The entries $r_{ij}$ have vanishing odd moments, and every positive even moment is $1/2$. To compute the moments of the sum, let $S_j={\sum }_{k=1}^j{r}_{ik}{w}_k$. For each positive integer $r$, independence and the binomial expansion give

Here $S_0=0$, with zeroth moment equal to $1$. For example, the surviving terms of the sixth moment have exponent patterns $(6)$, $(4,2)$, and $(2,2,2)$, so

The second and fourth moments are given in eq. 4.71. Collecting the terms above, and applying the same recurrence through order eight, gives the remaining exact identities

We retain these identities rather than replacing every moment by an upper bound: the coefficients of the polynomial alternate in sign.

We use the polynomial form in eq. 4.75, with coefficients determined below. The terms in eqs. 4.71 and A.1 depending only on ${\Vert \bar{w}\Vert }_2$ are

These are not the full binomial moments. We first seek positive weights and points such that

Matching coefficients requires

Each of the two points satisfies $a_{\pm }^2=(a_{-}+a_{+})a_{\pm }-a_{-}{a}_{+}$. Multiplying by further powers and taking the same weighted sums eliminates the weights:

Substitution of the four values of $m_r$ gives

Therefore the touching points are

The first two moment equations then give

Simplifying, all three weights are positive:

For a polynomial satisfying $F_4(a)\ge \exp (-sa)$, the right-hand side of eq. A.2 is smallest when the polynomial equals the exponential at $a_{-}$ and $a_{+}$. Applying the same tangency condition as in eq. 4.73, we impose

These four equations determine the four coefficients. The construction below also verifies that their solution is an upper bound on the whole nonnegative axis.

First construct a cubic $H$ that matches the two values and two slopes. Write $d=a_{+}-a_{-}>0$ and $E_{\pm }=\exp (-s{a}_{\pm })$, and start at the left point:

The two conditions at $a_{-}$ already hold. The conditions at $a_{+}$ are

Solving this two-by-two system gives

Expanding $H(a)=h_0+h_{1}a+h_2{a}^2+h_3{a}^3$, we obtain

The quartic term ${(a-a_{-})}^2{(a-a_{+})}^2$ and its derivative both vanish at the touching points. We can therefore add a multiple of this term without changing any of the four conditions. Choose that multiple to make the value at $0$ equal to $1$:

Using the sum and product of the touching points,

Consequently the coefficients in the original variable $a$ are

All these coefficients depend on the same free parameter $s$ through $E_{\pm }=\exp (-s{a}_{\pm })$; no numerical value of $s$ has been fixed.

The interpolation remainder has one factor for matching the value at $0$ and two factors for matching the value and slope at each positive point. For each $a\ge 0$ away from the touching points, the Hermite remainder is

for some $\xi$ in the interval containing $0,a,a_{-},a_{+}$. At the touching points equality holds by construction. Thus $\exp (-sa)\le F_4(a)$ for every $a\ge 0$ and $s\ge 0$.

We will also need three coefficient inequalities. First, applying the cubic remainder to $H$ at $0$ gives

so $D_4\ge 0$. The numerator in eq. 4.74 was shown to be nonnegative. Applying this with $x=sd\ge 0$, the cubic coefficient satisfies

Therefore

Finally, subtract the endpoint derivatives of the cubic:

Substituting eq. A.4 gives

Each inequality holds throughout $s\ge 0$, including $s=0$ when $F_4$ is identically $1$.

Substitution of the exact moments gives

where

The coefficient inequalities proved above imply

Hence

We use two inequalities between the coefficient sums. Term by term,

and consequently $2{\Vert \bar{w}\Vert }_2^2{\Vert \bar{w}\Vert }_6^6\le {\Vert \bar{w}\Vert }_2^4{\Vert \bar{w}\Vert }_4^4+{\Vert \bar{w}\Vert }_8^8$. Also the weights $p_j=w_j^2/{\Vert \bar{w}\Vert }_2^2$ sum to $1$. Jensen’s inequality for the cube function on nonnegative inputs gives

Thus ${\Vert \bar{w}\Vert }_8^8\ge {\Vert \bar{w}\Vert }_4^{12}/{\Vert \bar{w}\Vert }_2^4$. Since $D_4\ge 0$, these two inequalities preserve the upper-bound direction. Writing $\rho ={\Vert \bar{w}\Vert }_4^4/{\Vert \bar{w}\Vert }_2^4$, we obtain

This proves the comparison with the moment terms depending only on ${\Vert \bar{w}\Vert }_2$, uniformly in $\bar{w}$ and throughout $s\ge 0$.

Since $F_4(0)=1$ and $F_4(a_{\pm })=\exp (-s{a}_{\pm })$, eq. A.2 now yields

This holds for every $t\ge 0$. The moment matching minimizes the expression $1-A_4{m}_1+B_4{m}_2-C_4{m}_3+D_4{m}_4$ among degree-$4$ polynomial upper bounds. The separate argument for $\mathrm{\Delta }\le 0$ is what permits its use for the actual binomial variables.

Substituting eq. A.7 into eq. 4.69, and using $b\le {\Vert \bar{w}\Vert }_2$, bounds the small-coordinate probability by

We minimize this expression including its exponential prefactor. Put $u=(5-\sqrt{10})/2$ and $v=(5+\sqrt{10})/2$, so $L(t)={\lambda }_0+{\lambda }_{-}\exp (-ut)+{\lambda }_{+}\exp (-vt)$. Since $L(t)>0$, the sign of ${\mathrm{\Phi }}^{\prime }(t)$ is the sign of

Both $u,v>1/16$, so

Moreover $G(0)=1-16/2=-7$ and ${lim}_{t\to \mathrm{\infty }}G(t)=8/15>0$. Thus $G$ has exactly one zero, which is the unique global minimum of $\mathrm{\Phi }$ on the nonnegative axis. Numerically,

The rational choice $t=53/20$ gives essentially the same value:

For a certified inequality, rational interval bounds for $\sqrt{10}$ and the exponential give

The numerical minimum remains slightly above $2^{-145}$, and satisfies the target in lemma 4.14.10.