The rate splits into information and a mismatch

Two corners share each distribution, so the information between the example and the code equals the information between the sign and the code. That is $\mathcal H(q(z))-\mathcal H(q(z\mid\text{sign}))$ (Mutual information · 0.IT.04). The first entropy has no closed form; integrate it numerically.

Average KL. Every corner has $m^2=0.75$ and $v=0.25$, so each KL is $\tfrac12(0.75+0.25-1-\log0.25)=\log2=0.693147$, and so is the average.

Information. The aggregate posterior is the mixture $q(z)=\tfrac12\mathcal N(0.866025,0.25)+\tfrac12\mathcal N(-0.866025,0.25)$. Each component has entropy $\tfrac12\log(2\pi e\cdot0.25)=0.725791$. Numerical integration of $-q\log q$ gives the mixture’s entropy, and the difference is

It is below $\log2=0.693147$, the most a binary sign could carry, because the two components overlap.

Mismatch. $\KL(q(z)\,\|\,\mathcal N(0,1)) =-\mathcal H(q)+\tfrac12\log2\pi+\tfrac12\E_q[z^2]$, and $\E_q[z^2]=0.75+0.25=1$. Numerically it is $0.107208$.

The identity. $0.585939+0.107208=0.693147$, the average KL. Most of the rate pays for information about the example; about a sixth pays for the aggregate posterior not being the prior.

At $z=0$. The mixture’s density is $2\cdot\tfrac12\mathcal N(0;0.866025,0.25)=0.178032$, and the prior’s is $1/\sqrt{2\pi}=0.398942$. The mixture has the prior’s variance but not its shape. Codes drawn from the prior near zero, which are the commonest, are codes the encoder produced less than half as often. That is a hole, and the second term of (I.13.9) is its price.