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- Homework Statement
- Verify the chain rule holds for the 1-d Stratonovich integral.

- Relevant Equations
- $$df(X_t)=f'(X_t)\circ dX_t$$

where ##X_t## solves ##dX_t=bdt+\sigma\circ dW_t##.

Professor says one way to do this is to convert the equations to Itô form and back.

##dX_t=bdt+\sigma\circ dW_t## converted to Itô's SDE is

\begin{align*}

dX_t=&\left(b+\frac{1}{2}\sigma\frac{\partial}{\partial x}\sigma \right)dt+\sigma dW_t.

\end{align*}

We use Itô's formula to compute ##d(f(X_t)).##

\begin{align*}

d(f(X_t))=& \left(\frac{\partial}{\partial t}f+\left(b+\frac{1}{2}\sigma\frac{\partial}{\partial x}\sigma\right)\frac{\partial}{\partial x}f+\frac{\sigma^2}{2}\frac{\partial^2}{\partial x^2}f\right)dt+\sigma\frac{\partial}{\partial x}fdW_t\\

=& \left(\left(b+\frac{1}{2}\sigma\frac{\partial}{\partial x}\sigma\right)f'+\frac{\sigma^2}{2}f''\right)dt + \sigma f'dW_t.

\end{align*}

This is an SDE that's solved by ##f(X_t)##, so it can be converted to the Stratonovich form. To do this, we can see that the ##\frac{\sigma^2}{2}f''=\frac{\sigma}{2}\frac{\partial}{\partial x}(\sigma f') ## term is the additional 'drift' term that disappears in the Stratonovich form.

\begin{align*}

d(f(X_t))=&\left(b+\frac{1}{2}\sigma\frac{\partial}{\partial x}\sigma\right)f'dt+\sigma f'\circ dW_t\\

=& \left(\left(b+\frac{1}{2}\sigma\frac{\partial}{\partial x}\sigma\right)dt+\sigma \circ dW_t\right)f'\\

=& f' dX_t.

\end{align*}

I'm unsure how to get ##f'\circ dX_t## because I don't think I can factor out ##f'## in the second to last line. Since we have the Stratonovich integral symbol ##\circ##, what is the correct step to take? The professor says we can assume that ##f## is invertible. Also she said remember that ##\frac{\partial}{\partial f}=\left(\frac{\partial f}{\partial x}\right)^{-1}\frac{\partial}{\partial x}##, which I'm guessing means an operation that takes ##f## into ##1## . And I didn't use either of these hints the professor gave me, so I'm suspicious.

##dX_t=bdt+\sigma\circ dW_t## converted to Itô's SDE is

\begin{align*}

dX_t=&\left(b+\frac{1}{2}\sigma\frac{\partial}{\partial x}\sigma \right)dt+\sigma dW_t.

\end{align*}

We use Itô's formula to compute ##d(f(X_t)).##

\begin{align*}

d(f(X_t))=& \left(\frac{\partial}{\partial t}f+\left(b+\frac{1}{2}\sigma\frac{\partial}{\partial x}\sigma\right)\frac{\partial}{\partial x}f+\frac{\sigma^2}{2}\frac{\partial^2}{\partial x^2}f\right)dt+\sigma\frac{\partial}{\partial x}fdW_t\\

=& \left(\left(b+\frac{1}{2}\sigma\frac{\partial}{\partial x}\sigma\right)f'+\frac{\sigma^2}{2}f''\right)dt + \sigma f'dW_t.

\end{align*}

This is an SDE that's solved by ##f(X_t)##, so it can be converted to the Stratonovich form. To do this, we can see that the ##\frac{\sigma^2}{2}f''=\frac{\sigma}{2}\frac{\partial}{\partial x}(\sigma f') ## term is the additional 'drift' term that disappears in the Stratonovich form.

\begin{align*}

d(f(X_t))=&\left(b+\frac{1}{2}\sigma\frac{\partial}{\partial x}\sigma\right)f'dt+\sigma f'\circ dW_t\\

=& \left(\left(b+\frac{1}{2}\sigma\frac{\partial}{\partial x}\sigma\right)dt+\sigma \circ dW_t\right)f'\\

=& f' dX_t.

\end{align*}

I'm unsure how to get ##f'\circ dX_t## because I don't think I can factor out ##f'## in the second to last line. Since we have the Stratonovich integral symbol ##\circ##, what is the correct step to take? The professor says we can assume that ##f## is invertible. Also she said remember that ##\frac{\partial}{\partial f}=\left(\frac{\partial f}{\partial x}\right)^{-1}\frac{\partial}{\partial x}##, which I'm guessing means an operation that takes ##f## into ##1## . And I didn't use either of these hints the professor gave me, so I'm suspicious.

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