For a nonnegative integer-valued random variable, pointwise. Taking expectations gives
This is the indicator form of the Markov inequality.
Let be the indicator random variable that the unordered vertex pair has no common neighbour. For each of the other vertices, the two required edges are both present with probability . The edge pairs for different candidate neighbours are disjoint sets of independent random choices. Thus . Linearity of expectation yields
the missing common-neighbour count in a binomial random graph.
If , every pair of vertices has a common neighbour, so every pair has graph distance at most two. In particular the graph is connected and its graph diameter is at most two. By the Markov inequality and part (ii),
We include disconnected graphs among those with diameter greater than two. Therefore
by diameter bound for a fixed-density binomial random graph.

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