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

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

and

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

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.

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    $\begingroup$ 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
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    $\begingroup$ 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
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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

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  • $\begingroup$ 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

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