Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We all know that $\int_0^t dB(s) = B(t)$, where $B(t)$ is a standard Brownian Motion.

However, is the following identity true? Also, why or why not?

$\boxed{ \displaystyle \ \ \int_{t_1}^{t_2} dB(s) = B(t_2) - B(t_1)}$

share|cite|improve this question
This is clearly true, and in general for $a< b$ we have $$\int_{a}^{b} \xi \, dB_{s} = \xi \cdot (B_{b} - B_{a}) $$ for an $\mathcal{F}_{a}$-measurable random variable $\xi \in L^{2}(\Bbb{P})$. – Sangchul Lee Nov 10 '12 at 4:21
@sos440 Is your identity true even for $\displaystyle \ \ \int_a^b \mathscr{\epsilon}(s)dB(s)$? (I don't know how to make that funny symbol you made). – Jase Nov 10 '12 at 4:34
up vote 2 down vote accepted

If you recall the definition of the stochastic integral with respect to a Wiener process $(B_t, t \geq 0)$, it is plain that for $0 \leq a < b \leq T$ we have

$$ \int_{0}^{T} \xi \cdot \mathbf{1}_{[a, b)}(s) \, dB_s = \xi \cdot (B_{b} - B_{a})$$

for bounded $\mathcal{F}_a$-measurable random variable $\xi$. Extending to the $L^2$-case is immediate in view of the Itō isometry.

However, for a general process $X = (X_t : t \geq 0)$, we cannot say much about the integral

$$ \int_{0}^{T} X_{s} \, dB_{s}$$

even we assume a mild condition to $X$. For example, for a $C^2$ function $f$ we have

$$ f(B_t) - f(B_s) = \int_{s}^{t} f'(B_s) \, dB_s + \frac{1}{2} \int_{s}^{t} f''(B_s) \, ds,$$

which is the celebrated Itō formula.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.