I have been told that these two expressions of Itô's Lemma are the same, but written in different ways :

$$ f(t,X_t) = f(0, X_0) + \int_{0}^{t} \frac{\partial f}{\partial s} ds + \int_{0}^{t} \frac{\partial f}{\partial X_s} dX_s + \frac{1}{2}\int_{0}^{t} \frac{\partial^2 f}{\partial {X_s}^2}{\sigma_s}^2 ds$$


$$df(X_t, Y_t) = \frac{\partial f}{\partial X_t}dX_t + \frac{\partial f}{\partial Y_t}dY_t + \frac{1}{2} \frac{\partial^2 f}{\partial {X_t}^2}d<X_t>_t + \frac{1}{2} \frac{\partial^2 f}{\partial {Y_t}^2}d<Y>_t + \frac{\partial^2 f}{\partial X_t \partial Y_t}d<X, Y>_t $$

where $<X, Y>_t$ is the quadratic variation operator (and $<X>_t = <X,X>_t$).

I cannot figure out why they are similar. In my eyes, they seem pretty different. How can I pass from the first one to the second one? Actually I don't really understand how the second one can be interpreted and applied.

  • 1
    $\begingroup$ The second one is more general than the first one. Replace $X_{t}$ by $t$ then you have the first formula because $<.>_{t}=0$. Then take he integrated form $\endgroup$
    – glork
    Dec 1 '15 at 20:12
  • 1
    $\begingroup$ I think the two expressions refer to the integral and differential forms. The two you wrote are not for the same thing, that is, one is for $f(t, X_t)$ and the other is for $f(X_t, Y_t)$, then they are of course not the same. Try to write the second one for $f(t, X_t)$. $\endgroup$
    – Gordon
    Dec 1 '15 at 20:14
  • $\begingroup$ Thank you. But I don't understand why $<.>_t = 0$. And why $<X>_t = \sigma_t^2 $(by identification). And why $<X, Y>_t = 0$. As you can see, there are many things unclear... Could you please develop your answer a bit? $\endgroup$
    – MarinD
    Dec 1 '15 at 22:01

I assume you're confused between the integral and SDE writings of Ito's lemma, since the two equations you have are indeed different.

Let $X_t$ be an Ito process defined by $$ X_t = X_0 + \int_0^t \alpha_s \, ds + \int_0^t \sigma_s \, dW_s $$ for adapted processes $\alpha_s$ and $\sigma_s$ (and assuming some technical boundedness condition on the integrals). This equation may be written in shorthand as an SDE as $$ dX_t = \alpha_t dt + \sigma_t dW_t. $$ The SDE is not rigorous - it is simply a shortcut way of writing down the integrals above, and provides a bit of intuition behind the evolution of $X$ over "infinitesimally small" time intervals.

Now consider a measurable function $f: [0,T] \times \mathbb{R} \to \mathbb{R}$ such that $$ f(\cdot,x)\in C^1([0,T]) \quad \forall x \in \mathbb{R}, $$ and $$ f(t,\cdot) \in C^2(\mathbb{R}) \quad \forall t \in [0,T]. $$

I would argue the correct (mathematically rigorous) way of stating Ito's lemma is

$$ f(T,X_T) = f(0,X_0) + \int_0^T \frac{\partial}{\partial t}f(s,X_s) \, ds + \int_0^T \frac{\partial}{\partial x}f(s,X_s) \, dX_s \\ \qquad + \frac{1}{2}\int_0^T \frac{\partial^2}{\partial x^2}f(s,X_s) \, d<X,X>_s $$

The quantity $<X,X>_s$ is the quadratic variation accumulated by the Ito process $X$ up until time $s$. You can show (Shreve II, page 143-144, e.g.) that this is given by $$ <X,X>_s = \int_0^s \sigma^2_u \, du, $$ or, in differential (shorthand) form as $$ d<X,X>_s = \sigma^2_s \, ds. $$ Plugging this into Ito's lemma gives your first equation.

Now, just like the Ito process $X$ was written in shorthand as an SDE, so may $f$, since it, too, is an Ito process. That is, we also have

$$ df(t,X_t) = \frac{\partial}{\partial t}f(t,X_t)dt + \frac{\partial}{\partial x}f(t,X_t) dX_t + \frac{1}{2}\frac{\partial^2}{\partial x^2}f(t,X_t) d<X,X>_t. $$

The first boxed equation had precise mathematical meaning. The second boxed equation is just shorthand for the first.

Update: Your second equation is often called Ito's product rule. Ito's lemma is "usually" stated for functions of one Ito process as it was for my answer above. If you have a function of two Ito processes then both processes' quadratic variation and cross variation appear in Ito's lemma, aka Ito's product rule. See Shreve II, page 168, e.g. for a decent explanation

  • $\begingroup$ Thanks, it helps a lot! But what about my second equation? In which way should it be used? Or is it false? $\endgroup$
    – MarinD
    Dec 2 '15 at 8:58
  • $\begingroup$ @MarinD See update, does that help? $\endgroup$
    – bcf
    Dec 2 '15 at 15:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.