Transforming the Black–Scholes Payoff into a Diffusion Initial Condition

Summary

This exchange explains why a call payoff can look different after the change of variables used to convert the Black–Scholes partial differential equation into a diffusion equation. The question compares a payoff written using the stock price and strike with a transformed expression involving exponentials of the new spatial variable. The answer identifies an intermediate variable, v, whose initial condition is the normalized call payoff max(e^x − 1, 0), and then applies the exponential factor relating u to v. That factor shifts the exponents and gives the stated initial condition for u.

The explanation is an algebraic clarification, rather than a derivation of the full PDE transformation. It does not resolve every convention or notation difference between the question and the referenced discussion, and it gives no numerical example. Readers must ensure that the variables and scaling used in their own transformation match those assumed in the answer.

Key ideas

• The transformed initial condition depends on the scaling that relates u to v.
• A normalized call payoff can be written as max(e^x − 1, 0).
• Multiplying by the exponential prefactor shifts the exponents in the payoff.

Tags

# Black-Scholes to Diffusion Initial Condition


# Black-Scholes to Diffusion Initial Condition












I'm having troubles with the transformation from the Black-Scholes PDE and transforming it to the diffusion equation. I read this other stackexchange post (Here) and I understand most of the process, except where they changed the initial condition.

I got \begin{equation} \begin{split} u(x,0) &= e^{r\tau}C(S,T)\\ &=e^{r\tau}\text{max}(S-K,0)\\ &=e^{r\tau}\text{max}(e^y-K,0)\\ &=e^{r\tau}\text{max}(e^{x-(r-\sigma^2/2)\tau)}-K,0)\\ &=\text{max}(e^{x+\sigma^2\tau/2}-e^{r\tau}K,0)\\ \end{split} \end{equation}

Which is different from their equation of: \begin{equation} u(x,0) = u_0(x) = \text{max}(e^{\frac{1}{2}(a+1)x}-e^{\frac{1}{2}(a-1)x},0) \end{equation}

Where $a=2r/\sigma^2$

I would comment on the other post, however I don't have enough 'reputation' and this is a very specific question that I can't find elsewhere. Apparently it's in the textbook referenced in the original post, but the particular page referenced isn't freely available.

## Answer by user1157 (score 1)

https://quant.stackexchange.com/a/44829

$$u(x,0)=e^{\frac{k-1}{2}x}v(x,0)$$

and $$v(x,0)=max(e^{x}-1,0)$$ Hence $$u(x,0)=max(e^{\frac{k+1}{2}x}-e^{\frac{k-1}{2}x},0)$$

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