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

Question

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 ?

Answer 1

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.