Optimizing p for Asymmetric Ramsey Lower Bounds

Two Competing Obstructions

To show that the Ramsey number \(R(s,t)\) is large, it is enough to construct a graph with no clique of size \(s\) and no independent set of size \(t\). A random graph offers two competing ways to fail: its edges can create an \(s\)-clique, while its nonedges can create an independent \(t\)-set. The edge probability \(p\) controls both risks, but in opposite directions. Choosing \(p\) well is therefore central to the asymmetric first-moment argument.

Throughout, graphs are finite and simple. For integers \(s,t\ge 2\), \(R(s,t)\) is the least integer \(N\) such that every graph on \(N\) vertices contains a clique of size \(s\) or an independent set of size \(t\). We use \(G(n,p)\): each of the \(\binom n2\) possible edges is included independently with probability \(p\), where \(0\le p\le1\). We will choose \(n\ge\max(s,t)\), so both types of obstruction can occur.

For any fixed set of \(s\) vertices, all \(\binom{s}{2}\) required edges must be present. For any fixed set of \(t\) vertices, all \(\binom{t}{2}\) possible internal edges must be absent. Independence of edge choices therefore gives the exact expectations

The asymmetry is visible in both the binomial coefficients and the exponents. Increasing \(p\) raises the expected clique count and lowers the expected independent-set count. A probability that is effective for one obstruction may be poor for the other.

The First-Moment Criterion and the Balancing Choice

Proof. Put \(Z=X+Y\), the total number of obstructions in \(G(n,p)\). Since \(X,Y\) are nonnegative integer-valued, \(Z\ge1\) whenever \(Z\ne0\). Thus \(\mathbb{E}Z\ge\Pr(Z\ge1)\). The stated hypothesis and linearity of expectation give \(\mathbb{E}Z=\mathbb{E}X+\mathbb{E}Y<1\), so \(\Pr(Z\ge1)<1\). Consequently \(\Pr(Z=0)>0\): there is a graph on \(n\) vertices with neither obstruction. By the definition of \(R(s,t)\), this implies \(R(s,t)>n\). \(\square\)

Write \(a=\binom{s}{2}\), \(b=\binom{t}{2}\), \(A=\binom ns\), and \(B=\binom nt\). The two expected counts are \(f(p)=Ap^a\) and \(g(p)=B(1-p)^b\). A natural balancing rule is to choose \(p\) so that \(f(p)=g(p)\).

Proof. The function \(f(p)=Ap^a\) is continuous and strictly increasing on \([0,1]\), while \(g(p)=B(1-p)^b\) is continuous and strictly decreasing there. At \(p=0\), \(f(0)=0< B=g(0)\); at \(p=1\), \(g(1)=0<A=f(1)\). The intermediate value theorem gives an intersection, and strict monotonicity makes it unique. Let its common value be \(q\). If \(p<p_*\), then \(g(p)>g(p_*)=q\); if \(p>p_*\), then \(f(p)>f(p_*)=q\). At \(p_*\), the larger of the two terms equals \(q\). Hence no other choice makes their maximum smaller. \(\square\)

The equation defines a specific choice of \(p\); for general \(s,t,n\) it need not have a simple closed form. It can be solved numerically by bisection, since the difference \(f(p)-g(p)\) is strictly increasing and changes sign between 0 and 1. Equalizing the terms minimizes their maximum, not necessarily their sum. For the first-moment criterion, the equalized choice succeeds whenever \(2q<1\). More generally, any convenient \(p\) works if the sum itself is less than 1.

When \(s=t\), the two coefficients and exponents agree, so symmetry gives \(p_*=\tfrac12\). For unequal parameters, the optimal balance usually shifts away from \(\tfrac12\): the appropriate direction and amount depend on both the numbers of candidate vertex sets and the edge requirements within each set.

Worked Examples

How to Use the Balance in Practice

For fixed \(n,s,t\), a systematic choice is to compare \(f(0)=0\) with \(g(0)=B\), and \(f(1)=A\) with \(g(1)=0\), then locate the unique crossing \(p_*\). Evaluate the common expectation \(q\). If \(2q<1\), the first-moment criterion applies immediately. If not, the criterion has not proved the desired lower bound; it does not show that no suitable graph exists. One can test another \(n\), use a different probability, or seek a stronger method.

For hand calculations, an exact root is not always necessary. A rational trial value for \(p\) may make one expectation easy to compute and the other easy to bound. The examples used powers such as \((4/5)^{28}\) and bounded them by smaller powers. Since the base lies strictly between 0 and 1, raising it to a larger exponent makes the value strictly smaller. Such a bound is useful only when the resulting sum still lies strictly below 1.

A common mistake is to assign probability \(p\) to the absence of an edge. In \(G(n,p)\), an edge is present with probability \(p\), so an edge is absent with probability \(1-p\). For example, in the first worked example the independent 8-set probability is \((4/5)^{28}\), not \((1/5)^{28}\). The latter would correspond to a different random graph model. Always count the required present edges for a clique and the required absent edges for an independent set separately.

The method matters because it turns a parameter choice into a proof rather than a guess. The two expected counts express the competing structural risks, and a successful choice makes the expected total number of obstructions less than one. The inequality is strict for a reason: it forces positive probability that the integer-valued obstruction count is zero.