Deriving the Discounted Black–Scholes Equation by Change of Variables
Summary
The document shows how to transform the Black–Scholes partial differential equation into a simpler equation for the discounted option value as a function of the discounted underlying. It begins with the original equation and defines both discounted quantities using the risk-free rate. Applying the chain rule to the time and underlying-price derivatives gives the relationships needed to substitute back into the original equation.
After substitution, the interest-rate terms cancel, leaving the time derivative of the discounted option value equal to the negative diffusion term involving its second derivative with respect to the discounted underlying. The explanation provides an algebraic derivation rather than empirical evidence or a trading strategy. It assumes the Black–Scholes model and its usual smoothness conditions; the answer’s displayed derivative notation has a few inconsistencies, so care is needed to distinguish derivatives with respect to the original and discounted price variables.
Key ideas
- Discount both the underlying price and option value using the risk-free rate.
- Use the chain rule to relate derivatives in discounted and undiscounted variables.
- Substituting those relationships into the Black–Scholes equation cancels the rate terms.
- The resulting equation retains only the diffusion contribution to discounted option value.
- The derivation relies on the Black–Scholes model and consistent derivative notation.
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Full text
# Black-Scholes and Markovian contingent claim
# Black-Scholes and Markovian contingent claim
Background information:
Proposition 4.1 - For a European Markovian contingent claim, the Black-Scholes price satisfies $$\Theta(\tau,S) = -\frac{\sigma^2 S^2}{2}\Gamma(\tau,S) - rS\Delta(\tau,S) + rV(\tau,S)$$
Problem - Use Proposition 4.1, to show the above equality. That is let $\tilde{S}_t = e^{-rt}S_t$ and $\tilde{V}(t,\tilde{S}_t) = e^{-rt}V(t,S_t)$ be respectively the discounted underlying price and discounted option price. Then, we can show that $$\partial_t\tilde{V}(t,\tilde{S}) = -\frac{\sigma^2\tilde{S}^2}{2}\partial_{\tilde{S}\tilde{S}}V(t,\tilde{S})$$
I am confused where to begin, I think I can manage showing this if I have a formula or just a set up to work from, any suggestions is greatly appreciated.
## Answer by Gordon (score 1, accepted)
https://quant.stackexchange.com/a/24537
We assume the following Black-Scholes equation: \begin{align} \frac{\partial V}{\partial t} = -\frac{\sigma^2 S_t^2}{2}\frac{\partial^2 V}{\partial S_t^2} -r S_t \frac{\partial V}{\partial S_t} +r V.\tag{1} \end{align} From the assumption, \begin{align} V(t,\, S_t) = e^{rt}\tilde{V}(t,\, \tilde{S}_t).\tag{2} \end{align} Then \begin{align*} \frac{\partial V}{\partial t} &= re^{rt}\tilde{V}+e^{rt}\frac{\partial \tilde{V}}{\partial t} + e^{rt}\frac{\partial \tilde{V}}{\partial \tilde{S}_t}\frac{\partial \tilde{S}_t}{\partial t}\\ &= re^{rt}\tilde{V}+e^{rt}\frac{\partial \tilde{V}}{\partial t} -rS_t \frac{\partial \tilde{V}}{\partial \tilde{S}_t},\tag{3}\\ \frac{\partial V}{\partial S_t} &= e^{rt}\frac{\partial \tilde{V}}{\partial S_t}\\ &=e^{rt}\frac{\partial \tilde{V}}{\partial \tilde{S}_t}\frac{\partial \tilde{S}_t}{\partial S_t}\\ &=\frac{\partial \tilde{V}}{\partial \tilde{S}_t},\tag{4}\\ \frac{\partial^2 V}{\partial S_t^2} &=\frac{\partial^2 \tilde{V}}{\partial \tilde{S}_t^2}\frac{\partial \tilde{S}_t}{\partial S_t}\\ &=e^{-rt}\frac{\partial^2 \tilde{V}}{\partial \tilde{S}_t^2}.\tag{5} \end{align*} Now, plugin (2)-(5) into (1), we obtain that \begin{align*} \frac{\partial \tilde{V}}{\partial t} = -\frac{\sigma^2 \tilde{S}_t^2}{2} \frac{\partial^2 \tilde{V}}{\partial \tilde{S}_t^2}. \end{align*}
## Answer by mbison (score 0)
https://quant.stackexchange.com/a/24522
A good place to start is to calculate the $\Delta$ for your option $\tilde{V}$. Once you calc your delta, plug it back into the black scholes PDE. Now check which terms cancel each other out. After you remove the terms that cancel each other out you have the desired result.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.