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Applying Probability to Ramsey Theory

The following theorem, due to P. Erdős, is a true classic, and is presented here in a manner that is faithful to how it was first published.

Applying Probability to Ramsey Theory

The following theorem, due to P. Erdős, is a true classic, and is presented here in a manner that is faithful to how it was first published. As we shall see later, it was subsequently recastbut that's getting the cart ahead of the horse.

Theorem

If\(n\) is a positive integer. Then \[\begin{aligned}\end{aligned}\]

Proof

Let \(t\) be an integer with \(t>n\) and consider the set \(\cgF\) of all labeled graphs with vertex set \(\{1,2,\dots,t\}\). Clearly, there are \(2^{C(t,2)}\) graphs in this family. Let \(\cgF_1\) denote the subfamily consisting of those graphs which contain a complete subgraph of size\(n\). It is easy to see that \[\begin{aligned}\end{aligned}\] Similarly, let \(\cgF_2\) denote the subfamily consisting of those graphs which contain an independent set of size\(n\). It follows that \[\begin{aligned}\end{aligned}\]

We want to take the integer \(t\) as large as we can while still guaranteeing that \(|\cgF_1|+|\cgF_2|\le |\cgF|\). This will imply that there is a graph \(G\) in \(\cgF\) which does not contain a complete subgraph of size \(n\) or an independent set of size\(n\). So consider the following inequality: \[\begin{aligned}\end{aligned}\] Now we ask how large can \(t\) be without violating inequality? To answer this, we use the trivial inequality \(\binom{t}{n}\le t^n/n!\) and the use the Stirling approximation for \(n!\). After some algebra and taking the \(n^{\text{th} }\) root of both sides, we see that we need only guarantee that \[\begin{aligned}\end{aligned}\]

Now let's take a second look at the proof of . We consider a probability space \((S,P)\) where the outcomes are graphs with vertex set \(\{1,2,\dots,t\}\). For each \(i\) and \(j\) with \(1\le i \lt j\le t\), edge \(ij\) is present in the graph with probability \(1/2\). Furthermore, the events for distinct pairs are independent.

Let \(X_1\) denote the random variable which counts the number of \(n\)-element subsets of \(\{1,2,\dots,t\}\) for which all \(\binom{n}{2}\) pairs are edges in the graph. Similarly, \(X_2\) is the random variable which counts the number of \(n\)-element independent subsets of \(\{1,2,\dots,t\}\). Then set \(X=X_1+X_2\).

By linearity of expectation, \(E(X)=E(X_1)+E(X_2)\) while \[\begin{aligned}\end{aligned}\] If \(E(X)\lt 1\), then there must exist a graph with vertex set \(\{1,2,\dots,t\}\) without a \(K_n\) or an \(I_n\). And the question of how large \(t\) can be while maintaining \(E(X)\lt 1\) leads to exactly the same calculation we had before.

Condensed: the full section is in Keller & Trotter, Applied Combinatorics.

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Ερωτήσεις που κάνουν οι άνθρωποι

What is the difference between probability and statistics?

Probability goes from a known model to what the data should look like; statistics goes from data back to the model. Probability theory is the deductive half.

What does the law of large numbers promise?

That the average of many independent samples converges to the expected value. It says nothing about any single trial.

Μέρη αυτής της σελίδας προσαρμόζονται από Keller & Trotter, Applied Combinatorics (CC BY-SA 4.0). Πυκνωμένα και ανεξήγητα εδώ· τα λάθη είναι δικά μας.

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