Parallel Gaussian information density obeys a finite-variance law of large numbers

For $p_j>0$, completing the square gives $x^2\mathrm{/}{p}_j+(y-x{)}^2\mathrm{/}{\lambda }_j=((p_j+{\lambda }_j)\mathrm{/}(p_j{\lambda }_j))(x-p_{j}y\mathrm{/}(p_j+{\lambda }_j){)}^2+y^2\mathrm{/}(p_j+{\lambda }_j)$, so [F4] yields ${\int }_{\mathbb{R}}{\varphi }_{p_j}(x){\varphi }_{{\lambda }_j}(y-x)dx=(2\pi (p_j+{\lambda }_j){)}^{-1\mathrm{/}2}{e}^{-y^2\mathrm{/}(2(p_j+{\lambda }_j))}={\varphi }_{p_j+{\lambda }_j}(y)$. If $p_j=0$, then $X_j=0$ and $Y_j=Z_j$ has density ${\varphi }_{{\lambda }_j}$. Independence across coordinates and uses gives product output densities by [F5].

For $p_j>0$, substituting $Y_j=X_j+Z_j$ in the log-density ratio gives ${\mathrm{ı}}_j=\frac{1}{2}\ln (1+p_j\mathrm{/}{\lambda }_j)-Z_j^2\mathrm{/}(2{\lambda }_j)+(X_j+Z_j{)}^2\mathrm{/}(2(p_j+{\lambda }_j))$. By [F3] and [F9], the second moments are $p_j$ and ${\lambda }_j$ and the means are zero; [F5] factors the cross moment, so $\mathbb{E}{\mathrm{ı}}_j=\frac{1}{2}\ln (1+p_j\mathrm{/}{\lambda }_j)$. Its second moment is finite because $(a+b+c{)}^2\le 3(a^2+b^2+c^2)$, $(x+z{)}^4\le 8(x^4+z^4)$, and the Gaussian fourth moments in [F3] are finite. If $p_j=0$, the information density is identically zero, which agrees with the formula; for finitely many branches, $({\sum }_j{\mathrm{ı}}_j{)}^2\le d{\sum }_j{\mathrm{ı}}_j^2$, so the summed density has mean $I=\frac{1}{2}{\sum }_j\ln (1+p_j\mathrm{/}{\lambda }_j)$ and finite variance.

The pairs $(X_t,Y_t)$ are IID across uses because [A2] gives fresh independent noises and [F5] gives product coordinate laws. Their information densities are therefore IID with mean $I$ and finite variance $v$ by step 2.1. For $s\mathrm{\ne }t$, [F10] factors the expectation of the product of their centered information densities, which is zero; expanding the squared sum gives $\mathbb{E}[({\sum }_{t=1}^n\mathrm{ı}(X_t;Y_t)-nI{)}^2]=nv$. For every $\eta >0$, the normalized deviation event forces this square to be at least $n^2{\eta }^2$, so [F8] bounds its probability by $v\mathrm{/}(n{\eta }^2)\to 0$.