Suppose that $V(S,I,t)$ satisfies the equality

$$ \frac{\partial V}{\partial t}+\frac{1}{2}\sigma^2S^2 \frac{\partial^2V}{\partial S^2}+S\frac{\partial V}{\partial I}+rS\frac{\partial V}{\partial S}-rV=0 $$

Here $I=\int _0^t S dt$.

Now, let $R=S/I$ and $V(S,R,t)=I\,W(R,t)$. Is it true that $W$ satisfies the following equality?

$$\frac{\partial W}{\partial t}+\frac{1}{2}\sigma^2R^2\frac{\partial^2 W}{\partial R^2}+R(r-R)\frac{\partial W}{\partial R}-(r-R)W=0 $$

  • 1
    $\begingroup$ For arbitrary $f$? $\endgroup$ Commented Sep 7, 2021 at 20:46
  • $\begingroup$ Sorry about that .f(S,t) should be replaced by S.I have edited $\endgroup$
    – abc
    Commented Sep 7, 2021 at 21:33

1 Answer 1


Yes, but it looks like two things ought to be mentioned. First, the equality for $V$ needs to be clarified. I think it is a PDE for a function $V(S,I,t)$ of three independent variables. If that is so, the sentence "Here $I = \int_0^t Sdt$" doesn't fit, because it assumes $S$ and $I$ are functions of $t$, and the equality for $V$ then only holds along a thin curve instead of a region of $(S,I.t)$ space, which I think is what is intended.

The second point is that we ought to write $$ V(S,I,T) = IW(R,t), \qquad R = S/I $$ rather than $V(S,R,t)$. Then the result follows from ordinary chain rule. For example $$ V_I = W+IW_R\frac{-S}{I^2} = W-RW_R $$ and so forth.

If you really intend for $S$ to be a function of $t$ in the $V$ equation, you would still need it to hold in a region of $(S,I,t)$ space in order to use the chain rule. But if that is the case then you could specialize the $W$ PDE to the resulting curve in $(R,t)$ space. I'm not clear what the advantage of that would be since the PDEs have to hold in the entire region containing the curve anyway.

  • $\begingroup$ Thanks!.So you mean we need $\frac{\partial I}{\partial t}=0$ $\endgroup$
    – abc
    Commented Sep 9, 2021 at 12:32

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