Finite-input vector Gaussian mutual information is bounded by water-filling

Write $Y=X+Z$. Its density is the finite mixture $p_Y(y)={\sum }_x{P}_X(x){\prod }_j{\varphi }_{{\lambda }_j}(y_j-x_j)$, which is positive and bounded by [F1]. Fix any support point $x_0$ with $P_X(x_0)>0$. Then $p_Y(y)\ge P_X(x_0){\prod }_j{\varphi }_{{\lambda }_j}(y_j-x_{0j})$. The Gaussian upper bound and this lower bound show $\mathrm{\mid }\ln p_Y(y)\mathrm{\mid }\le C(1+\mathrm{\parallel }y{\mathrm{\parallel }}_2^2)$ for some finite $C$; $Y$ has finite second moment, so $h(Y)$ is absolutely finite by [F10]. The same bounds apply to each marginal $Y_j$, and each conditional density given a support point is a Gaussian translate, so all entropy integrals below are finite.

The joint output density has finite differential entropy by step 1.1. On the support of $X$, the joint-to-product density ratio in [F5] is ${\prod }_j{\varphi }_{{\lambda }_j}(y_j-x_j)\mathrm{/}{p}_Y(y)$. Expanding its logarithm and integrating gives $I(X;Y)=h(Y)-h(Y\mid X)=h(Y)-h(Z)$, because each conditional output density is a translate of the noise density. The KL divergence of $P_Y$ from the product of its marginals is finite by step 1.1 and nonnegative by [F3]; expanding it gives $h(Y)\le {\sum }_{j}h(Y_j)$.

Independence [F7] and the variance definition [F6] give $\mathrm{Var}(Y_j)=\mathrm{Var}(X_j)+{\lambda }_j$ and $\mathrm{Var}(X_j)\le \mathbb{E}{X}_j^2$. By [F2], $h(Y_j)\le \frac{1}{2}\ln (2\pi e(\mathrm{Var}(X_j)+{\lambda }_j))$. Direct integration of ${\varphi }_{{\lambda }_j}$ using [F1] and [F10] gives $h(Z_j)=\frac{1}{2}\ln (2\pi e{\lambda }_j)$, hence $h(Z)={\sum }_{j}h(Z_j)$ by independence. Tonelli [F8] on the integrable positive and negative parts from step 1.1 justifies the product integrals. Therefore $I(X;Y)\le \frac{1}{2}{\sum }_j\ln (1+\mathrm{Var}(X_j)\mathrm{/}{\lambda }_j)\le \frac{1}{2}{\sum }_j\ln (1+\mathbb{E}{X}_j^2\mathrm{/}{\lambda }_j)$. The powers $p_j=\mathbb{E}{X}_j^2$ sum to at most $P$, so [F4] proves the claim.