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Fundamental BridgeΒΆ

PropositionΒΆ

For any event AA, let πŸ™1A\indicator_A equal one on AA and zero outside AA. The fundamental bridge is

E[πŸ™1A]=P(A).\mathbb{E}[\indicator_A]=\mathbb{P}(A).

The expectation always exists because the indicator is bounded.


InterpretationΒΆ

The fundamental bridge says that the probability of an event equals the expected value of its indicator. It lets every probability be written as an expectation, and every expected count as a sum of probabilities.

Two probability bars of an indicator variable, 1 - P(A) at the value 0 and P(A) at the value 1, balanced at the point P(A).

In the figure P(A)=0.3\mathbb{P}(A)=0.3: the indicator equals 00 with probability 0.70.7, when AA did not occur, and 11 with probability 0.30.3, when it did. Treating the two bars as weights, their balance point, the expectation, lies at 0β‹…0.7+1β‹…0.3=0.3=P(A)0\cdot0.7+1\cdot0.3=0.3=\mathbb{P}(A).

The proof is one line, since πŸ™1A\indicator_A is a Bernoulli variable with parameter P(A)\mathbb{P}(A). The identity becomes useful together with linearity of expectation: if X=πŸ™1A1+β‹―+πŸ™1AnX=\indicator_{A_1}+\cdots+\indicator_{A_n} counts how many of the events occur, then E[X]=P(A1)+β‹―+P(An)\mathbb{E}[X]=\mathbb{P}(A_1)+\cdots+\mathbb{P}(A_n), whether or not the events are independent. For nn tosses with heads probability pp this gives the binomial mean npnp. If nn letters are put into nn envelopes at random, letter ii lands in its own envelope with probability 1/n1/n, so the expected number of correctly placed letters is nβ‹…1n=1n\cdot\tfrac{1}{n}=1 for every nn, although these events are dependent.

The bridge does not say that πŸ™1A\indicator_A equals P(A)\mathbb{P}(A). The indicator only takes the values 00 and 11; P(A)\mathbb{P}(A) is its average value, and it is usually not a value the indicator takes.


Key ResultsΒΆ

  • E[πŸ™1A]=1P(A)+0P(Ac)=P(A)\mathbb{E}[\indicator_A]=1\mathbb{P}(A)+0\mathbb{P}(A^c)=\mathbb{P}(A).
  • E[βˆ‘iπŸ™1Ai]=βˆ‘iP(Ai)\displaystyle \mathbb{E}[\sum_i\indicator_{A_i}]=\sum_i\mathbb{P}(A_i) for finitely many events, without independence.

ExamplesΒΆ

  • For an even result on a fair die, E[πŸ™1A]=1/2\mathbb{E}[\indicator_A]=1/2.
  • For 1010 components each failing with probability 0.020.02, the expected failure count is 0.20.2, even if failures are dependent.

PrerequisitesΒΆ

SourcesΒΆ

Aliases: indicator bridge, expectation of an indicator Keywords: probability-expectation bridge Symbols: sum

Last updated: 20 October 2026 03:02