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Let $G=(V,E)$ be a finite undirected simple graph, with $n=|V|$ vertices and $m=|E|$ edges. A cut of a graph assigns each vertex a label zero or one; an edge crosses precisely when its endpoints have different labels. Its size is the number of crossing edges. A maximum cut maximizes that size.

For the analysis, assign independent fair bits $X_v$ to the vertices. For each edge $uv$, the indicator random variable $I_{uv}=1_{\{X_u\ne X_v\}}$ has expected value $1/2$: exactly two of the four equally likely endpoint assignments cross. Thus linearity of expectation gives the random cut lower bound:Linearity of expectation does not require the indicator random variables to be independent. What matters here is the stipulated distribution of the endpoint bits.

The method of conditional probabilities turns an expected value guarantee into a deterministic construction. For a maximization objective function, fix one choice at a time, always keeping the conditional expectation of the final value at least as large as before. Once every choice is fixed, that conditional expectation is the actual value. The useful ingredient is a way to evaluate the conditional expectations efficiently.

No actual random choices are needed during this construction. The probability distribution describes hypothetical completions used in the analysis; the algorithm compares their conditional expectations and commits deterministically. For maximum cut, this converts the random cut lower bound into a cut of a graph containing at least half the edges. A maximum cut can still be larger, as the conditional expectations cut example shows.

For each edge $e$, let $I_e$ be the indicator random variable of the event that precisely one endpoint lies in $U$. Independent inclusion of the two endpoints with probability $1/2$ givesBecause $X={\sum }_{e\in E(G)}{I}_e$, linearity of expectation givesSome realization therefore has $X\ge m/2$. Retaining its crossing edges gives a subgraph that is bipartite, with vertex classes $U$ and $V(G)\setminus U$, proving the random cut lower bound.

Now suppose $n$ is an even number and instead choose $U$ uniformly among all subsets of size $n/2$. For a fixed edge $uv$,and the reverse orientation has the same probability. ThusAnother application of linearity of expectation givesAt least one such balanced choice attains this expectation, and its crossing edges give the required subgraph that is bipartite. This is the balanced random cut lower bound.