# Ito's lemma and Stratonovich calculus

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.

• Does this answer your question? transformation between Ito and Stratonovich calculus Feb 13 at 19:57
• 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?
– Fred
Feb 13 at 20:07

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.

• Your nice presentation (1) solved one of my question, Thanks a lot. Feb 13 at 22:03
• 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)?
– Fred
Feb 13 at 23:07
• @Fred please number all your equatios so that we can better address these problems. One I see already: your $\frac12\sigma f''$ should be $\frac12\sigma^2f''\,.$ Feb 14 at 6:31
• Thank you so much! I have corrected this typo and I will remember to number all my equations in the future.
– Fred
Feb 14 at 17:35

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.

• Really nice! Every coin has two sides and you found one that shines. Feb 15 at 10:55

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}

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