# Integration of Wiener process: $\int_{t_1}^{t_2} dB(s)$

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?

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

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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

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.

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