I'm teaching myself Malliavin calculus and Skorohod integrals and with this kind of math I find myself following the logic through but lacking solid intuition about what is going on.

In particular take the Skorohod integral which generalises the Ito integral to possibly anticipating integrands (it reduces to an Ito integral when the integrand is non-anticipating)

So for example take the Skorohod integral, of the Brownian motion $B(t)$, $B(0)=0$ $$ \int_0^T B(T)\delta B(t) = B^2(T)-T$$ (see for example http://www.nhh.no/Files/Filer/institutter/for/dp/1996/wp0396.pdf)

It is emphasised that $$ \int_0^T B(T)\delta B(t) \neq B(T)\int_0^T\delta B(t)=B(T)\int_0^T dB(t)=B^2(T)$$

and I am trying to understand exactly why this is the case. I appreciate that one can't just view it as a Riemann sum which is what gives the intuition that the above is trying to ward off. So the only other kind of intuition I have is that it is because the integrand and integrator are correlated in the same way that the integrand and integrator are correlated in the Stratonovich integral which means it is not a Martingale. I.e. analogously to $$E[f(B)\circ dB]\neq E[f(B)]E[dB]=E[f(B)]\cdot 0=0\quad\left(=E[f(B)dB]\right)$$

However, in which case I don't understand why the difference between the naive and correct interpretations above is $T$ so that the difference grows independently of the correlation.

In other words I would expect that (given $T_2>T_1$) $$\lim_{(T_2-T_1)\to\infty}\int_0^{T_1}B(T_2)\delta B(t)=B(T_2)\int_0^{T_1}\delta B(t)=B(T_2)\int_0^{T_1}dB(t).$$

Or that I would expect the difference $$\int_0^T B(T)\delta B(t) - B(T)\int_0^T\delta B(t)=\int_0^T\xi(t-T)dt$$

where $\xi(t-T)\to 0$ as $T\to\infty$.

But all of this is completely at odds with the definition of the Skorohod integral above. Can someone explain this in a relatively simple with intuitive ideas?

A secondary question it raises is how to numerically simulate such an integral (say given a pre computed Wiener process)