In normal calculus we can write $d\ln{x} = \frac{dx}{x}$ since there is no quadratic variation to deal with. This isn't true for stochastic processes, and Ito's Lemma is used to calculate $d\ln{X}$. So when I was reading about volatility/realized volatility, I saw that there were two expressions used for realized variance $\sigma ^2 dt$, one is $\frac{dX}{X}^2$, while the other is $d\ln{X}^2$. So although these are equal I was wondering how come taking the square root doesn't lead to a valid relationship. I know that certain terms are 'negligible' thus leading the square operation to give the same results, but I don't see how to 'reverse' this using the square root.
Here's my work (it's just using Ito's Lemma and simplifying quadratic terms):
For an Ito Process of form: $\frac{dX_t}{X_t} = \mu(t,X_t) dt + \sigma(t,X_t) dW_t$
\begin{equation} (\frac{dX}{X})^2 = (\mu dt + \sigma dW)^2 = \sigma^2dt \end{equation} while \begin{align} d\ln{X} = \frac{1}{X}dX - \frac{1}{2X^2}*dX^2 = \frac{dX}{X} - \frac{1}{2} \sigma ^2dt = (\mu - \frac{1}{2} \sigma ^2 ) dt + \sigma dW \\ (d\ln{X})^2 = ((\mu - \frac{1}{2} \sigma ^2 ) dt + \sigma dW)^2 = (\sigma dW) ^2 = \sigma ^2 dt \end{align}
so $d\ln{X} \neq \frac{dX}{X}$ and $(d\ln{X})^2 = (\frac{dX}{X})^2$.
In general, if $x^2 = y^2$, then $x = \pm y$, but $d\ln{X} \neq \pm \frac{dX}{X}$ since $d\ln{X}$ has the $- \frac{1}{2} \sigma ^2 dt $ term due to the quadratic variation of a Brownian motion.
So basically I am wondering why this is invalid, and what other sorts of operations on stochastic processes are invalid. Thanks!
