Is $\int_{0}^{t} W_s ds = \int_{0}^{t} s dW_s$?

Question

Apologies if this is a really silly question, I’m more or less trying to teach myself stochastic calculus on my own .

Is $\int_{0}^{t} W_s ds = \int_{0}^{t} s dW_s$?

Also, is this expression solvable as a function of $\tau$?:

$\int_{t-\tau}^{t} (\int_{0}^{u} W_s ds)^2 du$

Can I write it as

$\int_{t-\tau}^t (\int_{0}^{u} s dW_s)^2 du$

$= \int_{t-\tau}^t (\int_{0}^{u} s^2 ds) du$

$=\int_{t-\tau}^t \frac{u^3}{3} du$

$= \frac{t^4-(t-\tau)^4}{12}$

Is this correct ???

Simulations seem to indicate the answer is probably closer to

$\frac{\tau^2}{9}$ independent of $t$.

And similarly is

$\int_{t-\tau}^{t} (\int_{0}^{u} W_s ds) du$

equal to $0$?

Answer 1

You are asking too many questions in the same post. For your first question:

From Ito-lemma:

$$ d(tW(t))=W(t)dt+tdW(t)\\ \Rightarrow tW(t)=\int_0^tW(u)du+\int_0^tudW(u) $$