Applying a Log-Price Change of Variables to the Black–Scholes PDE
Summary
This explanation derives the first and second derivatives of an option value after changing variables from the underlying price S to its logarithm ξ = log(S). It represents the value as a discounted function of the transformed variable, then applies the chain and product rules. The resulting second derivative contains both the transformed function’s second derivative and a negative first-derivative term, each scaled by the inverse square of the underlying price.
The derivation also explains why the discount factor can be omitted when substituting into the Black–Scholes PDE: it is common to every term and can be divided out. The note gives an algebraic walkthrough rather than a full derivation of the PDE transformation. It relies on the stated discounted representation and does not discuss boundary conditions, payoff specifics, or alternative transformations.
Key ideas
- Under ξ = log(S), the derivative with respect to S uses dξ/dS = 1/S.
- The second derivative includes a negative term from differentiating 1/S.
- The second derivative also includes the transformed function’s second derivative, scaled by 1/S².
- A common discount factor across the PDE terms can be divided out.
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# Change of variables - Black-Scholes
# Change of variables - Black-Scholes
While manipulating the Black-Scholes equation, Paul Wilmott (Paul Wilmott on Quantitative Finance, chapter 7, page 110) makes the following change of variables and I am having trouble understanding how he evaluated the second derivative:
## Answer by Pleb (score 1, accepted)
https://quant.stackexchange.com/a/60291
I do not own the book you're referring to. However, we know that the payoff of an option has the following representation in order to satisfy the Black-Scholes PDE (this is a consequence of the Feynman-Kac formula):
$$V(S,t)= e^{-r(T-t)}\cdot U(S,t) = X \cdot U(S,t),$$
where $X = e^{-r(T-t)}$ for convenience. Now, using a change of variable, $\xi = log(S)$, the above equation becomes:
$$ V(\xi,t)= X \cdot U(\xi,t)$$
and therefore we get the following derivatives by applying the chain-rule and the quotation-rule (under the second-order derivative):
\begin{align} \frac{\partial V}{\partial S} &= X \cdot \frac{\partial U}{\partial \xi} \cdot \frac{\partial \xi}{\partial S} \\ &= X \cdot \frac{\partial U}{\partial \xi} \frac{1}{S} \\ &= X \cdot e^{-\xi} \cdot \frac{\partial U}{\partial \xi}\\ \end{align} where $S=e^{\xi}$ from the change of variable. Moreover, we get: \begin{align} \frac{\partial^2V}{\partial S^2} &= \frac{\partial}{\partial S} \left[X \cdot \frac{\partial U}{\partial \xi} \frac{1}{S}\right]\\ &= X \cdot \frac{\partial U}{\partial \xi} \cdot \frac{-1}{S^2} + X \cdot \frac{\partial}{\partial S} \left(\frac{\partial U}{\partial \xi}\right) \cdot \frac{1}{S}\\ &= X \cdot \frac{\partial U}{\partial \xi} \cdot \frac{-1}{S^2} + X \cdot \frac{\partial^2 U}{\partial \xi^2} \cdot \frac{\partial \xi}{\partial S} \cdot \frac{1}{S}\\ &= X \cdot \frac{\partial U}{\partial \xi} \cdot \frac{-1}{S^2} + X \cdot \frac{\partial^2 U}{\partial \xi^2} \cdot \frac{1}{S^2}\\ &=-X\cdot e^{-2\xi} \cdot \frac{\partial U}{\partial \xi} + X \cdot e^{-2\xi} \cdot \frac{\partial^2 U}{\partial \xi^2}. \end{align} Now, under the Black-Scholes PDE all of the terms have common factor $X$ and thus you can conveniently divide out the common factor. I believe, that is why $X$ is not included in your above specified derivations, since it does not contribute to the overall solution of the PDE.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.