# Using Ito's lemma for Brownian motion

I am a little confused by Ito's lemma. I reviewed the basic application for geometric brownian motion. I'm now trying to apply it to a different functional form to make myself better. My mind leaps to the following:

$$dx=x (\mu dt+(\sigma+\delta^t) dz)$$ \begin{equation*} \delta<1 \end{equation*} \begin{equation*} dz=\epsilon \sqrt{dt}, \epsilon = N(0,1) \end{equation*} \begin{equation*} F=ln(x) \end{equation*}

I tried to put my work below: \begin{equation*} dF \approx \frac{dF}{dt}dt+\frac{dF}{dx}dx+\frac{1}{2}\frac{d^2F}{d^2dx}dx^2 \end{equation*}

\begin{equation*} dF \approx 0 +\frac{1}{x}x(\mu dt+(\sigma+\delta^t)dz)-\frac{1}{2}\frac{1}{x^2}x^2(\mu dt+(\sigma+\delta^t)dz)^2 \end{equation*}

\begin{equation*} dF \approx \mu dt+(\sigma+\delta^t)dz-\frac{1}{2} (\mu^2 dt^2+(\sigma+\delta^t)^2 dz^2+ 2 \mu (\sigma+\delta^t)dz dt) \end{equation*}

$$dF \approx \mu dt+(\sigma+\delta^t)dz-\frac{1}{2}(\sigma+\delta^t)^2dz^2$$

I'm wondering, is it true that df/dt=0 now that the functional form is different? There's clearly a t inside x at this point, so F(x(t)). Previously, df/dt was zero, and I thought it was because Brownian motion isn't time-differentiable. But here, I think that it df/dt might be $\frac{1}{x}*x*\delta^t*ln(\delta)*dz$ by the chain rule. Is df/dt still 0 for this functional form? Why/Why not?

I've always preferred the similar form of Ito's Lemma present on wikipedia. Let's use that and verify if your solution is correct. We start with your process $$d x_t = x_t\mu\;dt + x_t(\sigma + \delta^t) \; dZ_t$$

where $Z_t$ is brownian motion.

$$\mu_t = x_t \mu$$ $$\sigma_t = x_t(\sigma + \delta^t)$$ $$f(x) = F(x) = \ln{x}$$

So,

$$\frac{\delta f}{\delta t} = 0$$ $$\frac{\delta f}{\delta x} = \frac{1}{x}$$ $$\frac{\delta^2 f}{\delta x^2} = \frac{-1}{x^2}$$

Plugging this into Ito's formula:

$$d F(x_t) = \left( 0 + \mu_t \frac{1}{x_t} - \frac{1}{2}\sigma_t^2\frac{1}{x_t^2} \right)dt + \sigma_t \frac{1}{x_t} dZ_t$$

Simplifying:

$$d F(x_t) = \left( \mu - \frac{1}{2}(\sigma + \delta^t)^2 \right)dt + (\sigma + \delta^t) dZ_t$$

Which is the same as you have up to $dZ_t^2$ (which is $dt$). The benefit of using this version of Ito's formula is in its simplicity and lack of needing to expand squares and drop differential terms.

• If we note that dz^2=dt, we're still on the same page. Any idea why df/dt is 0, since F(x(t)) clearly has t's in it? My initial thought was that I needed to chain rule. – RegressForward Dec 30 '13 at 19:39
• Ah, so the functional $F$ does not care about the time dependency of $x$. i.e. we are interested in $F(x,t)$, not $F(x_t, t)$. So $F(x,t) = ln(x)$, not $ln(x_t)$. – Cam.Davidson.Pilon Dec 30 '13 at 19:48
• Could you elaborate a bit? I think I get what you're saying, but I'm not confident. – RegressForward Dec 30 '13 at 19:58
• We ignore the fact that $x_t$ is time dependent when calculating the derivatives. From the point of view of $F$, the functional, the input is a time-component, $t$, and a space-component $x$, with no relation to each other. This makes sense from stochastic process's point of view: the next movement ($dF(x_t)$) is a function of where I currently am ($x$) and how much time has passed ($t$), plus noise. – Cam.Davidson.Pilon Dec 30 '13 at 20:18