Quadratic variation of $e^{B_t}$ where $B_t$ is the standard brownian motion.

Question

Consider $f(x) = e^x$ , then show that $<f(B)>_t$ is equal to $\int^t_0e^{2B_t}dt$.

I know from jensen's inequality that $e^{B_t}$ is a submartingale. How is quadratica variation defined for that? For a square integrable martingale $X_t$ it is the natural increasing process of the D-M decomposition of $X^2_t$
Do i somehow have to use ito's formula to get the answer? i tried, but could not get anything.

For $X_t=e^{B_t}$ you get $dX_t=X_t,dB_t+\frac12X_tdt$ from the Ito formula. This gives, informally, $(dX_t)^2=X_t^2,(dB_t)^2=e^{2B_t},dt$ for the quadratic variation. — Lutz Lehmann, Jun 24 '22 at 08:21
how to arrive at this informal formula from the integral version of the ito's formula? — confusedsam, Jun 26 '22 at 12:08

score 1 · Answer 1 · answered Aug 12 '23 at 18:30

As mentioned in here Quadratic Variation of Diffusion Process and Geometric Brownian Motion, you simply apply Ito's formula to $Y_{t}=f(B_{t})$ get

$$dY_{t}=f'(B_{t})dB_{t}+\frac{1}{2}f''(B_{t})dt=Y_{t}dB_t+\frac{1}{2}Y_{t}dt$$

and so

$$<Y_{t},Y_{t}>=\int^{t} Y_{s}^{2}d[B]_{s}=\int^{t} e^{2B_{s}}ds.$$

1 Answers1