For $W_t$ - standard Wiener process - let: $$Y_t=\int_0^t \left( \int_0^s (s-u) dW_u \right) dW_s$$ a) Show that $Y_t$ is well defined for all $t>0$.
b) Calculate $\mathbb EY_t$ and $\mathbb EY_t^2$.
If I understand correctly $Y_t$ will be well defined if it is integrable and predictable (correct me if I'm wrong).
Integrable means $\mathbb EY_t<\infty$. However I don't know what I can use to estimate $$\mathbb E \left(\int_0^t \left( \int_0^s (s-u) dW_u \right) dW_s \right)$$ or how to calculate a specific value $\mathbb EY_t$ and $\mathbb EY_t^2$
Usually, Stochastic Integral $\int_{0}^{}f(s)\,dW_{s}$ is defined for functions $f$ which are $L^{2}(\lambda\vert_{[0,t]}\otimes \mathbb{P})$. So you have to show that the integrand satisfies the above.
$$\int_{0}^{s} (s-u)\,dW_{u}=sW_{s}-\int_{0}^{s}u\,dW_{u}=sW_{s}-sW_{s}+\int_{0}^{s}W_{u}\,du$$ using integration by parts for stochastic integrals.
So $\displaystyle Y_{t}=\int_{0}^{t}\bigg(\int_{0}^{s}W_{u}\,du\bigg)\,dW_{s}$
Now we wish to note that whether $\int_{0}^{s}W_{u}\,du$ is $L^{2}\bigg([0,t]\otimes\mathbb{P}\bigg)$ or not.
First notice that $\displaystyle\mathbb{E}(\int_{0}^{s}W_{u}\,du)=0$ due to an application of Fubini (I'll go to the justification later) and we have $\displaystyle Var(\int_{0}^{s}W_{u}\,du)=Var(\int_{0}^{s}(s-u)\,dW_{u})=\int_{0}^{s}(s-u)^{2}\,du=\frac{s^{3}}{3}$ due to Ito's Isometry.
And we have that $\int_{0}^{t}\frac{s^{3}}{3}ds=\frac{t^{4}}{12}$ is finite. Thus $\int_{0}^{s}W_{u}\,du$ is in $L^{2}\bigg([0,t]\otimes\mathbb{P}\bigg)$ and hence $Y_{t}$ is well defined.
Thus we have that $\displaystyle\int_{0}^{t}\bigg(\int_{0}^{s}W_{u}\,du\bigg)dW_{s}$ is a stochastic integral of an $L^{2}\bigg([0,t]\otimes\mathbb{P}\bigg)$ process and hence is a martingale.
Thus $\mathbb{E}(Y_{t})=\mathbb{E}(Y_{0})=0$
For $Var(Y_{t})$, we have by Ito's Isometry that it would equal
$$\mathbb{E}\bigg(\int_{0}^{t} \bigg(\int_{0}^{s}W_{u}\,du\bigg)^{2}\,ds\bigg)$$ .
But by Fubini again, you have that the above equals $$\int_{0}^{t} Var(\int_{0}^{s}W_{u}\,du)\,ds=\int_{0}^{t}\frac{s^{3}}{3}=\frac{t^{4}}{12}$$