How can I show $dX_t=X_t(\tfrac12 dt+dB_t)$ for $x_t=e^{B_t}$

Question

How can I show $dX_t=X_t(\tfrac12 dt+dB_t)$ for $x_t=e^{B_t}$?
when $B_t$ is standard Brownian motion .I can show that by Ito derivation formula ,but my professor says use direct method to show that. Can anyone help me ?

Answer 1

you can take $x_t=2e^{B_t}$ and use definition of $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$
Note that $dx_t=x_{t+d t}-x_t\\d(2e^{B_t})=2e^{B_{t+dt}}-2e^{B_t}$

$$\quad{d(2e^{B_t})=2e^{B_{t+dt}}-2e^{B_t}=2e^{B_{t}}e^{B_{dt}}-2e^{B_t}=2e^{B_t}(e^{dB_t}-1)\\\text{ use } e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots \text { put x=}B_{dt}\\so \to \\d(2e^{B_t})=2e^{B_t}\Bigl(1+dB_t+\frac12\underbrace{(dB_t)^2}_{\to dt}+O((dB_t)^3)-1\Bigr)\\d(2e^{B_t})=2e^{B_t}\Bigl(\not1+dB_t+\frac12\underbrace{(dB_t)^2}_{\to dt}+\underbrace{O((dB_t)^3)}_{tends\to 0}-\not1\Bigr) \\d(2e^{B_t})=2e^{B_t}(dB_t+\tfrac12 dt)=e^{B_t}dt+2e^{B_t}dB_t\\so \\dX_t=X_t(\tfrac12 dt+dB_t)}$$