Question About Ito's Chain Rule and SDE's

Question

I'm having a little trouble understanding the setup for Ito's chain rule, and how to expand it to finding SDE's. In class, I was taught that an example of Ito's chain rule was the following: $$ df = (f_t+1/2f_{BB})\,dt + f_B\,dB $$ where $f$ is a function of $t$ and the random variable $B$, typically governed by a Brownian Motion. Now, I have this problem, where I'm asked to solve the SDE for $Z = e^{B(t)^2}$ where the hint to the question is to let a random variable $X = (B(t))^2$ and apply the Ito chain rule to $f(x) = e^x$. I interpreted this as asking to substitute in this equation to let $Z = e^X$, where $(B(t))^2$ is written as the random variable $X$ in the original equation. But now, I'm confused on how to take the SDE. I'm unclear how to separate $Z$ into having a drift and diffusion term, but more importantly I'm unclear on how to use Ito's chain rule in this situation, as I don't see a function f which doesn't have the parameters of $t$(time) and $B$(random variable), I only see $Z$ as having these traits, and so I'm not sure how the chain rule would apply. I'm not looking for an answer, I just want to understand how to set up my problem, so I can understand how to solve it. Thanks.

Answer 1

Ito's formula is valid for semi-martingales. In this case we have $X_t = B_t^2$, which is a semi-martingale as may be observed by applying Ito's formula,

$$ dX_t = 2B_t dB_t + 2 dt $$

So since $Z_t = \exp(X_t)$ we should have

$$ Z_t = Z_t dX_t + 1/2 Z_t d[X]_t $$

where $[.]$ stands for the quadratic variation. Now the quadratic variation of $X_t$ can be read directly off of the differential form. It is just the square of the $2B_t dB_t$ bit where we use the algebraic rule that $dB_tdB_t = dt$. (The justification for this is that the quadratic variation of the Ito integral is known to be $[\mathcal{I}_{B_t}(\Phi)] = \int \Phi_t^2d t$ when the latter is finite). That is, $d[X]_t = 4B_t^2 dt$.

So we have

$$ Z_t = Z_t dX_t + 2 B_t^2 Z_t dt = 2Z_t B_t dB_t + 2(1+B_t^2)Z_t dt $$

As Lutz mentioned, you can also just apply Ito's formula directly without any substitutions,

$$ dZ_t = 2B_t Z_t dB_t + 1/2(2Z_t + 4B_t^2 Z_t)dt $$

It clearly comes to the same thing and in this case is probably simpler. Still, convincing yourself of each step in the other approach is a valuable exercise.