Goal. Minimize \(\mathcal{L}_\beta(\proposal) = \KL{\vposterior}{\proposal} + \beta\,\KL{\proposal}{\prior}\) over all distributions \(\proposal\) on the support of the prior, and show that the minimizer is the mixture \(\alpha_\beta\,\vposterior + (1 - \alpha_\beta)\,\antiposterior\), with the weight \(\alpha_\beta\) given by \(\eqref{eq:alpha-fixed-point}\) of the main text.

Step 1: reduce to the segment. Write \(\alpha \defeq \proposal(\valid)\). Both KL terms decompose over the partition \(\{\valid, \validc\}\) (the chain rule for KL: the divergence between the bin masses, plus the mass-weighted divergences within bins):

For the forward KL, the sum runs over \(\valid\) only, and there the proposal factors into bin mass times within-bin shape:

\[ \begin{aligned} \KL{\vposterior}{\proposal} &= \sum_{\str \in \valid} \vposterior(\str)\, \log\frac{\vposterior(\str)}{\proposal(\str)} && \explain{$\vposterior$ puts mass $1$ on $\valid$} \\[6pt] &= \sum_{\str \in \valid} \vposterior(\str)\, \log\frac{\vposterior(\str)}{\alpha\,\proposal(\str \mid \valid)} && \explain{$\proposal(\str) = \alpha\,\proposal(\str \mid \valid)$ on $\valid$} \\[6pt] &= \log\frac{1}{\alpha} + \KL{\vposterior}{\proposal(\cdot \mid \valid)} && \explain{split the log; $\vposterior(\valid) = 1$} \end{aligned} \]

For the anchor, both bins contribute; factor each side into bin mass times within-bin shape (\(\prior = \Z\,\vposterior\) on \(\valid\) and \((1{-}\Z)\,\antiposterior\) on \(\validc\)):

\[ \begin{aligned} \KL{\proposal}{\prior} &= \sum_{\str \in \valid} \alpha\,\proposal(\str \mid \valid)\, \log\frac{\alpha\,\proposal(\str \mid \valid)}{\Z\,\vposterior(\str)} \;+ \sum_{\str \in \validc} (1{-}\alpha)\,\proposal(\str \mid \validc)\, \log\frac{(1{-}\alpha)\,\proposal(\str \mid \validc)}{(1{-}\Z)\,\antiposterior(\str)} \\[6pt] &= \alpha \log\frac{\alpha}{\Z} + (1{-}\alpha) \log\frac{1{-}\alpha}{1{-}\Z} + \alpha\,\KL{\proposal(\cdot \mid \valid)}{\vposterior} + (1{-}\alpha)\,\KL{\proposal(\cdot \mid \validc)}{\antiposterior} && \explain{split the logs; each conditional sums to $1$} \\[6pt] &= d(\alpha \,\|\, \Z) + \alpha\,\KL{\proposal(\cdot \mid \valid)}{\vposterior} + (1{-}\alpha)\,\KL{\proposal(\cdot \mid \validc)}{\antiposterior} && \explain{the leading terms are a Bernoulli KL} \end{aligned} \]

with \(d(\alpha \,\|\, \Z) \defeq \alpha \log\frac{\alpha}{\Z} + (1-\alpha)\log\frac{1-\alpha}{1-\Z}\), the KL between the Bernoulli distributions \((\alpha, 1{-}\alpha)\) and \((\Z, 1{-}\Z)\). Collecting the objective:

\[ \begin{aligned} \mathcal{L}_\beta(\proposal) &= \KL{\vposterior}{\proposal} + \beta\,\KL{\proposal}{\prior} && \explain{the anchored objective} \\[6pt] &= \log\frac{1}{\alpha} + \beta\, d(\alpha \,\|\, \Z) \\ &\qquad + \KL{\vposterior}{\proposal(\cdot \mid \valid)} + \beta\,\alpha\, \KL{\proposal(\cdot \mid \valid)}{\vposterior} + \beta\,(1{-}\alpha)\, \KL{\proposal(\cdot \mid \validc)}{\antiposterior} && \explain{substitute the two decompositions} \\[6pt] &= f(\alpha) + \underbrace{\KL{\vposterior}{\proposal(\cdot \mid \valid)} + \beta\,\alpha\, \KL{\proposal(\cdot \mid \valid)}{\vposterior}}_{\ge\, 0} + \underbrace{\beta\,(1{-}\alpha)\, \KL{\proposal(\cdot \mid \validc)}{\antiposterior}}_{\ge\, 0} && \explain{define $f(\alpha) \defeq \log\frac{1}{\alpha} + \beta\, d(\alpha \| \Z)$} \end{aligned} \]

The trailing terms vanish exactly when \(\proposal(\cdot \mid \valid) = \vposterior\) and \(\proposal(\cdot \mid \validc) = \antiposterior\) — choices that don’t touch \(f(\alpha)\). So, for each fixed weight \(\alpha\), the best proposal is the segment point \(\alpha\,\vposterior + (1 - \alpha)\,\antiposterior\), and its objective value is \(f(\alpha)\): the forward KL charges the log-loss of the valid event, and the anchor charges the KL between the two ways of splitting mass across the bins.

Step 2: minimize over the weight. What’s left is a one-variable problem: minimize \(f\) over \(\alpha \in (0, 1]\).

\[ \begin{aligned} f'(\alpha) &= -\frac{1}{\alpha} + \beta\,\log\frac{\alpha\,(1 - \Z)}{\Z\,(1 - \alpha)} && \explain{$\tfrac{d}{d\alpha}\, d(\alpha \| \Z) = \log\frac{\alpha(1-\Z)}{\Z(1-\alpha)}$} \\[6pt] f''(\alpha) &= \frac{1}{\alpha^{2}} + \beta\,\Bigl(\frac{1}{\alpha} + \frac{1}{1 - \alpha}\Bigr) \;>\; 0 && \explain{$f$ strictly convex} \end{aligned} \]

Since \(f'(\alpha) \to -\infty\) as \(\alpha \to 0^{+}\) and \(f'(\alpha) \to +\infty\) as \(\alpha \to 1^{-}\), there is a unique interior minimizer \(\alpha_\beta\), the root of \(f'\). Both facts are visible live below for any finite \(\beta\): \(f\) falls and then rises (top panel), so the strictly increasing \(f'\) climbs through zero exactly once (bottom panel) — at the marked point, \(\alpha_\beta\), strictly between \(\Z\) and \(1\). Move \(\beta\) and watch the balance shift: small \(\beta\) leaves the minimizer pinned near the posterior’s \(\alpha = 1\); large \(\beta\) drags it toward the prior’s \(\alpha = \Z\). (It is the same \(\beta\) — and the same valid mass \(\Z\) — as in the visualization further below; the marker is draggable. At the \(\beta = 0\) endpoint \(f\) collapses to \(\log\frac{1}{\alpha}\), falling all the way to the boundary minimizer \(\alpha = 1\) (\(f'\) never reaches zero); at the \(\beta = \infty\) endpoint the panels show the normalized limit \(f/\beta = d(\alpha \,\|\, \Z)\), whose minimum sits at \(\alpha = \Z\).)

Setting \(f'(\alpha_\beta) = 0\) and rearranging describes that marked point as a fixed point:

\[ \begin{aligned} \frac{1}{\alpha_\beta\, \beta} &= \log\frac{\alpha_\beta\,(1 - \Z)}{\Z\,(1 - \alpha_\beta)} && \explain{$f'(\alpha_\beta) = 0$, divided by $\beta$} \\[6pt] \alpha_\beta &= \frac{\Z}{\Z + e^{-1/(\alpha_\beta \beta)}\,(1 - \Z)} && \explain{exponentiate; solve for $\alpha_\beta$} \end{aligned} \]

Two sanity checks: \(f'(\Z) = -1/\Z < 0\), so \(\alpha_\beta > \Z\) for every finite \(\beta\) — the anchored optimum always conditions strictly harder than the prior; and as \(\beta \to 0\), \(f \to \log\frac{1}{\alpha}\), minimized at \(\alpha = 1\) — the posterior.

Conclusion. The minimizer of \(\mathcal{L}_\beta\) is exactly \[ \proposal^\star_\beta \;=\; \alpha_\beta\,\vposterior \,+\, (1 - \alpha_\beta)\,\antiposterior, \qquad \alpha_\beta = \frac{\Z}{\Z + e^{-1/(\alpha_\beta \beta)}\,(1 - \Z)} \;\in\; (\Z, 1) \] describing the segment point at the fixed-point weight, as claimed.

Equation \(\eqref{eq:alpha-fixed-point}\) is transcendental—no ‘elementary’ closed form solution for \(\alpha_\beta\) exists, but it does untangle into the Lambert \(W\) function, the inverse of \(w \mapsto w\,e^{w}\):

\[ \begin{aligned} \Bigl(\frac{1}{\alpha_\beta} - 1\Bigr)\, e^{1/(\alpha_\beta \beta)} &= \frac{1-\Z}{\Z} && \explain{rearrange $\eqref{eq:alpha-fixed-point}$} \\[6pt] \bigl(\frac{1}{\alpha_\beta} - 1\bigr)\frac{1}{\beta}\; e^{(\frac{1}{\alpha_\beta} - 1)\frac{1}{\beta}} &= \frac{1-\Z}{\Z\,\beta}\; e^{-1/\beta} && \explain{divide both sides by $\beta\, e^{1/\beta}$} \\[6pt] s\, e^{s} &= \frac{1-\Z}{\Z\,\beta}\; e^{-1/\beta} && \explain{let $s \defeq \bigl(\frac{1}{\alpha_\beta} - 1\bigr)\frac{1}{\beta}$} \\[6pt] s &= W_0\!\Bigl(\frac{1-\Z}{\Z\,\beta}\, e^{-1/\beta}\Bigr) && \explain{invert $s \mapsto s\,e^{s}$} \\[6pt] \alpha_\beta &= \frac{1}{1 + \beta\, W_0\!\Bigl(\dfrac{1-\Z}{\Z\,\beta}\, e^{-1/\beta}\Bigr)} && \explain{solve for $\alpha_\beta$} \end{aligned} \]

The argument of \(W\) is positive, so this lands on the principal branch \(W_0\), where the solution is unique (the same uniqueness that strict convexity of \(f\) gave above).

It slides monotonically from \(1\) at \(\beta = 0\) (recovering \(\vposterior\)) down to \(\Z\) as \(\beta \to \infty\) (recovering \(\prior\), which sits on the same segment at weight \(\Z\)) — but very asymmetrically. Equation \(\eqref{eq:alpha-fixed-point}\) says the odds of invalidity under the optimum are the prior odds, discounted: \[ \frac{1 - \alpha_\beta}{\alpha_\beta} \;=\; e^{-1/(\alpha_\beta \beta)}\; \frac{1-\Z}{\Z}. \] For small \(\beta\) (where \(\alpha_\beta \approx 1\)) the discount factor is \(\approx e^{-1/\beta}\), exponentially small — at \(\beta = \tfrac{1}{2}\) the invalid set keeps at most \(e^{-2} \approx 0.14\) of its prior odds; at \(\beta = 0.1\), only \(e^{-10} \approx 5 \times 10^{-5}\) — giving the leading-order approximation \[ 1 - \alpha_\beta \;\approx\; \frac{1 - \Z}{\Z}\, e^{-1/\beta} \qquad (\beta \to 0). \] At the other end, the approach to the prior is slow: \(\alpha_\beta - \Z \approx (1 - \Z)\frac{1}{\beta}\) as \(\beta \to \infty\).

Both regimes also fall out of the closed form of the previous aside in a single stroke: the argument of \(W_0\) tends to \(0\) at both ends of the \(\beta\) axis, so the one expansion \(W_0(u) \approx u\) covers them — as \(\beta \to 0\), \(1 - \alpha_\beta \approx \beta\,W_0(\cdot) \approx \frac{1-\Z}{\Z}\, e^{-1/\beta}\), recovering the approximation above; and as \(\beta \to \infty\), \(\beta\,W_0(\cdot) \to \frac{1-\Z}{\Z}\), so \(\alpha_\beta \to \Z\), with \(\alpha_\beta - \Z \approx (1 - \Z)\frac{1}{\beta}\) at the next order.

We minimize over all distributions on the support of \(\prior\), so introduce a Lagrange multiplier for normalization (nonnegativity turns out to be slack — see the end):

\[ \begin{aligned} \mathcal{J}(\proposal, \lambda) &\defeq \sum_{\str} \posterior(\str)\,\log\frac{\posterior(\str)}{\proposal(\str)} + \beta \sum_{\str} \proposal(\str)\,\log\frac{\proposal(\str)}{\prior(\str)} + \lambda\,\Bigl(\sum_{\str} \proposal(\str) - 1\Bigr) && \explain{Lagrangian of $\mathcal{L}_\beta$} \\[6pt] 0 = \frac{\partial \mathcal{J}}{\partial\, \proposal(\str)} &= -\frac{\posterior(\str)}{\proposal(\str)} + \beta\,\Bigl(\log\frac{\proposal(\str)}{\prior(\str)} + 1\Bigr) + \lambda && \explain{stationarity, term by term} \\[6pt] \frac{\posterior(\str)}{\proposal(\str)} &= \beta\,\log\frac{\proposal(\str)}{\prior(\str)} + c && \explain{rearrange; $c \defeq \beta + \lambda$} \end{aligned} \]

Note this pointwise condition makes no use of the binary assumption — it holds for any potential. For the binary case it splits into two branches. Off the valid set it pins the ratio to the prior:

\[ \begin{aligned} 0 &= \beta\,\log\frac{\proposal(\str)}{\prior(\str)} + c && \explain{$\posterior(\str) = 0$ for $\str \in \validc$} \\[3pt] \proposal(\str) &= e^{-c/\beta}\,\prior(\str) && \explain{a constant ratio to the prior on $\validc$} \end{aligned} \]

On the valid set, write \(\rho(\str) \defeq \proposal(\str)/\posterior(\str)\); since \(\posterior = \prior/\Z\) there, \(\proposal/\prior = \rho/\Z\), and the condition becomes

\[ \begin{aligned} \frac{1}{\rho(\str)} &= \beta\,\log\frac{\rho(\str)}{\Z} + c && \explain{substitute $\posterior = \prior/\Z$ on $\valid$} \end{aligned} \]

The left side is strictly decreasing in \(\rho(\str)\) and the right side strictly increasing, so this equation has exactly one solution — the same one for every \(\str \in \valid\): the ratio \(\rho\) is constant on \(\valid\). Summing \(\proposal = \rho\,\posterior\) over \(\valid\) identifies the constant:

\[ \begin{aligned} \proposal(\str) &= \alpha_\beta\,\posterior(\str) \quad \text{for } \str \in \valid, \qquad \alpha_\beta \defeq \proposal(\valid) && \explain{since $\posterior(\valid) = 1$} \end{aligned} \]

It remains to combine the two branches and eliminate \(c\):

\[ \begin{aligned} 1 - \alpha_\beta &= e^{-c/\beta}\,(1 - \Z) && \explain{normalization: sum the $\validc$ branch} \\[6pt] c &= \frac{1}{\alpha_\beta} - \beta\,\log\frac{\alpha_\beta}{\Z} && \explain{the $\valid$ branch, at $\rho = \alpha_\beta$} \\[6pt] \varepsilon_\beta \defeq \frac{e^{-c/\beta}}{\alpha_\beta/\Z} &= e^{-1/(\alpha_\beta \beta)} && \explain{ratio of the two $\proposal/\prior$ levels; substitute $c$} \end{aligned} \]

So the stationary \(\proposal\) is proportional to the prior at one level on \(\valid\) (\(\proposal/\prior = \alpha_\beta/\Z\)) and at the relative level \(\varepsilon_\beta\) below it on \(\validc\) — exactly \(\proposal^\star_\beta \propto \prior \cdot \max\{\potential, \varepsilon_\beta\}\) — and normalizing this form gives back \(\alpha_\beta = \Z/(\Z + \varepsilon_\beta (1 - \Z))\): with \(\varepsilon_\beta = e^{-1/(\alpha_\beta \beta)}\) just derived, exactly the fixed point found in the binary section.

Finally, this stationary point is the global minimum: \(\mathcal{L}_\beta\) is strictly convex in \(\proposal\) (the forward KL is convex in its second argument, and the reverse KL strictly convex in its first), the feasible set is convex, and the optimum is interior — as any \(\proposal(\str) \to 0\) the derivative tends to \(-\infty\) on both branches, so no mass is pushed to the boundary.

Assume the binary potential. On \(\valid\) we have \(\max\{\potential, \varepsilon_\beta\} = 1\), so \(\proposal^\star_\beta \propto \prior\) there, with unnormalized block mass \(\Z\); on \(\validc\) we have \(\max\{\potential, \varepsilon_\beta\} = \varepsilon_\beta\), giving unnormalized block mass \(\varepsilon_\beta\,(1 - \Z)\). Normalizing, \[ \proposal^\star_\beta(\valid) = \frac{\Z}{\Z + \varepsilon_\beta\,(1 - \Z)} = \alpha_\beta, \] (the fixed point once more), while within each block the conditional shape is untouched — it is the prior’s: \(\proposal^\star_\beta(\cdot \mid \valid) = \prior(\cdot \mid \valid) = \vposterior\) and \(\proposal^\star_\beta(\cdot \mid \validc) = \antiposterior\). Hence \[ \proposal^\star_\beta = \alpha_\beta\,\vposterior + (1 - \alpha_\beta)\,\antiposterior. \]