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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.

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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 $g$: $$ \frac{\partial g}{\partial t} -\frac12 \sigma^2 \left( \frac{\partial g}{\partial y} - \frac{\partial^2 g}{\partial y^2}\right) = rg $$

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  • $\begingroup$ Indeed, it's about notation. Thanks a lot! $\endgroup$
    – J. Lin
    Feb 1, 2022 at 13:50

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