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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 $0<p<1$.