CounterexampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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Uncorrelated finite random variables need not be independent

Statement refuted

If $Cov (X, Y) = 0$ , then the finite random variables $X$ and $Y$ are independent.

Facts & Assumptions

Given: The uniform space $Ω = {- 1, 0, 1}$ , the identity variable $X (ω) = ω$ , and $Y = X^{2}$ .

[L1]

Uniform probabilities are cardinality ratios (The uniform probability space on a nonempty finite set).

[L2]

Independence requires every joint attained-value probability to factor (Pairwise and mutual independence of finite-valued random variables).

[L3]

Covariance is defined by centred products and equals $E [X Y] - E [X] E [Y]$ (Variance, standard deviation, and covariance on a finite probability space, $Var (X) = E [X^{2}] - E [X]^{2}$ and $Cov (X, Y) = E [X Y] - E [X] E [Y]$ ).

Counterexample

technique · constructive

1.1

On the stated three-point space, symmetry gives $E [X] = 0$ and $E [X Y] = E [X^{3}] = 0$ , while $E [Y] = 2 / 3$ .

givenL1algebraconstruct

1.2

But $P (X = 0, Y = 0) = 1 / 3$ , whereas $P (X = 0) P (Y = 0) = (1 / 3) (1 / 3) = 1 / 9$ .

L1algebra

2.1

Hence $Cov (X, Y) = 0$ .

step 1.1L3algebra

3.1

Thus $X, Y$ are uncorrelated but not independent.

step 2.1step 1.2L2discharge-construct∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 36 results over 12 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

H. Pishro-Nik, Introduction to Probability, Statistics, and Random Processes, Section 5.3.1 (standard reference, not scraped)