produces a minus sign, and that cancels the one we got from the first differentiation. – Daniel Fischer Feb 19 '14 at 16:33
  • @Raphael But the differentiation produces the integrand evaluated at the integral limit, that is $K$ here. By the way, something is wrong in the question, you write $$C(S,K) = \int_0^\infty \max (S-K,0)\phi(S),ds,$$ which doesn't make sense. If $s = S$, the integral doesn't depend on $S$, so it ought to be $C(K)$ alone. If $s\neq S$, what is $S$, and what $s$, something in the integral must depend on $s$ then. – Daniel Fischer Feb 19 '14 at 16:45
  • let us continue this discussion in chat –  Feb 19 '14 at 16:51
  • Sorry, wrote the answer too hastily and made a sign mistake at the end. Thanks Daniel Fischer for the edit. – Sargera Feb 22 '14 at 01:36
  • @Raphael. The part where I mention "...as long as $\phi$ and $S\phi$ are integrable we're fine" is the justification for differentiation under the integral sign (see the link for Leibniz's rule). – Sargera Feb 22 '14 at 01:45
  • @Taylor thanks, understood. –  Feb 22 '14 at 08:40