Random profile lifts

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This gives a probabilistic counterpart to Convex profile amplification. It does not provide an unrestricted amplifier or prove convergence of $f(k,r)/k$.

Uniform exponential bound

Let $\mathcal H$ be a finite nonempty $P_r$ family of nonempty sets on $n$ points, where $r\ge3$, and put

$$P=\operatorname{conv}\{\mathbf1_E:E\in\mathcal H\}.$$

Replace each point by a block of $L$ points. For prescribed profiles $p^{(1)},\ldots,p^{(r)}\in P$ with integral $Lp^{(j)}$, independently realize each row by uniformly choosing $Lp^{(j)}_v$ points in block $v$, independently between blocks. With $\delta=1/(n+1)$,

$$\Pr(\text{the rows have no two-cover}) \le2(L+1)^{r-1}(1-\delta^{r-1})^L. \tag{1}$$

Caratheodory's theorem expresses each profile using at most $n+1$ seed edges. Some edge $E_j$ has coefficient at least $\delta$, so $p^{(j)}_v\ge\delta$ on $E_j$. A seed two-cover partitions the rows into two nonempty groups of sizes at most $r-1$, with a designated point common to each group's seed edges. If one seed point covers every row, split the rows into groups of sizes $r-1$ and one using the same designated block.

For $s$ independent uniform subsets of an $L$-point block, of integral densities $p_i\ge\delta$, the probability of empty common intersection is at most

$$(L+1)^s(1-\textstyle\prod_i p_i)^L \le (L+1)^s(1-\delta^s)^L.$$

Indeed, first use independent Bernoulli subsets. Their intersection is empty with probability $(1-\prod_i p_i)^L$. Conditioning each cardinality on its integral mean gives the uniform model. That cardinality is a mode of its binomial distribution, and has probability at least $1/(L+1)$. Dividing by the conditioning probability proves the bound. Apply this to the two designated groups and take a union bound to prove (1). The same proof handles any number of rows from two through $r$.

A sufficient alteration inequality

Suppose, more generally, that $e^{o(L)}$ allowed profiles on $nL$ points satisfy the following uniform estimates, for constants $s>0$, $B\ge0$, and $I>0$:

• every point set of size at most $sL$ is avoided by a random realization of some profile with probability at least $e^{-(B+o(1))L}$;
• independently sampled realizations of any prescribed $r$ profiles form a bad tuple with probability at most $e^{-(I-o(1))L}$.

If

$$I>rB, \tag{2}$$

there are $P_r$ lifted families with transversal number greater than $\lfloor sL\rfloor$ for all sufficiently large $L$.

Choose $B<\alpha<(I-B)/(r-1)$ and sample $\lceil e^{\alpha L}\rceil$ rows per profile. For each candidate transversal, its designated profile supplies at least $e^{(\alpha-B-o(1))L}$ avoiding rows, simultaneously for all candidates: Chernoff's inequality gives a doubly exponential failure estimate, which beats their number $2^{nL}$. Meanwhile the expected number of bad $r$-tuples of distinct sample positions is at most $e^{(r\alpha-I+o(1))L}$. Markov's inequality, with any sufficiently small fixed exponential slack, bounds their number by less than the preceding avoidance count. Delete every sampled row participating in a bad tuple. The strict inequality $r\alpha-I<\alpha-B$ ensures that each candidate transversal still misses a row. A smaller bad subfamily among the remaining rows would extend to a bad $r$-tuple, since the surviving transversal number tends to infinity. Removing duplicate rows is harmless.

Why this is not yet an amplifier

The positive rate in (1) need not satisfy (2) at the desired transversal scale. At the ideal homogeneous Fano boundary $p=4/7$, deleting $3L/7$ points in each of $n$ blocks gives avoidance exponent

$$B=n h(4/7),\qquad h(u)=-u\log u-(1-u)\log(1-u).$$

For $7\mid L$, balanced assignments of the seven Fano caps in every block give bad tuples with exponent

$$I\le n\bigl(7h(4/7)-\log7\bigr)<7B.$$

This calculation concerns the limiting homogeneous model, not a claim that a particular fixed $P_7$ seed contains that exact boundary profile. It shows why uniform exponential rarity alone does not recover the boundary coefficient by this alteration argument. It is consistent with the separate Bernoulli thinning obstruction, and does not exclude more structured sampling or deletion methods.