After some calculations I ended up with (there each $V_k : \mathbb{R} \to \mathbb{R}$)
\sum_d \widetilde{P}_D(d) \left( \sum_{f' \in \mathcal{F}} P_{F|\;D}(f'|\;d) V_j(f',d) \right) \left( \sum_{f \in \mathcal{F}} P_{F|\;D}(f|\;d) V_i(f,d) \right)
Since
$ E_{P_{F|\;D}}[V_i|\;d] = \sum_{f \in \mathcal{F} } P_{F|\;D}(f|\;d) V_i(f,d) $
I get
$ \sum_d \widetilde{P}_D(d) E_{P_{F|\;D}}[V_j|\;d] E_{P_{F|\;D}}[V_i|\;d] $
Which I believe equals
$ E_{\widetilde{P}_d P_{F|\;D}}[V_i V_j] $
and just rewriting the first expression it is equal to:
\sum_d \widetilde{P}_D(d) \sum_{f \in \mathcal{F}} \sum_{f' \in \mathcal{F}} P_{F|\;D}(f|\;d) P_{F|\;D}(f'|\;d) V_j(f',d) V_i(f,d)
Which seems strange to me, that the expectation value of $V_i V_j$ would take into consideration the value when they are fed the same $d$ but different $f$.
I thought it would be written as
$ \sum_d \widetilde{P}_D(d) \sum_{f \in \mathcal{F}} P_{F|\;D}(f|\;d) V_j(f,d) V_i(f,d) $
Is it correct that the expression at the top equals $E_{\widetilde{P}_d P_{F|\;D}}[V_i V_j]$?