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Write $p=p(n)$ and let $T$ be the triangle count in a binomial random graph. For every three-element set $S$ of vertices, let $I_S$ be the indicator random variable that its three edges are present. ThenFor two distinct triples $S,R$, their indicator random variables are independent random variables unless they share an edge. Sharing just one vertex still leaves the sets of relevant edges disjoint. If they share an edge, their union has five edges, and thereforeThere are $\left(\genfrac{}{}{0ex}{}{n}{2}\right)\left(\genfrac{}{}{0ex}{}{n-2}{2}\right)$ unordered pairs of distinct triangles in a graph sharing an edge: choose that edge, and then choose their two additional vertices. The variance of a sum consequently gives the exact formulaFor sufficiently large $n$, the hypothesis $np\to \infty$ ensures $p>0$, so $\mu >0$. Dropping negative terms and simplifying the binomial coefficients,Here $n^{2}p=n(np)\to \infty$ as well. The event $T\ge 2\mu$ implies $|T-\mu |\ge \mu$. Applying the Chebyshev inequality from part (i) gives the requested conclusion:Thus the triangle count is concentrated around its expectation throughout this range of the binomial random graph.

For each unordered triple of distinct vertices, let $I_A$ be the indicator variable that its three edges are present. This is a binomial random graph, so edge independence gives $\mathbb{E}[I_A]=p^3$. The triangle count in a binomial random graph is $T={\sum }_A{I}_A$. There are $\left(\genfrac{}{}{0ex}{}{n}{3}\right)$ such triples; all are triangles in the complete graph.

The combinatorial maximum and expectation areThe maximum is attainable with positive probability if $p>0$; for $p=0$, the random variable is identically zero. The linearity of expectation requires no independence between different triangle indicators.

The Markov inequality states that for a nonnegative random variable $Z$ and $a>0$,Because $T$ is integer-valued, apply it with $a=1$. With $p=n^{-\alpha }$,ThereforeThis is the first moment method: if the expected triangle count in a binomial random graph tends to zero, triangles occur with vanishing probability.

Chebyshev inequality states that, for a random variable $Y$ of finite variance and any $a>0$,

Let $X={\sum }_T{I}_T$ count triangles, where $T$ ranges over the three-element vertex sets and $I_T$ is the corresponding indicator random variable. ThenTwo distinct triangle indicators are independent unless their triangles share an edge. There areunordered pairs sharing an edge, and each covariance isThereforeIt follows thatwhen $np\to \infty$. Chebyshev's inequality now givesHenceas summarized by the triangle count in a binomial random graph.

Markov inequality says that a nonnegative random variable $Y$ satisfiesChebyshev inequality says that a random variable with finite variance satisfies

Let $X$ count triangles in $G(n,p)$. The triangle count in a binomial random graph calculation givesSince $p\gg n^{-1}$,For sufficiently large $n$, $\mu >200$, and Chebyshev inequality gives