Intuitive difference between $\int_0^T W_s ds $ and $\int_0^T W_t dW_t $

Question

What is the intuitive difference between the integrals $\int_0^T W_s ds $ and $\int_0^T W_t dW_t$?

The first cant be solved analytically (ie when integrating with respect to a time? variable), but when you integrate with respect to the Brownian motion it has an analytical solution.

Clearly, I am missing the difference between what these integrals are representing.

Answer 1

By way of example of what is being done here, if you have a differentiable $f(x)$ with $f(0) = 0$, you can define (e.g. in the Riemann-Stieltjes sense) $$ I = \int_0^x f(s) df(s) = \int_0^x f(s) f'(s) ds = \frac{f(x)^2}{2} , $$ which is easy to see using a substitution $u = f(x)$. It would also be easy to see that $$ J = \int_0^x f(s) ds $$ would be well-defined, but would not necessarily have a nice analytic result you can associate it with (for example consider $f(x) = e^{-x^2}$.

A similar thing is happening here, except of course $W_t$ is continuous everywhere and not differentiable anywhere, so you need a different notion of the integral. When you define the stochastic integral, e.g. using Ito's definition (there is also a Stratonovich definition with different properties), it is not that surprising that $\int_0^t W_s dW_s$ has a closed form and $\int_0^t W_s ds$ does not.

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UPDATE

Thanks for clarifying what your questions really are. As for the need for the new integral, typically, for Riemann-Stieltjes (R-S), you need $f$ to be monotone, which Brownian motion is not. There are generalizations of R-S which take any $f$ of bounded variation, but it certainly must be deterministic. Here, $W_t$ is a stochastic process, and in that sense, Ito and Stratonovich both generalize R-S.

As for why it is or is not surprising, generally, $\int_0^t f(s) ds$ involves crossing the natures of 2 different things: $f(x)$ and $g(x)=x$, whereas $\int_0^t f(t)df(t)$ would only involve one thing, so in my mind, it makes more sense for that to have a hope of analytic form in terms of $f$ itself, like it works out both in my deterministic example above and in the stochastic one you are posting.