# Why is it more interesting to define Itô integral rather to use $f(t)B_t$?

SDE's are normally of the form $$dX_t=f(X_t,t)dt+\sigma (X_t,t)dB_t,$$ but the sense of that is $$X_t=X_0+\int_0^t f(X_s,s)ds+\int_0^t \sigma (X_s,s)dB_s,\quad a.s.$$

Now, I was wondering why is it more interesting to develop a machinerie to gives sense to $$\int_0^t \mu(X_s,s)dB_s$$ rather than to consider the process $$X_t=X_0+\int_0^t f(X_s,s)ds+\sigma (X_t,t)B_t.$$

Because the idea of the Itô integral is to generalize in somehow the Brownian motion, no ?

• @saz: yes thank you :) May 13, 2019 at 6:54

Suppose $$(B_t)$$ is a standard Brownian motion, i.e. $$B_t\sim \mathcal N(0,t)$$.
• The main idea behind the Itô process is to generalize in somehow the Brownian motion in a suitable way. Just avoid for the moment the dependence of $$\sigma$$ and $$f$$ of $$X_t$$, i.e. look at $$\,\mathrm d X_t=f(t)\,\mathrm d t+\sigma (t)\,\mathrm d B_t.\tag{*}$$
• The main idea behind $$(*)$$ is that $$X_{t+h}-X_t\approx f(t)h+\sigma (t)(B_{t+h}-B_t),$$ in a suitable sense. So, in some sense, the increament are independents, $$t\mapsto X_t$$ is a.s. continuous, $$X_{t+h}-X_t\sim \mathcal N(f(t)h,\sigma ^2(t)h)),$$ and $$(X_t)_t$$ is Gaussian. So it's really the fact that $$(X_t)$$ is a sort of Brownian motion but instead to gravitate around $$0$$ with Variance $$t$$, it will create a Brownian motion that gravitate around a the curve $$\int_0^tf(s)\,\mathrm d s$$ and with variance $$\int_0^t \sigma (t)^2dt-\left(\int_0^t f(t)\right)^2$$.
• An important remark is : what we expect of a Brownian motion is not the fact that $$B_t\sim \mathcal N(0,t)$$ but really the fact that $$B_{t}-B_s\sim \mathcal N(0,t-s)$$ or equivalently $$B_{t+h}-B_t\sim \mathcal N(0,h)$$. So, in some sense, the increments will be quite reasonable in the sense that the variance will grow linearly. And this is exactly what will offert Itô integral when $$\sigma$$ is nice enough.
• So, from this point of view (i.e. seeing an SDE as a generator of Brownian motion that gravitate around a curve of speed $$f(t)$$), your equation wouldn't make any sense since $$\sigma (t+h)B_{t+h}-\sigma (t)B_t$$ wouldn't have the wished property of being normal with linear variance since it's going to be $$\mathcal N(0,\sigma ^2(t+h)h+(\sigma (t+h)-\sigma (t))^2t)$$. As you can see, the variance is really not what we expect from a Brownian motion.