# How to Understand the Extra Term in Ito Integral of Brownian Mition

I am now studying stochastic process and Ito Integral of Brownian motion. We know $$\int_0^t{B_s}dB_s=\frac{1}{2}B^2_t-\frac{1}{2}t, \text{in } L^2(\Omega)$$

I know the derivation but still really wonder how to "intuitively" explain there is an extra $$\frac{1}{2}t$$ compared to deterministic integral $$\int{x}dx=\frac{1}{2}x^2$$.

My first guess is this term is coming from the quadratic variation of Brownian motion. In usual cases (e.g. differentiable functions), the quadratic variation is zero, and hence there is no extra term. Is my guess right?

• I found an answer here: mathoverflow.net/a/206022 and I think this agrees my guess. Jan 27 at 8:43

As you conclude, the extra deterministic term comes from the quadratic variation of brownian motion. Nevertheless, how to explain its presence intuitively, in other words, how to interpret it ? In fact, it has to be seen as a correction to the brownian motion in order to keep the increments and the (past) values independent of each other. Indeed, since $$\mathbb{E}[B_t\mathrm{d}B_t] = \mathbb{E}[B_t]\mathbb{E}[\mathrm{d}B_t] = 0$$, because the increments $$\mathrm{d}B_t$$ are independent of $$B_t$$, one should have $$\mathbb{E}\left[\int_0^t B_s \,\mathrm{d}B_s\right] = 0$$, which wouldn't agree with $$\mathbb{E}\left[\frac{1}{2}B_t^2\right] = \frac{1}{2}t$$.