Independence and product probabilities
Recognize independence as factorization of measurable information, build product laws, and distinguish mutual independence from pairwise tests.
On this page
The question for today
What you’ll understand
- Test independence through measurable rectangles in a joint distribution.
- Construct a pairwise-independent triple that is not mutually independent.
- Use product integration while keeping its integrability hypotheses visible.
Independence states that learning one collection of events does not change the probabilities assigned to another collection. It is a property of a joint model, not a property of individual histograms. In practice, independence assumptions often enter before any calculation and control whether count models, variance formulas, and asymptotic arguments are valid. A careful treatment therefore tests entire collections of observable events rather than relying on zero correlation or on a small number of favorable numerical coincidences.
Factorization of information
Two events are independent when their intersection probability is the product of their separate probabilities. Two random variables are independent when this factorization holds for every pair of measurable output sets. Equivalently their generated sigma-algebras are independent. For real variables, checking a suitable generating family such as pairs of lower half-lines is enough, with a measure-uniqueness argument extending the result. Checking just one pair of events is generally insufficient to establish independence of the full observations.
If an event B has positive probability, event independence is equivalent to the conditional probability of A given B being its unconditional probability. The product definition remains meaningful when B has probability zero, while the elementary ratio definition does not. This is one reason to use factorization as the foundation. Also independence is stable under measurable transformations performed separately on the variables: output questions about transformed values simply pull back to questions that were already independent.
Product measures and expectation factorization
The product of two probability measures assigns each measurable rectangle the product of its side masses and extends uniquely to the product sigma-algebra. On that product space the coordinate observations are independent by construction. For a nonnegative measurable function, Tonelli permits iterated integration without first proving a finite integral. For a signed function, Fubini's usual finite-valued conclusion needs absolute integrability. These theorems justify replacing a joint integral by successive averages; they do not license rearranging undefined differences of infinities.
For independent integrable real variables, the product is integrable and its expectation factors. Indeed the product integral of the absolute values is the product of their finite absolute expectations. For square-integrable variables this implies zero covariance. The converse fails: covariance tests only a particular pair of functions, while independence tests all measurable events. A symmetric variable and its square provide a familiar example: the covariance can vanish by symmetry even though the square is determined completely by the original variable.
Pairs are weaker than a whole family
A family is mutually independent when every finite subfamily and every choice of one event from each corresponding sigma-algebra satisfy the full product identity. Pairwise independence asks only about subfamilies of size two. These requirements agree for two variables but differ as soon as a third variable enters. To see the distinction, use two independent fair bits and let a third record their parity. Each pair looks independent, while all three together satisfy a deterministic relation.
The distinction has practical consequences. Pairwise independence suffices to eliminate covariance terms in the variance of a finite square-integrable sum. It does not justify multiplying an arbitrary collection of event probabilities or treating all coordinates as a product measure. Some limit theorems admit pairwise-independent variants, but such extensions need their own hypotheses and proofs. When a calculation invokes an independent sequence, interpret the default assumption as mutual independence of every finite subcollection, rather than merely an absence of visible pairwise associations.
Work it through
A parity bit hides a three-way constraint
Let A and B be independent fair bits, and C their sum modulo two. Check pairwise independence and compare the probability that all three equal zero with the mutually independent prediction.
The four equally likely triples are (0,0,0), (0,1,1), (1,0,1), and (1,1,0). Each coordinate is zero in two of the four triples, so all three marginals are fair.
For any pair of coordinates, the four possible bit pairs each occur once. Every pair therefore has the uniform product law, establishing pairwise independence.
The all-zero triple occurs with probability 1/4. Mutual independence of three fair bits would instead assign (1/2)³=1/8. The mismatch disproves mutual independence.
The three bits are pairwise independent but not mutually independent; the deterministic parity constraint survives every pairwise test.
The living notebook
Explore the model
Interactive lab · probability
Inspect the assumption behind binomial bars
Set n=3 and p=0.5. Predict the all-success bar as 1/8 before checking it. This binomial law assumes three mutually independent trials with the same probability. Compare that product prediction with the parity-triple calculation: matching every pair does not reproduce the full three-coordinate law.
- Mean · np
- 5
- Variance · np(1−p)
- 2.5
- Total probability
- 1
These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forThe exact binomial panel builds independence into its formula. Its bars cannot test whether an external sequence of trials actually satisfies that assumption.
Pause & reason
Check your understanding
Your turn
Try it yourself
Let X be uniform on {−1,0,1}, and Y=X². Compute E[X], E[Y], and E[XY]. Show that zero covariance here does not imply independence.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Use symmetry for the odd moments, then compare the event X=0 with the event Y=0.
Compare with the solution
The means are E[X]=0 and E[Y]=2/3. Since XY=X³, symmetry gives E[XY]=0, hence covariance zero. But P(X=0,Y=0)=1/3 while P(X=0)P(Y=0)=1/9. These probabilities differ, so independence fails. The square determines whether X is zero even though the linear covariance cannot detect that information.
Keep the idea
Take it with you
- Independence factorizes a whole joint measure, not merely a covariance.
- Product integration requires nonnegativity or appropriate absolute integrability.
- Pairwise independence does not generally imply mutual independence.
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