1
$\begingroup$

I am having a hard time understanding the application of Ito's formula. In my lecture notes, it is given that $X_t = X_0 + \int_0^t b_sds + \int_0^t \sigma_sdb_s$ is an Ito Process.

What I don't understand is how can I apply Ito's formula to represent the following $(X_t)_{t \geq 0} $ as an Ito Process for some $(b,\sigma)$.

a) $X_t = B_t^2 $

Solution : $X_t = 0 + \int_0^t 2B_sdB_s + \int_0^t dt $

My workings: $\frac{\partial X_t}{\partial B_t} = 2B_t ; \frac{\partial X_t}{\partial t} = 0 ; \frac{\partial X_t^2}{\partial^2 B_t} = 2 $

How can I use the partial derivatives to make it look like the solution given? Also, where did the $0$ in the solution come from?

b) $X_t = \exp(\frac{-\sigma^2t}2 +\sigma B_t) $

Solution : $X_t = 1 + \int_0^t \sigma\exp(\frac{-\sigma^2s}2 +\sigma B_s) dB_s $

Would appreciate if someone could guide me through these two examples. Thank you!

$\endgroup$

1 Answer 1

1
$\begingroup$

You have some (typo ?) issue in your definition. An Itô process $(I_t)$ is a stochastic process which can be written as $$X_t = X_0 + \int_{s=0}^t \mu_s ds + \int_{s=0}^t\sigma_sdB_s $$ Where $(B_t)$ is a standard Brownian motion and $(\mu_t)$ (drift) and $(\sigma_t)$ (diffusion) are two stochastic processes adapted to $(B_t)$.

As two extremely elementary examples of Itô processes, you can think of the Brownian motion $(B_t)$, which can be written as $B_t =0 + \int_{s=0}^t 0\ ds + \int_{s=0}^t1\ dB_s$, or the deterministic process $(t)$, which writes $ t = 0 + \int_{s=0}^t 1\ ds + \int_{s=0}^01\ dB_s $.

Now Itô's lemma states that, given an Itô process $(X_t)$ with drift $ (\mu_t)$ and diffusion $ (\sigma_t)$, and a twice differentiable function $f(\cdot,\cdot)$, the process $(f(t,X_t)) $ is also an Itô process with drift $ (\mu_t^f) = \left({\frac {\partial f}{\partial t}}(t,X_t)+\mu _{t}{\frac {\partial f}{\partial x}}(t,X_t)+{\frac {\sigma _{t}^{2}}{2}}{\frac {\partial ^{2}f}{\partial x^{2}}}(t,X_t)\right)$ and diffusion $(\sigma_t^f) = \sigma _{t}{\frac {\partial f}{\partial x}}(t,X_t)$.
In other words, $$f(t,X_t) = f(X_0) + \int_{s=0}^t \mu_s^f ds + \int_{s=0}^t\sigma_s^f\ dB_s \tag1$$ To apply that in the setting of example a), we identify :
$(B_t)$ is an Itô process with drift $0$ and diffusion $1$, $(X_t) = (B_t^2) = f(t,B_t)$ with $f(t,x) := x^2$. As you did, we compute every derivative term that appear in the formula of Itô's lemma : $$\frac{\partial f}{\partial t}(t,B_t) =0,\; \frac{\partial f}{\partial x}(t,B_t) = 2B_t,\; \frac{\partial^2 f}{\partial x^2}(t,B_t) =2$$ Now we can plug the values in to get $\mu^f$ and $\sigma^f$ : $$\mu^f_t = 0 + 0 + 1 = 1 \;\text{ and } \sigma^f_t = 1\times2B_t = 2B_t $$ We can finally plug in these expressions in $(1)$ to get the solution : $$\begin{align}B_t^2 = f(t,B_t) &= B_0^2 + \int_{s=0}^t \mu_s^f ds + \int_{s=0}^t\sigma_s^f\ dB_s\\ &=0+ \int_{s=0}^t 1\ ds + \int_{s=0}^t2B_s dB_s\\ &=\int_{s=0}^t \ ds + \int_{s=0}^t2B_s dB_s \ \; \; \; \; \blacksquare \end{align} $$

If you understood this, you can proceed the same exact way to solve example b). I let you do the calculations yourself ;)

$\endgroup$
1
  • $\begingroup$ Thank you so much for the explanation! $\endgroup$ Commented Oct 30, 2021 at 5:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .