# Doob Meyer decomposition for exponentials

Let $B$ be a Brownian motion, show that $X_t = e^{B_t}$ is a submartingale and find the Doob-Meyer decomposition of $X$.

Proving submartingality is straight forward. For Doob Decomposition. I looked into the discrete time predictable process and choosing {t^n_i}=i2^{-n},\space i=0,1,...,2^n\begin{align*} A^n_1=\sum_iE[e^{B_{t^n_{i+1}}}-e^{B_{t^n_i}} \mid F_{t^n_i}] &=\sum_iE[e^{B_{t^n_{i+1}}} \mid F_{t^n_i}]-e^{B_{t^n_i}} \\ &=\sum_i \left(e^{B_{t^n_{i}}}e^{{t^n_{i+1}}-{t^n_{i}}\over2}-e^{B_{t^n_i}}\right) \\ &=\sum_i{e^{B_{t^n_{i}}}(e^{{{t^n_{i+1}}}-{t^n_{i}}\over2}-1)}\end{align*} Any hints on how to proceed? • I think it should read\frac{t_{i+1}^n-t_i^n}{2}$instead of$\frac{t_i^n-t_{i+1}^n}{2}$. – saz Dec 1, 2017 at 18:05 ## 1 Answer Since$e^x-1 \approx x$for$x$close to$0$it is reasonable to expect that $$S_n := \sum_i e^{B_{t_i^n}} \left( \exp \left[ \frac{t_{i+1}^n-t_i^n}{2} \right]-1 \right) \approx \frac{1}{2} \sum_i e^{B_{t_i^n}} (t_{i+1}^n-t_i^n)$$ and the right-hand side converges to $$\frac{1}{2} \int_0^1 e^{B_s} \, ds.$$ To make this rigorous, write $$S_n = \sum_i e^{B_{t_i^n}} \left( \exp \left[ \frac{t_{i+1}^n-t_i^n}{2} \right]-1 - \frac{t_{i+1}^n-t_i^n}{2} \right)+ \sum_i e^{B_{t_i^n}} \frac{t_{i+1}^n-t_i^n}{2}$$ and show that the first term on the left-hand side converges to$0\$ almost surely.

Remark: If you know Itô formula, then it is much easier to apply Itô's formula to obtain the Doob-Meyer decomposition.

• It makes sense now. Thank you @saz Dec 1, 2017 at 18:17
• @user375432 You are welcome.
– saz
Dec 1, 2017 at 18:55