I am new to this stuff. Can some one explain how I could compute the stochastic integral of the form $\int_0^t W_sds$, where $W_t$ is Brownian process?

  • $\begingroup$ I wouldn't expect such integral admits a simpler form. $\endgroup$ – Ilya Jan 4 '13 at 8:49
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    $\begingroup$ That is not a stochastic integral. It is a standard Lebesgue integral. $\endgroup$ – Learner Jan 4 '13 at 8:49
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    $\begingroup$ @Learner: this can be even considered as a Riemann integral as $W_s(\omega)$ is continuous on $[0,t]$ for any $\omega$. However, the term "stochastic integral" here may refer to methods that can be applied to find its value. E.g. one has $$ \int\limits_0^t W_s\mathrm ds = tW_t - \int\limits_0^t s\mathrm dW_s $$ which however we can't have for any Lebesgue/Riemann integral. Neither this latter form is simpler, though. $\endgroup$ – Ilya Jan 4 '13 at 9:03
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    $\begingroup$ @Ilya I do not disagree with any of your points. Still, the title is misleading. $\endgroup$ – Learner Jan 4 '13 at 9:08

What to compute the integral means is unclear in this context but one can say this: for every $t\geqslant0$, the random variable $$ X_t=\int_0^tW_s\mathrm ds $$ is centered normal vith variance $\sigma_t^2$ where $$ \sigma_t^2=\mathbb E(X_t^2)=2\int_0^t\int_0^s\mathbb E(W_sW_u)\mathrm du\mathrm ds=\int_0^t\int_0^s2u\mathrm du\mathrm ds=\frac{t^3}3. $$ The process $(X_t)_{t\geqslant0}$ is called integrated Brownian motion and is the subject of some active research, for a sample see this paper and the list of references therein.

  • $\begingroup$ Thanks. How do you prove that $X_t$ has normal distribution? $\endgroup$ – John Peter Jan 8 '13 at 10:30
  • $\begingroup$ Linear combinations of gaussian families are normal and (Riemann-Stieltjes) integrals are pointwise (hence in distribution) limits of linear combinations. $\endgroup$ – Did Jan 8 '13 at 10:36

As Learner pointed out, the integral $\omega \mapsto \int_{0}^t W_s(\omega) \, ds$ is not a stochastic integral, it's a pathwise Lebesgue integration.

But anyway: If we would like to obtain another expression for this integral, we can apply Itô's formula:

$$f(W_t)-f(W_0) = \int_0^t f'(W_s) \, dW_s + \frac{1}{2} \int_0^t f''(W_s) \, ds \tag{1}$$

Since we are looking for the "$ds$-part", it would be nice to have

$$f''(W_s) = 2 W_s$$

i.e. $f''(x)=2x$. We obtain this by choosing $f(x) := \frac{x^3}{3}$. By applying $(1)$:

$$\frac{W_t^3}{3} - 0 = \int_0^t W_s^2 \, dW_s + \frac{1}{2} \int_0^t 2 W_s \, ds \\ \Rightarrow \int_0^t W_s \, ds = \frac{W_t^3}{3} - \int_0^t W_s^2 \, dW_s$$

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    $\begingroup$ certainly, another expression - but how does it help OP? $\endgroup$ – Ilya Jan 4 '13 at 9:05
  • $\begingroup$ @Ilya Since the OP mentioned that he's new to this stuff, it's probably a transformation like this he is looking for. If not so, I'll delete my answer. $\endgroup$ – saz Jan 4 '13 at 9:14

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