# Apply Ito's formula to represent $X_t$ as an Ito's proces

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!

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}