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Deriving the Call-Price Density with the Correct Strike Derivative Sign

Article Quant Q&A · Author: user30614
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Summary

The document explains how to obtain the terminal asset-price density from the second derivative of a European call price with respect to strike. It starts from the discounted payoff integral and its first strike derivative, which is minus the discounted probability of finishing above the strike. The sign issue arises when differentiating the Heaviside function: its argument is asset price minus strike, so the chain rule contributes a negative sign when differentiating with respect to strike.

Applying the Dirac delta property then cancels the leading minus sign and gives the discounted density evaluated at the strike. The answer also offers the Leibniz integral rule as a direct way to check the derivative of the tail probability. This is a concise calculus explanation rather than a treatment of broader option pricing assumptions; it presumes the stated pricing integral and density representation.

Key ideas

  • Differentiating the call price once with respect to strike gives the negative discounted tail probability.
  • The derivative of a Heaviside function depends on the derivative of its argument.
  • Because the argument is asset price minus strike, its strike derivative is negative one.
  • The second strike derivative of the call price equals the discounted terminal density at the strike.
  • The Leibniz integral rule provides an independent check of the sign.

Tags

OptionsDerivatives pricingStatistics
Full text
# Probability Density Function sign problem when using Call Price


# Probability Density Function sign problem when using Call Price












Given the call price $C_t = e^{-\int_t^Tr(s)ds}\int (s-K)^{+}\phi_{S_T}(T,s)ds $

we know that $$\frac{dC}{dK}=-e^{-\int_t^Tr(s)ds}\int_K^{\infty} \phi_{S_T}(T,s)ds$$

Now when I use dirac delta function property $ \int f(t)\delta(t-T)dt = f(T) $when taking the second derivative with respect to $K$ I find an akward sign minus which I don't know where my calclulus is wrong :

$$\frac{d^2}{dK^2}C=-e^{-\int_t^Tr(s)ds}\int \frac {d}{dK} H(s-K) \phi_{S_T}(T,s)ds$$

with $H$ the heavy side function, which gives : $$\frac{d^2}{dK^2}C= -e^{-\int_t^Tr(s)ds}\int \delta(s-K) \phi_{S_T}(T,s)ds $$

$$\frac{d^2}{dK^2}C= -e^{-\int_t^Tr(s)ds}\phi_{S_T}(T,K)$$

and the right result should be without the minus sign here. Where did I go wrong? I know how to find the right result using just the indicator function, but I wanna use the delta function property here to derive it. Any help?

## Answer by Quantuple (score 2, accepted)

https://quant.stackexchange.com/a/42126

Note that with $H(\cdot)$ the Heaviside function $$\frac{d}{ds} H(s-K) = \delta(s-K)$$ but $$\frac{d}{dK} H(s-K) = \color{red}{-}\delta(s-K)$$

You can also use the Leibniz integral rule to write that $$ \frac{d}{dK} \int_K^\infty \phi_{S_T}(T,s) ds = -\phi(S_T,K) $$

## Answer by Magic is in the chain (score 1)

https://quant.stackexchange.com/a/42128

Just the chain rule;

$\frac{d}{dK} H \left (S-K\right)=\delta \left (S-K\right) \frac{d}{dK} \left (S-K\right)=-\delta \left (S-K\right) $

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.