What am I doing wrong here?
Let $W_s$ for $s\in[0,T]$ be the standard wiener process and define two stochastic processes given by $dX_t = x_sdW_s$ and $dY_s=y_sdW_s$ where $x_s,y_s$ are adapted and square-integrable. Now take the following expectation from the perspective of period $t$: $$ E_t\Bigg[\int_t^Tx_s Y_sdW_s\bigg] $$ My guess is the above expression equals zero because $x_sY_S$ is an adapted process and $x_sY_SE_S[dW_s]=0$. But we also can rewrite the above equation as $$ E_t\Bigg[\int_t^T x_s\bigg\{\int_0^sy_k dW_k\bigg\}dW_s\bigg] $$ which by changing the order of integration gives: $$ E_t\Bigg[\int_t^T y_s\bigg\{\int_s^Tx_k dW_k\bigg\}dW_s\bigg] $$ which now no longer seems it is equal to zero. Not sure where I went wrong.
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Here may be this will elaborate more:
From changing the order of integration, we had
$$
E_t\Bigg[\int_t^Tx_s Y_sdW_s\bigg]
$$
$$
=E_t\Bigg[\int_t^T y_s\bigg\{\int_s^Tx_k dW_k\bigg\}dW_s\bigg]
$$
$$
=E_t\Bigg[\int_t^T y_s\bigg(X_T-X_s\bigg)dW_s\bigg]
$$
$$
=E_t\Bigg[X_T\bigg(Y_T-Y_t\bigg)\bigg]-E_t\Bigg[\int_t^T y_sX_sdW_s\bigg]
$$
Assume$$
E_t\Bigg[\int_t^Tx_s Y_sdW_s\bigg]=0
$$
for the reason I mentioned above which then means by a similar argument, we have
$$
E_t\Bigg[\int_t^T y_s X_sdW_s\bigg]=0
$$.
Therefore, we get
$$
E_t\Bigg[\int_t^Tx_s Y_sdW_s\bigg]
$$
$$
=E_t\Bigg[X_T\bigg(Y_T-Y_t\bigg)\bigg]
$$
$$
=E_t\Bigg[X_TY_T\bigg] -Y_tX_t
$$
Using Ito's lemma, $d[X_sY_S]=X_sy_sdW_s+Y_sx_sdW_s+x_sy_sds$
and thus,
$$
E_t\Bigg[X_TY_T\bigg] -Y_tX_t
$$
$$
=E_t\Bigg[\int_0^T d[X_sY_s]\bigg] -Y_tX_t
$$
$$
=E_t\Bigg[\int_t^T d[X_sY_s]\bigg]+X_tY_t -Y_tX_t
$$
$$
=E_t\Bigg[\int_t^T X_sy_sdW_s+\int_t^T Y_sx_sdW_s+ \int_t^T x_sy_sds\bigg]
$$
$$
=E_t\Bigg[\int_t^T x_sy_sds\bigg]\neq 0
$$
Replacing your $t$ by $0$ and your $T$ by $1$ to simplify the setting, you are basically asserting that $$\int_0^1Y_tdX_t=\int_0^1\left(\int_0^tdY_s\right)dX_t$$ is equal to $$\int_0^1\left(\int_s^1dX_t\right)dY_s=\int_0^1(X_1-X_s)dY_s=X_1Y_1-\int_0^1X_tdY_t$$ But it happens that $$X_1Y_1=\int_0^1Y_tdX_t+\int_0^1X_tdY_t+\int_0^1d\langle X,Y\rangle_t$$ hence the step equating $$\int_0^1\left(\int_0^tdY_s\right)dX_t\quad\text{and}\quad\int_0^1\left(\int_s^1dX_t\right)dY_s$$ is invalid. One can note that in the formula $$\int_0^1\left(\int_s^1dX_t\right)dY_s$$ the process $$Z_s=\int_s^1dX_t=X_1-X_s$$ is not progressively measurable, hence a definition of the object $$\int_0^1Z_sdY_s$$ is lacking.