# On Geometric Brownian motion and Itô's formula

Let $$S_t$$ be a geometric brownian motion such as $$d S(t) = rS(t)dt +\sigma S(t)dW(t),$$ where $$W$$ is a standard Brownian motion.

With Itô's lemma and formulas $$(dt)^2=dtdW_t=dW_tdt=0$$ and $$(dW_t)^2=dt$$, we can show that

$$(dS_t)^2=\sigma^2 S_t^2 dt$$ Problem:

At the beginning of a demonstration, the author of an article uses the following equality: $$(\int^T_0 dS_t)^2=\int^T_0 (dS_t)^2$$ I can't see how he got such a result knowing that I find with Itô's lemma that : $$(\int^T_0 dS_t)^2=\int^T_0 (dS_t)^2+2(S_0^2-S_0S_T+\int^T_0 S_tdS_t)$$ Help me see a little more clearly. Thank you in advance.

• This is Ito isometry. Take a look at en.wikipedia.org/wiki/Itô_isometry. I think you are missing the expectation operations. – ilovevolatility Mar 25 at 14:57
• Exactly, I think the author forgot to put the average operator. – M. A. Kacef Mar 25 at 15:04