How to obtain this result using Ito's Lemma?

Question

My book writes that when:

$dx = a(x,t)dt + b(x,t)dz$

$x' := x + dx$

Then using Ito's lemma: $E[F[x+ \Delta x, t + \Delta t \mid x)] = F(x,t) + [F_t(x,t) + a F_x + \frac{1}{2} b^2 F_{xx}] \Delta t + o(\Delta t)$.

Now I am not sure about two things:

1. How can Ito's lemma be used to say something about expectations - I thought it was just for functions

2. Where does the first term [$F(x,t)$] come from? I don't see it in Ito's lemma.

Thanks in advance!

Answer 1

It's simpler than you think. The expectation $\mathbb{E}\left[F(x+\Delta x,t+\Delta t)|x,t\right]$ is an expectation over all possible Brownian motions $z$. It will not depend on $z$ since we average over it realizations during the time $\Delta t$. So it should only depend on $\Delta t$. To obtain it, one has to expand $F$ to first order in time and to the second order in space. Taylor expansion yields, for any $\delta x$, $\delta t$ $$F(x+\delta x, t+\delta t)=F(x,t)+\partial_tF(x,t)\delta t+\partial_xF(x,t)\delta x+\frac12\partial_x^2F(x,t)\delta x^2+o(\delta t)+o(\delta x^2).$$ Now plug the $\Delta x$ obtained from the equation and compute the expectation. We have $\mathbb{E}[\Delta z]=0$ because $z$ is a Brownian motion. for the same reason $\mathbb{E}[(\Delta z)^2]=\Delta t$. Using $\Delta x=a \Delta t+b\Delta z$, you get the desired result. The $F(x,t)$ remains in first position because it is a constant with respect to $z$ so its average is equal to $F(x,t)$. The $(\Delta t)^2$ coming from $(\Delta x)^2$ is absorbed in the $o(\Delta t)$.

Answer 2

Ito's Lemma states

$$dF(x,t)=\frac{\partial F}{\partial t}dt+\frac{\partial F}{\partial x}dx+\frac12 \frac{\partial^2 F}{\partial x^2}(dx)^2$$

We can write the increment of $F$ as

$$\Delta F=F(x+\Delta x,t+\Delta t)-F(x,t)=\frac{\partial F}{\partial t}dt+\frac{\partial F}{\partial x}dx+\frac12 \frac{\partial^2 F}{\partial x^2}(dx)^2+o(\Delta t) \tag 1$$

Using $E\left(dx\right)=a(x,t)dt$ and $E\left((dx)^2\right)=b^2(x,t)dt+O(\Delta t)^{2}$ in $(1)$ gives

$$\begin{align} E\left(F(x+\Delta x,t+\Delta t)\right)&=E\left(F(x,t)\right)+E\left(\frac{\partial F}{\partial t}dt+\frac{\partial F}{\partial x}dx+\frac12 \frac{\partial^2 F}{\partial x^2}(dx)^2+o(\Delta t)\right)\\\\ &=F(x,t)+\frac{\partial F}{\partial t}dt+a(x,t)\frac{\partial F}{\partial x}+\frac12 b^2(x,t)\frac{\partial^2 F}{\partial x^2}+o(\Delta t) \end{align}$$

as was to be shown!