Finite chaos projection, Parseval identity, mean, and variance

For each $\beta \in \mathrm{\Lambda }$, bilinearity and [F2] give $⟨u-P_{\mathrm{\Lambda }}u,{\psi }_{\beta }⟩=c_{\beta }-{\sum }_{\alpha \in \mathrm{\Lambda }}{c}_{\alpha }{\delta }_{\alpha \beta }=0$. Thus the residual is orthogonal to every basis vector and hence to their finite span.

Write $u=P_{\mathrm{\Lambda }}u+(u-P_{\mathrm{\Lambda }}u)$. By step 1.1 the two terms are orthogonal, so bilinearity and symmetry from [F3] give $\mathrm{\parallel }u{\mathrm{\parallel }}_2^2=\mathrm{\parallel }{P}_{\mathrm{\Lambda }}u{\mathrm{\parallel }}_2^2+\mathrm{\parallel }u-P_{\mathrm{\Lambda }}u{\mathrm{\parallel }}_2^2$. Expanding the finite sum with [F2] gives $\mathrm{\parallel }{P}_{\mathrm{\Lambda }}u{\mathrm{\parallel }}_2^2={\sum }_{\alpha ,\beta \in \mathrm{\Lambda }}{c}_{\alpha }{c}_{\beta }{\delta }_{\alpha \beta }={\sum }_{\alpha \in \mathrm{\Lambda }}{c}_{\alpha }^2$. This proves the norm identity directly from the integral pairing.

Since ${\psi }_0=1$, $c_0=⟨u,1⟩=\mathbb{E}[u]$. The centered variable decomposes as $u-\mathbb{E}[u]={\sum }_{\alpha \in \mathrm{\Lambda }\setminus \{0\}}{c}_{\alpha }{\psi }_{\alpha }+(u-P_{\mathrm{\Lambda }}u)$. The two summands are orthogonal by step 1.1, and expansion of the integral norm square as in step 2.1, followed by [F2], gives the displayed variance identity.