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Source scope. Published Section 3, p. 272 (PDF), asks for the weighted version of the deletion proof. This page supplies the numerical step, with a parameter depending on the bias.

Statement. Fix $0<p\le1/2$ and put $t=p/(1-p)$. For $0<\delta\le t/40$, nonempty families have one of the following slice choices: either a growth step with product at least $1+\delta$ times the old product, or the widening step $(\mathcal F_1,\mathcal G_0\cap\mathcal G_1)$ with product at least

$$b_{p,\delta}=1-\frac\delta t-\frac{8\delta^2}{t^2}>0$$

times the old product. The growth choices are the two $1$-slices or $(\mathcal F_0,\mathcal G_0\cup\mathcal G_1)$, with the same forbidden interval changes as in definitions. The families may first be interchanged.

Proof. Write $f=\mu_p(\mathcal F)$ and $g=\mu_p(\mathcal G)$; slice measures have one fewer ambient coordinate. If $f_1g_1>(1+\delta)fg$, use the first growth choice. Otherwise interchange the families so $a=f_1/f\le b=g_1/g$. Thus $a\le\sqrt{1+\delta}$. If $f_0u>(1+\delta)fg$, where $u=\mu_p(\mathcal G_0\cup\mathcal G_1)$, use the second growth choice. We bound the remaining case.

Set $y=a-1$, $z=b-1$, and $x=u/g-1$. The slice identity and inclusion-exclusion give

$$\frac{f_0}f=1-ty,\quad \frac{g_0}g=1-tz,\quad \frac{\mu_p(\mathcal G_0\cap\mathcal G_1)}g =1+(1-t)z-x. \tag{1}$$

Since $u\ge\max(g_0,g_1)\ge g$, we have $x\ge0$. Failure of the second growth test implies $1-ty\le1+\delta$, hence $y\ge-\delta/t$. Also $y\le\sqrt{1+\delta}-1\le\delta/2$ and $z\ge y$. The inequalities $u\ge g_1,g_0$ give

$$(1-ty)(1+z)\le1+\delta,\qquad (1-ty)(1-tz)\le1+\delta.$$

Consequently,

$$-\delta/t\le z\le2\delta,\quad 0\le x\le\frac{\delta+ty}{1-ty}\le2\delta,\quad x\le\delta+ty+2\delta^2,\quad y+z\ge-\delta/t+t yz. \tag{2}$$

For the third inequality, subtract $\delta+ty$ from the fraction: if $y\le0$ the difference is nonpositive; if $y>0$ it is at most $(\delta/2)2\delta\le\delta^2$. All denominators are positive under the stated bound on $\delta$.

Using (1) and then (2), the widening product divided by $fg$ is

$$\begin{aligned} (1+y)(1+(1-t)z-x) &\ge1-\delta+(1-t)(y+z)+(1-t)yz-xy-2\delta^2\\ &\ge1-\delta/t+(1-t^2)yz-xy-2\delta^2\\ &\ge1-\delta/t-8\delta^2/t^2. \end{aligned}$$

The last line uses $|y|\le\delta/t$, $|z|\le2\delta/t$, $0\le x\le2\delta$, and $0<t\le1$. This also proves that the chosen product is positive. $\square$

Dependencies. definitions.