# How to calculate this easy stochastic integral?

I have a relatively simple homework for stochastic calculus that I recently started to learn. I cannot seem to calculate the following integral: $$\int_0^t s dW_s$$ In principle, it should be solved by guessing some primitive function $f(W_t,t)$, applying Ito's lemma and magically obtaining the integral itself. For example, this it what I came up with: $$f(W_t,t) = tW_t \Rightarrow df = t dW_t + W_t dt \Rightarrow f = \int_0^t s dW_s + \int_0^t W_t dt,$$ from which we can express $\int_0^t s dW_s$. However, this doesn't help much, since I cannot calculate $\int_0^t W_t dt$. Can anyone suggest a better candidate for $f(W_t,t)$?

In general, it is not possible to calculate stochastic integrals explicitly. Here we mean by an "explicit calculation" that we can write it as a function of time and the Brownian motion itself, i.e.

$$F(t,W_t) = x + \int_0^t f(s,W_s) \, dW_s \tag{1}$$

where $f$ is a given function.

By Itô's formula,

$$F(t,W_t)-F(0,W_0) = \int_0^t \frac{\partial}{\partial x} F(s,W_s) \, dW_s + \int_0^t \left( \frac{1}{2} \frac{\partial^2}{\partial x^2} F(s,W_s) + \frac{\partial}{\partial t} F(s,W_s) \right) \, ds.$$

In order to calculate $\int_0^t s \, dW_s$ -in the sense of $(1)$-, we need a function $F$ such that

\begin{align*} \frac{\partial}{\partial x} F(s,W_s) &= s \tag{2} \\ \left( \frac{1}{2} \frac{\partial^2}{\partial x^2} F(s,W_s) + \frac{\partial}{\partial t} F(s,W_s) \right) &= 0. \tag{3} \end{align*}

From $(2)$ we see that

$$F(s,x) = s \, x + c(s)$$

for some (differentiable) function $c$ which does only depend on the time $s$. Plugging this into $(3)$, we see that

$$x+ c'(s) \stackrel{!}{=} 0.$$

Since $c$ does only depend on $s$, there does not exist a function $F$ satisfying $(2)$ and $(3)$. Therefore, we cannot calculate the integral explicitly.

• Thank you! Very informative answer. Commented Mar 22, 2014 at 11:07
• @saz I might be missing something trivial, but why is $\frac{\partial ^2 F}{\partial x^2} = x$? I thought equation (2) would imply its zero Commented Jan 15, 2019 at 15:33
• @dleal Where did I claim/use that $\frac{\partial^2}{\partial x^2 }F=x$...? I can't find the spot...
– saz
Commented Jan 15, 2019 at 15:40
• @saz I'm just a bit confused. When you say "Plugging this into (3), we see that, $x + c'(s) != 0$". Are you using $F(s,x) = sx + c(s)$ plugging it into (3). Would that imply that you have to take the second partial derivative wrt x of the latter plus the time derivative. But the second derivate wrt x is zero, isnt it? Commented Jan 15, 2019 at 15:44
• @dleal Well, the whole point is that there does not exist a function $F$ which satisfies both (2) and (3). I'm saying that any function $F$ which satisfies (2) has to be of the form $F(s,x) = sx + c(s)$ and then I show that any such function cannot satisfy (3). You are totally right that there might be other ways to come to the same conclusion.
– saz
Commented Jan 15, 2019 at 15:55

At least, it is easy to compute the law of this stochastic integral. $$\int_{0}^{t} s \,dWs = \lim_{n\to\infty} \sum_{k=1}^{n} t_k(W_{t_{k+1}} - W_{t_k})$$ such that it is a limit of gaussian random variables by Brownian Motion being a gaussian process. A gaussian process is closed under weak convergence. Therefore this limit is gaussian. To characterize the law of a gaussian random variable, one can compute its mean and its variance.

We can show this with L1 convergence by showing that first and second moments of partial sums from the limit are uniformly bounded due to to centered process and to properties related to quadratic variation of brownian motion respectively.

$$\int_{0}^{t} s \,dWs \sim N(0,\frac{t^3}{3})$$