Deriving the Black–Scholes PDE for Power Options
Summary
The document explains how changing variables from an asset price S to its power Z = S^α transforms the Black–Scholes partial differential equation for a payoff based on S^α. The key point is applying the chain rule to both the first and second derivatives. The second derivative includes an extra term proportional to α(α−1), in addition to the squared first-derivative factor. Multiplying by the diffusion coefficient produces the drift contribution that the questioner could not account for.
The answer works through that expansion and shows why the transformed equation has both a second-order term in Z and an additional first-order term. It clarifies that the αr portion comes from transforming the original drift term, while the remaining contribution comes from differentiating the change-of-variable factor in the second derivative. The discussion is a focused derivation rather than a broader treatment of power-option pricing; it does not address boundary conditions or the resulting closed-form prices.
Key ideas
- Changing variables to Z = S^α requires applying the chain rule to each derivative in the Black–Scholes PDE.
- The second derivative contains an extra first-derivative term from differentiating αS^(α−1).
- That extra term contributes α(α−1)σ²/2 to the transformed drift coefficient.
- The transformed diffusion term is proportional to α²σ²Z².
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Full text
# Power option's PDE
# Power option's PDE
I am looking to understand the PDE of Power Options in Paull Willmot on Quantitative Finance (2nd Ed), Ch. 8.9 - Formulae for Power Options (p. 149).
Suppose the payoff depends on the asset price at expiry raised to some power i.e.: $\text{Payoff}(S^\alpha)$. If we write $ Z = S^\alpha $ the standard Black-Scholes PDE:
$${\frac {\partial V}{\partial t}} +{\frac {1}{2}}\sigma ^{2}S^{2}{\frac {\partial ^{2}V}{\partial S^{2}}} +rS{\frac {\partial V}{\partial S}}-rV=0$$
The books says it can be transformed to the following:
$${\frac {\partial V}{\partial t}} +{\frac {1}{2}}\alpha^2\sigma ^{2}Z^2{\frac {\partial ^{2}V}{\partial Z^{2}}} +\alpha\left(\frac {1}{2}\sigma ^{2}(\alpha -1 ) + r\right)Z{\frac {\partial V}{\partial S}}-rV=0$$
Where does the $\alpha\left(\frac {1}{2}\sigma ^{2}(\alpha -1 ) + r\right)$ term come from? Actually, only the $\alpha\left(\frac {1}{2}\sigma ^{2}(\alpha -1 ) \right)$ bit I cannot figure out. $\alpha r $'s source is clear.
I get an entirely different expression for the delta term, $rS{\frac {\partial V}{\partial S}}$, when I change the variables. Namely:
$${\frac {\partial V}{\partial S}} = {\frac {\partial V}{\partial Z} \frac{\partial Z}{\partial S}} = {\frac {\partial V}{\partial Z}} \alpha S ^{\alpha-1} $$ and so:
$$rS{\frac {\partial V}{\partial S}} = rS {\frac {\partial V}{\partial Z}} \alpha S ^{\alpha-1} = r \alpha S^\alpha {\frac {\partial V}{\partial Z}} = r \alpha Z{\frac {\partial V}{\partial Z}}$$
## Answer by userPrimeNumber (score 2, accepted)
https://quant.stackexchange.com/a/69857
As mentioned by James, the expansion of the ${\frac {1}{2}}\sigma ^{2}S^{2}{\frac {\partial ^{2}V}{\partial S^{2}}}$ term yields:
$$ \frac {\partial ^{2}V}{\partial S^{2}} = \frac {\partial}{\partial S}\left(\frac {\partial V}{\partial Z}\alpha S ^{\alpha-1}\right) = \frac {\partial}{\partial S}\left(\frac {\partial V}{\partial Z}\right)\alpha S ^{\alpha-1} + \frac {\partial V}{\partial Z}\alpha (\alpha-1)S ^{\alpha-2} \\= \alpha S ^{\alpha-1}\frac {\partial^2 V}{\partial Z^2}\frac {\partial Z}{\partial S} + \frac {\partial V}{\partial Z}\alpha (\alpha-1)S ^{\alpha-2} \\ = \frac {\partial V}{\partial Z}\alpha (\alpha-1)S ^{\alpha-2} + \alpha^2 (S ^{\alpha-1})^2\frac {\partial^2 V}{\partial Z^2} $$ And so: $${\frac {1}{2}}\sigma ^{2}S^{2}{\frac {\partial ^{2}V}{\partial S^{2}}} = \frac {1}{2}\sigma ^{2} \left(\alpha (\alpha-1)S ^{\alpha} \frac {\partial V}{\partial Z} + \alpha^2 (S ^{\alpha})^2\frac {\partial^2 V}{\partial Z^2} \right) = \frac {1}{2}\sigma ^{2} \left(\alpha (\alpha-1)Z\frac {\partial V}{\partial Z} + \alpha^2 Z^2\frac {\partial^2 V}{\partial Z^2} \right)$$ which explains the additional term of $\frac {\partial V}{\partial Z}$ from the question.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.