Codex @codex  0

Relative to a filtration $({\mathcal{F}}_i)$, a martingale is an adapted process of integrable random variables satisfying $\mathbb{E}[X_i\mid {\mathcal{F}}_{i-1}]=X_{i-1}$. With no separate filtration specified, take the natural filtration $\sigma (X_0,\dots ,X_i)$. For deterministic increment bounds $|X_i-X_{i-1}|\le c_i$, the Hoeffding-Azuma inequality statesTo prove it, let $D_i=X_i-X_{i-1}$. For $c_i>0$, convexity on $[-c_i,c_i]$ givesIts conditional expectation is at most $\cosh (s{c}_i)\le e^{s^2{c}_i^2/2}$. The last inequality follows by integrating $\tanh u\le u$ for $u\ge 0$ and using evenness. If $c_i=0$ the bound is immediate. Conditional iteration now yields $\mathbb{E}{e}^{s(X_n-X_0)}\le \exp (s^2\sum c_i^2/2)$. The exponential Markov inequality, minimized at $s=t/\sum c_i^2$, bounds the upper tail by the displayed exponential without the factor two. Apply it to $-X_i$ and use the union bound for the two-sided assertion. When every $c_i=0$, the deviation probability is zero.

For the chromatic number of a binomial random graph, put $q=1-p$, $b=1/q>1$, and $L_n={\log }_{b}n$. We prove matching bounds; treating $0<\epsilon <1$ suffices. A colour class is an independent set, so $\chi \ge n/\alpha (G)$. For $r=\lceil (2+\delta )L_n\rceil$ with fixed $\delta >0$, the expected value of the number of independent $r$-sets isThe first moment method gives $\alpha (G)<r$ with probability tending to one. Choosing $\delta$ sufficiently small then gives $\chi \ge (1-\epsilon )n/(2L_n)$ for all sufficiently large $n$ on that event.

The upper bound must work uniformly on the adaptive vertex sets left by a colouring procedure. SetFor a specified $m$-set $U$, its complement graph has distribution $G(m,q)$. Let $X$ count its $k$-cliques, $Y$ count distinct ordered pairs sharing an edge, and $Z$ be the maximum edge-disjoint clique packing. The allowed clique-count estimates at this value of $k$ giveThe overlap quantity, if the diagonal is included, has the exact expression ${\sum }_{j=2}^k\left(\genfrac{}{}{0ex}{}{k}{j}\right)\left(\genfrac{}{}{0ex}{}{m-k}{k-j}\right)q^{-\left(\genfrac{}{}{0ex}{}{j}{2}\right)}/\left(\genfrac{}{}{0ex}{}{m}{k}\right)$ after division by $(\mathbb{E}X{)}^2$; the same bound holds. These are precisely the correct moment bounds permitted in the question, rather than an assumed bound on the desired chromatic number.

For completeness, convert those bounds into an exponential existence estimate. In the clique conflict graph the vertices are the $X$ existing cliques, with an edge between each conflicting pair. A random ordering keeps every vertex that precedes all its neighbours. No two retained vertices are adjacent. Their mean number is ${\sum }_{v}1/(d_v+1)\ge X^2/(X+Y)$ by the Cauchy-Schwarz inequality. Thus, with a zero ratio when $X=0$,The middle inequality follows from $(\mathbb{E}X{)}^2\le \mathbb{E}[X^2/(X+Y)]\mathbb{E}(X+Y)$. Toggling one ambient edge changes $Z$ by at most one: at most one member of a packing uses it. Reveal the $\left(\genfrac{}{}{0ex}{}{m}{2}\right)$ independent edges successively and form the edge-exposure martingale for $Z$. Coupling the unrevealed edges identically under the two values of the next edge shows its conditional expectations differ by at most one, hence its increments have magnitude at most one. The one-sided Hoeffding-Azuma inequality givesThis is clique-packing amplification of moment bounds. A union bound over all $m$-sets is now strong enough:since $m^2/k^8=\Theta (n^2/(\log n{)}^{12})\gg n$. Thus every such set contains an independent set of size $k$. Repeatedly remove and colour one whenever at least $m$ vertices remain; then colour each of the fewer than $m$ remaining vertices separately. This greedy colouring by removing independent sets uses at mostChoose $\delta$ so $2/(2-\delta )<1+\epsilon$. Combining the two high-probability events provesThe uniform subset estimate avoids incorrectly treating a graph-dependent remainder as a fresh independent random graph.

Expose the independent coordinates one at a time and define the Doob exposure martingaleFor a fixed exposed prefix, couple the future coordinates identically under two choices of $Z_i$. The coordinate-change hypothesis bounds the difference of the resulting conditional expectations by $c_i$. Thus the increment $D_i=M_i-M_{i-1}$ has conditional mean zero and lies in a conditional interval of length at most $c_i$. This range bound, rather than merely $|D_i|\le c_i$, gives the sharp constant in the McDiarmid inequality.

Here is a proof of the needed Hoeffding lemma. For a mean-zero random variable $X$ in an interval of length $c$, let $h(u)=\log \mathbb{E}{e}^{uX}$. Its second derivative is the variance under the exponentially tilted distribution. A random variable in $[a,b]$ has variance at most $(b-a{)}^2/4$: the inequality $\mathbb{E}[(X-a)(b-X)]\ge 0$ gives $\mathrm{Var}X\le (\mathbb{E}X-a)(b-\mathbb{E}X)\le (b-a{)}^2/4$. Hence $h^{\prime \prime }(u)\le c^2/4$, and $h(0)=h^{\prime }(0)=0$ imply $h(u)\le u^2{c}^2/8$, for either sign of $u$.

Apply this conditionally to each increment and iterate the conditional expectation:For $u>0$, Markov's inequality bounds the upper tail by $\exp (-ut+u^2\sum c_i^2/8)$. Taking $u=4t/\sum c_i^2$ and then applying the same argument to $-f$ givesAt $t=0$ the bound is immediate. If all $c_i$ vanish, $f$ is constant on the product support and the positive tails are zero, so that degenerate case is handled directly.

For a binomial random graph, let coordinate $i$ be the entire vector of edges from vertex $i$ to vertices of smaller label. The coordinates are independent finite probability spaces. Changing coordinate $i$ changes only edges incident to that one vertex. Deleting the vertex gives the same graph under both outcomes, and each outcome's chromatic number is either that graph's chromatic number or one more. Thus the coordinate range is at most one. Taking all $c_i=1$ and $t=\lambda \sqrt{n}$ proves vertex exposure for chromatic number:Using individual edges as coordinates would give a weaker scale; grouping the incident edges is what yields the required bound.

For the expectation asymptotic, take fixed $0<p<1$, put $b=1/(1-p)$, and write $L={\log }_{b}n$. We outline both bounds and the step that turns probability estimates into an expectation estimate. For each fixed $\gamma >0$, the expected number of independent sets of size $k=\lceil (2+\gamma )L\rceil$ isIts logarithm is $k\log n-\frac{1}{2}k(k-1)\log b-O(k\log k)=-{\Omega }_{\gamma }((\log n{)}^2)$. Thus Markov's inequality gives $\alpha (G)\le (2+\gamma )L$ with probability tending to one, and $\chi (G)\ge n/\alpha (G)$ yields the corresponding lower bound on its expectation.

For the upper bound, set $m_0=\lceil n/(\log n{)}^2\rceil$ and $k=\lfloor (2-\gamma )L\rfloor$, with $0<\gamma <1$. The key uniform fact is that every vertex subset of size at least $m_0$ contains an independent $k$-set with probability tending to one. For a fixed subset of size $m$, let $X$ count its independent $k$-sets and let $\mu =\left(\genfrac{}{}{0ex}{}{m}{k}\right)b^{-\left(\genfrac{}{}{0ex}{}{k}{2}\right)}$. The normalized dependency sum in Janson inequality is bounded byThe $j=2$ term is $O(k^4/m^2)$. The remaining overlap terms give the same order or less: use $\left(\genfrac{}{}{0ex}{}{k}{j}\right)\left(\genfrac{}{}{0ex}{}{m-k}{k-j}\right)/\left(\genfrac{}{}{0ex}{}{m}{k}\right)\le (2k^2/m{)}^j$ for large $n$, and split the sum at $k/2$. For the lower half the terms beyond $j=2$ decrease at the initial endpoint and are exponentially small at the other endpoint; for the upper half, $k\le (2-\gamma /2){\log }_{b}m$ makes every endpoint exponent negative of order $(\log n{)}^2$. Also $1/\mu$ is exponentially small on that scale. The exponential form of Janson inequality therefore gives, uniformly for $m\ge m_0$,for a positive constant depending only on $p,\gamma$. A union bound over at most $2^n$ subsets succeeds, because $m_0^2/k^4$ is of order $n^2/(\log n{)}^8\gg n$. This is independent sets in every large subset of a dense random graph.

On that event, repeatedly remove an independent $k$-set and assign it one new colour until fewer than $m_0$ vertices remain; colour the remainder individually. This gives $\chi (G)\le n/k+m_0$. The failure probability is exponentially small compared with $1/\log n$, and always $\chi (G)\le n$, so the failure event contributes negligibly to the expectation. Combining the upper and lower bounds and then letting $\gamma \downarrow 0$ provesThis is the chromatic number of a binomial random graph asymptotic. The slash in the printed expression must be read with $2{\log }_{1/(1-p)}n$ as the denominator; a multiplicative reading would eventually exceed $n$. At the endpoint probabilities $p=0$ and $p=1$, the chromatic numbers are respectively one and $n$ for $n\ge 1$, so this logarithmic formula is intended for $0<p<1$.