I saw an example where the following Itô integral was solved by Itô calculus: $\int^{T}_{0}W(t)dW(t)$. They say: let's take the stochastic process $X(t) = W(t)$, which means that $dX(t) = 0 dt + 1 dW(t)$. If we now apply Itô's lemma to g(X(t)) = $X^2(t)$, this would result in the following:
$dg(X) = \frac{\partial g}{\partial X}dX(t) + \frac{1}{2}\frac{\partial ^2 g}{\partial X^2}(dX(t))^2$. If you now fill in $X(t) = W(t)$ I can see how you get the result, but I don't understand the formula for $dg(X)$. I would think that Itô's lemma: $dY(t) = (\frac{\partial g}{\partial t} + \mu(t, X)\frac{\partial g}{\partial X} +\frac{1}{2}\frac{\partial ^2 g}{\partial X^2}\sigma^2(t, X))dt + \frac{\partial g}{\partial X}\sigma(t, X)dW(t)$ and since $dX(t) = 0 dt + 1 dW(t)$ with $\mu = 0$ and $\sigma = 1$, you would get: $dg(X) = (0 + 0 \cdot 2X(t) + 0.5 \cdot 2 \cdot 1)dt + 2X(t) \cdot 1 \cdot dW(t)$. Which is a completely different formula for $dg(X)$. Could someone explain to me what I'm missing
Thank you!
The expansion of $dg(X(t)) = \frac{\partial g}{\partial X}dX(t) + \frac{1}{2}\frac{\partial^2 g}{\partial X^2}(dX(t))^2$ follows what is called a standard form representation of Ito's lemma. After all we know that $\frac{\partial g}{\partial t} = 0$.
It is equivalent to your other form because $\frac{\partial g}{\partial X}dX(t) = 2 X(t) dX(t)$ and $\frac{1}{2}\frac{\partial^2 g}{\partial X^2}(dX(t))^2 = dt$ because $(dX(t))^2 = (dW(t))^2 = dt$ where the last equality follow from the Weiner process having a constant quadratic variance.