I am self studying stochastic calc and in my book it says:
Given that $$ Y(t)=Y(0) + \int_{o}^{t} \mu (s) ds \ +\int_{o}^{t} \sigma (s) dB(s) $$ and therefore; $$dY(t)= \mu (t)dt+\sigma(t)dB(t)$$ We can show that $$B^{2}(t)=t+2 \int_{o}^{t} B(s)dB(s)$$
By setting $Y(t) = B^{2}(t)$ and therefore $\mu (s) = 1$ and $\sigma (s) = 2B(s)$
I was wondering how the mu and sigma values were derived. The book skips over this bit and gives it as a statement, then jumps straight to $d(B^{2}(t))=2B(t)dB(t)+dt$
Any help would be appreciated
I have the book you mentioned, so I went to check out what exactly is written there. It is page 108 in the 3rd edition of Klebaner. I think you are overthinking this. The point the author is trying to make is that $B^2(t)$ is an Ito process as it can be written in the required form.
So here is how you should read this page. First, we have the definition of Ito process as a process that can be expressed as \begin{align} Y(t) = Y(0) + \int_0^t \mu(s) ds + \int_0^t \sigma(s)dB(s), \quad 0\leq t \leq T, \quad(\ast) \end{align} where $\mu(s)$ and $\sigma(s)$ satisfy suitable integrability conditions. Then, we are told that the above process can be written in a differential form as \begin{align} dY(t)=\mu(t)dt+\sigma(t)dB(t) \end{align} The two forms really say the same things, though only the first equation has a proper meaning as Klebaner remarks.
Then, as an example Klebaner shows that $B^2(t)$ is an Ito process. This requires us to write $B^2(t)$ in the form of Equation $(\ast)$.
To do so, note that in an example in the previous section, which is a simple application of Ito Lemma, it was shown that \begin{align} B^2(t)=2 \int_0^t B(s)dB(s) + t \end{align} To see that this defines an Ito process, we note that the above equation can be written as \begin{align} B^2(t)=B(0) + \int_0^t 1 ds+ \int_0^t 2B(s)dB(s) \end{align} Therefore, by setting $\mu(s)=1$ and $\sigma(s)=2B(s)$ we see that this last equation maps exactly into the definition of an Ito process.