# Basic stochastic integral

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?

Thanks!

-
I wouldn't expect such integral admits a simpler form. – Ilya Jan 4 at 8:49

1
That is not a stochastic integral. It is a standard Lebesgue integral. – Learner Jan 4 at 8:49

1
@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. – Ilya Jan 4 at 9:03

1
@Ilya I do not disagree with any of your points. Still, the title is misleading. – Learner Jan 4 at 9:08

## 2 Answers

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.

-

Thanks. How do you prove that $X_t$ has normal distribution? – John Peter Jan 8 at 10:30
Linear combinations of gaussian families are normal and (Riemann-Stieltjes) integrals are pointwise (hence in distribution) limits of linear combinations. – Did Jan 8 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 $(1)$:

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

-

1
certainly, another expression - but how does it help OP? – Ilya Jan 4 at 9:05
@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. – saz Jan 4 at 9:14
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 14, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9382899403572083, "perplexity_flag": "head"}