what would be power series of $x_t = e^{\beta_t} $ if $\beta_t$ is a Brownian motion process?

Question

In general the power series of $e^x =1+x/1!+x^2/2!+x^3/3!+...$ but because the process is random we can't apply the direct differentiation than how can i write it's power series.In the book stochastic processes and filtering theory byAndrew H jazwinski it is written as $\delta x_t=e^{\beta_t+\delta\beta_t}$ $e^\beta$ $x_t(\delta\beta _t+\delta\beta _t^2/2+..)$ , $\delta\beta _t^2 \cong (\delta\beta _t)^2$ i am unable to understand it.

Answer 1

What you've written is the result that you get when you perform the simplest approximation of Ito's formula. In this case Ito's formula reads

$$e^{B_t} = 1 + \int_0^t e^{B_s} dB_s + \int_0^t \frac{1}{2} e^{B_s} ds.$$

For small $t$, $\int_0^t e^{B_s} dB_s \approx B_t$ (replacing the integrand by its value at $0$, which is $1$) and $\int_0^t \frac{1}{2} e^{B_s} ds \approx \frac{t}{2}$ (doing the same). So

$$e^{B_t} \approx 1 + B_t + \frac{t}{2}.$$

Also, $e^{B_t} = \sum_{n=0}^\infty \frac{B_t^n}{n!}$ is still correct at each $t$ (at least with probability $1$). The only catch is that the $n$th term is not $O(t^n)$ but rather $O(t^{n/2})$.