3
$\begingroup$

I tried converting Ito's lemma to Stratonovich form, but I got inconsistent results.

Consider a 1-D SDE in both Ito and Stratonovich sense: $$ dX_t = \mu(X_t) dt + \sigma(X_t) dW_t = \bar{\mu}(X_t) dt + \sigma(X_t) \circ dW_t, $$ where $$ \bar{\mu} = \mu - \frac{1}{2}\sigma \sigma'. \qquad(1) $$

Let $f=f(x)$ be a smooth function of $x$. For simplicity, $f, \mu$, and $\sigma$ have no time dependence. By Ito's lemma, we get $$ df = \left( \mu f' + \frac{1}{2} \sigma^2 f'' \right) dt + \sigma f' dW_t. $$

Now, I want to transform the above equation into Stratonovich form, but I got inconsistent results. Here are my derivations.

1. Using transformation formula

From Eq.(1), we have: $$ \begin{align} df &= \left[ \mu f' + \frac{1}{2} \sigma^2 f'' - \frac{1}{2} \sigma f' \frac{\partial (\sigma f')}{\partial X} \right] dt + \sigma f' \circ dW_t\\ &= \left[ \mu f' + \frac{1}{2} \sigma^2 f'' - \frac{1}{2} \sigma \sigma' (f')^2 - \frac{1}{2} \sigma^2 f' f'' \right] dt + \sigma f' \circ dW_t. \qquad \qquad (2) \end{align} $$

2. Using the chain fule for Stratonovich SDE

Since Stratonovich SDE follows the regular chain rule, we have $$ \begin{align} df &= f' \circ dX_t = f' (\bar{\mu} dt + \sigma\circ dW_t ) \\ &= \left( \mu f' - \frac{1}{2} \sigma \sigma' f' \right) dt + \sigma f' \circ dW_t. \qquad \qquad (3) \end{align} $$

It seems like these two results can't be identical, but I'm not sure where it goes wrong.


Edit: Corrected a few typos according to Kurt's comment.

$\endgroup$
2
  • $\begingroup$ Does this answer your question? transformation between Ito and Stratonovich calculus $\endgroup$
    – Kurt G.
    Feb 13 at 19:57
  • $\begingroup$ Thanks for your reply, but I didn't really see how it helps the current question. Do you think I need to derive a discrete summation in order to figure out which form is correct? $\endgroup$
    – Fred
    Feb 13 at 20:07

3 Answers 3

4
$\begingroup$

In this answer I show that for any process $Y_t$ whose stochastic integrals are defined we have $$ \tag{1} \underbrace{\int_0^tY_s\circ\,dW_s}_{\text{Stratonovich}}=\underbrace{\int_0^tY_s\,dW_s}_{\text{Ito}}+\tfrac{1}{2}\langle W,Y\rangle_t\, $$ (the notation for $Y_s$ was $f(s)$ in the link).

In your case we have an Ito SDE $$ dX_t=\mu\,dt+\sigma\,dW_t $$ and we know \begin{align}\tag{Ito} df(X_t)&=f'(X_t)\,dX_t+\tfrac 12f''(X_t)\,d\langle X\rangle_t\\[2mm] &=f'(X_t)\sigma(X_t)\,dW_t+f'(X_t)\mu(X_t)\,dt+\tfrac 12f''(X_t)\sigma^2(X_t)\,dt\,.\tag{2} \end{align} Claim: $$\tag{3} df=f'\left(\mu-\frac{\sigma\sigma'}{2}\,dt\right)+f'\sigma\circ dW_t\,. $$ In particular, if $\mu\equiv 0$ and $\sigma\equiv 1$ then $X_t$ is a Brownian motion and the Stratonovich calculus follows the rules of ordinary calculus: $$\tag{Stratonovich} \boxed{\quad df(W_t)=f'(W_t)\circ\,dW_t\,.\phantom{\Big|}\quad} $$ Proof. Writing $$ \alpha(X_t)=\sigma(X_t)f'(X_t) $$ we get from Ito $$ d\alpha=\alpha'\,dX_t+\frac{\alpha''}{2}\,d\langle X\rangle_t\,. $$ Using $Y_t=\alpha(X_t)$ and (2) this means $$\tag{4} d\langle Y,W\rangle_t=\alpha'(X_t)\,\sigma(X_t)\,dt=\Big(\sigma\,\sigma'\,f'+\sigma^2\,f''\Big)(X_t)\,dt\,. $$ Using now (1), (2) and (4) we get (3). $$\tag*{$\Box$} \quad $$ To answer the question where your mistake was:

  • In short: in your equation (2) you have too many $f'\,.$

  • Up to a typo you have correctly from Ito $$\tag{5} df=\left(\mu f'+\frac12\sigma^{\color{red}2}f''\right)\,dt+\sigma f'\,dW_t\,. $$

  • When going to $\sigma f'\circ dW_t$ you follow the pattern $$ \mu\,dt+\sigma\,dW_t=\left(\mu-\frac{\sigma\sigma'}{2}\right)\,dt+ \sigma\circ dW_t $$ which is correct for $dX_t$ but not for $df(X_t)\,.$

  • This incorrect approach leads to \begin{align}\tag{incorrect} df&=\left(\mu f'+\frac12\sigma^2 f''-\frac{\sigma f'(\sigma f')'}2\right)\,dt+\sigma f'\circ dW_t\\ &=\left(\mu f'+\frac12\sigma^2 f''-\frac{\sigma\sigma'(f')^\color{red}{2}+\sigma^2f' f''}2\right)\,dt+\sigma f'\circ dW_t\\ \end{align} where $\sigma^2$-terms don't cancel and the $\sigma\sigma'$ term has one $f'$ too many.

  • The correct step from (5) to (3) is to use the expression (4) which accounts via (1) for another drift change when going from $\sigma f'dW_t$ to $\sigma f'\circ dW_t\,.$

  • I have to say that this was new and surprising to me. Every time I look at Stratonovich I learn something new.

$\endgroup$
4
  • $\begingroup$ Your nice presentation (1) solved one of my question, Thanks a lot. $\endgroup$
    – Khosrotash
    Feb 13 at 22:03
  • $\begingroup$ Thanks for your answer! I think I understand your derivation. However, could you point out which part is wrong in the derivation of my Eq.(2)? $\endgroup$
    – Fred
    Feb 13 at 23:07
  • $\begingroup$ @Fred please number all your equatios so that we can better address these problems. One I see already: your $\frac12\sigma f''$ should be $\frac12\sigma^2f''\,.$ $\endgroup$
    – Kurt G.
    Feb 14 at 6:31
  • $\begingroup$ Thank you so much! I have corrected this typo and I will remember to number all my equations in the future. $\endgroup$
    – Fred
    Feb 14 at 17:35
1
$\begingroup$

Based on Kurt's answer and comment, the mistake I made in Eq. (2) in the post can be resolved in an alternative way below.

From Ito's lemma, we have $$ df = \left( \mu f' + \frac{1}{2} \sigma^2 f'' \right) dt + \sigma f' dW_t. \qquad\qquad(1) $$

To calculate its Stratonovich form, what we actually need is: $$ \begin{align} df &= \left[ \mu f' + \frac{1}{2} \sigma^2 f'' - \frac{1}{2} \sigma f' \frac{\partial (\sigma f')}{\partial {\color{red}{f}}} \right] dt + \sigma f' \circ dW_t. \qquad(2) \end{align} $$ Therefore, we get $$ \begin{align} \sigma f' \frac{\partial (\sigma f')}{\partial {f}} &= \sigma f' \frac{\partial (\sigma f')}{\partial X} \frac{\partial X}{\partial f} = \sigma f' (\sigma' f' + \sigma f'') / f' = \sigma (\sigma' f' + \sigma f''). \qquad(3) \end{align} $$ Plug it back to Eq.(2), we get: $$ df = \left( \mu f' - \dfrac{1}{2}\sigma \sigma' f' \right) dt + \sigma f' \circ dW_t, $$ which is consistent with the other method.

$\endgroup$
1
  • $\begingroup$ Really nice! Every coin has two sides and you found one that shines. $\endgroup$
    – Kurt G.
    Feb 15 at 10:55
0
$\begingroup$

It is not very clear from Kurt's answer to Fred's answer why $\frac{\partial}{\partial f}$ is legitimate. I write a detailed explanation here. The Ito $\Leftrightarrow$ Stratonovich conversion is only valid when $dX_t$ is an expression of the following form $$dX_t=\mu(X_t,t)dt+\sigma(X_t,t) dW_t\Leftrightarrow dX_t=\bar{\mu}(X_t,t)+\sigma(X_t,t)\circ dW_t$$

Clearly $$df(X_t)=\left(\mu(X_t,t)f'(X_t)+\frac12\sigma^2(X_t,t)f''(X_t,t)\right)dt+\sigma(X_t,t)f'(X_t)dW_t$$ is not in this form. So the conversion formula does not apply. If we let $Y_t=f(X_t)$, the coefficients of the righthand side should involve $Y_t$ only and then we can apply the conversion formula. To see this, assuming $f$ is invertible. Then we have $$x=f^{-1}(y):=s(y)$$ $$f'(x)=\frac{1}{(f^{-1})'(y)}:=g(y)$$ $$f''(x)=-\frac{(f^{-1})''y}{((f^{-1})'(y))^3}:=h(y)$$ Via these conversions, we can rewrite the righthand side of the expression involving $Y_t$ only as

$$dY_t=\left(\mu(s(Y_t),t)g(Y_t)+\frac12\sigma^2(s(Y_t),t)h(Y_t)\right)dt+\sigma(s(Y_t),t)g(Y_t)dW_t$$ This form is consistent with Ito $\Leftrightarrow$ Stratonovich conversion, so we can apply the rule

$$dY_t=\left(\mu(s(Y_t),t)g(Y_t)+\frac12\sigma^2(s(Y_t),t)h(Y_t)-\frac12\left(\sigma(s(Y_t),t)g(Y_t)\right) \frac{\partial}{\partial y}\left(\sigma(s(Y_t),t)g(Y_t)\right)\right)dt+\left(\sigma(s(Y_t),t)g(Y_t)\right)\circ dW_t$$

This expression is exactly the same as Fred's correction

$$ \begin{align} df &= \left[ \mu f' + \frac{1}{2} \sigma^2 f'' - \frac{1}{2} \sigma f' \frac{\partial (\sigma f')}{\partial {\color{red}{f}}} \right] dt + \sigma f' \circ dW_t. \qquad(2) \end{align} $$

$\endgroup$
1
  • $\begingroup$ Thanks! I just made a hand-waving there, but your answer proves that doing such is legitimate! $\endgroup$
    – Fred
    Mar 21 at 20:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .