Transforming the Black–Scholes PDE into Forward Log Space
Summary
The document examines how to transform the Black–Scholes pricing equation from log spot coordinates to log forward coordinates. It starts with the risk-neutral log-price process and the associated pricing PDE, then introduces a forward coordinate whose drift excludes the interest-rate term. The original attempt applies the spatial chain rule but overlooks that the coordinate transformation also depends on time.
The answer distinguishes the pricing functions in the two coordinate systems and relates them using the fact that the terminal spot and forward values coincide under the stated setup. Applying the chain rule to this time-dependent relationship adds a drift contribution to the time derivative. Combining that term with the original PDE yields the forward-coordinate equation. This is a mathematical derivation for the Black–Scholes setting with constant volatility and rate as presented; it does not address numerical solution methods or extensions to more general processes.
Key ideas
- Changing from log spot to log forward coordinates also changes the time derivative because the transformation depends on time.
- The pricing functions in the two coordinate systems are related through their shared terminal value.
- The spatial first and second derivatives transform directly under the stated coordinate shift.
- Including the time-chain-rule term recovers the forward-coordinate pricing equation.
- The derivation is limited to the Black–Scholes assumptions used in the document.
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# Black Scholes PDE in forward log space
# Black Scholes PDE in forward log space
In BS world, we have the stock process in log space $dS_t=(r-\frac{1}{2}\sigma^2)dt+\sigma dW$. Let's say we want to price $f(t,x)=\mathbb{E}_{t,x}[h(S(T)]$. Using Feynman-kac, we get \begin{equation} \frac{\partial f}{\partial t} + (r-\frac{1}{2}\sigma^2)\frac{\partial f}{\partial x}+\frac{1}{2} \sigma^2 \frac{\partial^2 f}{\partial x^2}-rV=0 \end{equation}
On the other hand, if we consider the forward process (again in log space) $F_t=S_t+r(T-t)$, we have the forward process $dF_t=-\frac{1}{2}\sigma^2 dt+\sigma dW$ and the price becomes $f(t,y)=\mathbb{E}_{t,y}[h(F(T)]$. Using F-K again, we get \begin{equation} \frac{\partial f}{\partial t} - \frac{1}{2}\sigma^2\frac{\partial f}{\partial y}+\frac{1}{2} \sigma^2 \frac{\partial^2 f}{\partial y^2}-rV=0 \end{equation}
Somehow I fail to transform the first PDE to the second by change of variable directly from $S_t$ to $F_t$. Since $y=x+r(T-t)$, by chain rule, $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial x}\frac{\partial y}{\partial x}=\frac{\partial f}{\partial y}$, i.e., the first order is the same and so as the second order. So I end up with \begin{equation} \frac{\partial f}{\partial t} +(r-\frac{1}{2}\sigma^2)\frac{\partial f}{\partial y}+\frac{1}{2} \sigma^2 \frac{\partial^2 f}{\partial y^2}-rV=0 \end{equation} which is obviously wrong and I couldn't figure out why.
## Answer by user34971 (score 4, accepted)
https://quant.stackexchange.com/a/69713
It's probably best to use different notation. So first of all $$ f(t,x) := E_t (h(S_T)) $$ and $$ g(t, y) := E_t (h(F_T)). $$ Since $S_T = F_T$, by no arbitrage we must have $$ g(t,y) = f(t,x) = f(t, y - r(T-t)). $$
This means that $$ \frac{\partial g}{\partial t} = \frac{\partial f}{\partial t} + r \frac{\partial f}{\partial x}. $$ As you've already pointed out $$ \frac{\partial g}{\partial y} = \frac{\partial f}{\partial x}. $$ Using this and the PDE satisfied by $f$ you will then obtain the following PDE for $g$: $$ \frac{\partial g}{\partial t} -\frac12 \sigma^2 \left( \frac{\partial g}{\partial y} - \frac{\partial^2 g}{\partial y^2}\right) = rg $$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.