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Finite Bernoulli difference laws and parity of moments

Statement

Let $0 \leq p \leq 1$ and let $B, B^{'}$ be independent Bernoulli $(p)$ random variables on a finite probability space. Put $W = B - B^{'}$ . Then $P (W = 1) = P (W = - 1) = p (1 - p)$ and $P (W = 0) = p^{2} + (1 - p)^{2}$ . For every positive integer $k$ , $E [W^{k}]$ is zero if $k$ is odd and $2 p (1 - p)$ if $k$ is even; $E [W^{0}] = 1$ , with $W^{0}$ the constant one.

More generally, let $E$ be a finite set of unordered edges, meaning two-element subsets of a finite vertex set, and suppose that the family $(B_{e}, B_{e}^{'})_{e \in E}$ of all its Bernoulli $(p)$ coordinates is mutually independent. Then the differences $W_{e} = B_{e} - B_{e}^{'}$ are mutually independent. This hypothesis holds for the coordinate variables on the finite product of two Bernoulli spaces per edge. Defining $W_{i j} = W_{{i, j}}$ for distinct vertices makes $W_{i j} = W_{j i}$ : reflection is the same random variable, not another independent copy.

Statement references

Bernoulli random variables

Mutual independence of finite-valued random variables

Finite product probability spaces

Finite expectation

Finite probability spaces and event probabilities

Finite real random variables and their distributions

Facts & Assumptions

[F1]

By Bernoulli random variables and binomial random variables as sums of independent Bernoulli trials, a Bernoulli $(p)$ variable has probabilities $p$ and $1 - p$ at $1$ and $0$ , respectively.

[F2]

By Pairwise and mutual independence of finite-valued random variables, mutual independence means that every simultaneous assignment to any nonempty subfamily has probability equal to the product of its coordinate probabilities.

[F3]

By The finite product of finite probability spaces, a product outcome has weight equal to the product of its coordinate weights; the empty product has weight one.

[F4]

By Expectation of a real random variable on a finite probability space, expectation is the sum of a random variable's values weighted by outcome probabilities.

[F5]

By Finite probability spaces, outcome weights, events, and event probabilities, a finite probability space $(Ω, w)$ has nonnegative outcome weights summing to one, and $P (A) = \sum_{ω \in A} w (ω)$ for each event $A \subseteq Ω$ . Summing over disjoint events therefore sums their probabilities.

[F6]

By Real random variables on finite probability spaces and their finite distributions, a finite real random variable is a function $Ω \to R$ , and its law assigns each value the probability of its fibre. Differences and nonnegative integer powers here are formed pointwise as functions on the same finite outcome set, with zeroth powers defined as the constant one.

Proof

Given: The finite probability spaces and variables in the statement.

Goal: Establish the difference law, its moments, and independence across edge blocks.

1.1F1F2F5F6

By independence, the four assignments $(B, B^{'}) = (1, 0), (0, 1), (1, 1), (0, 0)$ have respective probabilities $p (1 - p), p (1 - p), p^{2}, (1 - p)^{2}$ . Their differences are respectively $1, - 1, 0, 0$ , giving the asserted law, including at $p = 0$ and $p = 1$ .

2.1F4F5F6step 1.1

For $k > 0$ , the weighted sum over these values is $E [W^{k}] = p (1 - p) (1 + (- 1)^{k})$ , since the zero term vanishes. This gives the two parity cases. The zeroth power is the constant one, whose expectation is one by normalization.

2.2F2F5F6step 1.1

For a nonempty $J \subseteq E$ and prescribed values $a_{e} \in {- 1, 0, 1}$ , sum the joint coordinate-assignment probabilities over the disjoint assignments satisfying $b_{e} - b_{e}^{'} = a_{e}$ for each $e \in J$ . Mutual independence factors every summand into its edge-block factors, and finite distributivity factors the sum into $\prod_{e \in J} P (W_{e} = a_{e})$ . Thus the differences are mutually independent; impossible values have probability zero, and the empty family is independent by convention.

3.1F1F2F3F5F6step 2.2∎

On the stated finite product, summing product weights over all unprescribed coordinates gives one for each such coordinate, so every prescribed coordinate assignment has the product of its marginal weights. The coordinates are therefore mutually independent. Finally, ${i, j} = {j, i}$ , so the two reflected entries are literally the same difference variable.

Depends on

Used by · 4

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