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?

  • $\begingroup$ I think it should read $\frac{t_{i+1}^n-t_i^n}{2}$ instead of $\frac{t_i^n-t_{i+1}^n}{2}$. $\endgroup$
    – saz
    Dec 1, 2017 at 18:05

1 Answer 1


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.

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

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