How do we evaluate the average of the square of a Wiener signal?
Standard case:
Typically, the signal average is $S(t)=\frac{1}{T}\int_{0}^{T}s(t)dt$, where we can write the integral in Ito form $S(t)=\frac{1}{T}\int_{0}^{T}dS$, where $dS=s(t)dt=\mu dt+\sigma dW$.
This is easy to evaluate $$ S(t)=\frac{1}{T}\int_{0}^{T}dS=\frac{1}{T}\int_{0}^{T}\left(\mu dt+\sigma dW\right)=\mu+\frac{\sigma}{\sqrt{T}}W_{\sigma^{2}=1}$$
Squared case:
We want to evaluate $$ S_2(t)=\frac{1}{T}\int_{0}^{T}s(t)^2 dt $$
I guess the challenge is to write this in proper Ito form. It isn't obvious to me how to use Ito lemma here, since $S_{2}\neq S^{2}$ or some function $f(S)$. Is $S_2$ not an Ito process anymore?
Naive approach:
From the standard case, $s(t) = \frac{dS}{dt}=\mu+\sigma\xi(t)$, where $\xi (t)=\frac{dW}{dt}$.
So $s(t)^2 = (\frac{dS}{dt})^2=\mu^2+\sigma^2\xi(t)^2+2 \mu\sigma\xi(t)$.
Integrating $\int_{0}^{T}s(t)^2 dt = \int_{0}^{T} (\frac{dS}{dt})^2 dt=\int_{0}^{T} \frac{dS}{dt}dS $, which I have no idea how to evaluate, or weather this is the right approach.
Without an additional regularisation (say with colored noise), the integral you are trying to compute is strongly divergent.
The integrand is dominated by $\xi^2=\left(\frac{dW}{dt}\right)^2$ which is positive and of order $\frac{1}{dt}$ and there is thus no hope to give it a meaning in the stochastic integral sense. You are just integrating infinite positive terms. Sometimes you may indeed see integrals of white noise squared exponentiated, e.g. : $$ e^{-\int_0^T \xi(t)^2 dt + \cdots} $$ which are a physicist way of writing the Wiener measure in path integrals. However, they make no sense in the Ito formalism.