How to calculate this easy stochastic integral?

Question

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 $f(W_t,t)$?

Answer 1

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.