= Independent random variables
{title2=$\mathbb P_{(X_1,\ldots,X_n)}=\mathbb P_{X_1}\otimes\cdots\otimes\mathbb P_{X_n}$}
= Independently
{synonym}
= Independent
{synonym}
= Independent random variable
{synonym}
= Independence
{synonym}
= Independence of random variables
{synonym}
<Random variables> $X_1,\ldots,X_n$ are independent when, for every choice of <Borel sets> $A_1,\ldots,A_n$,
$$
\mathbb P\!\left(\bigcap_{i=1}^n\{X_i\in A_i\}\right)
=\prod_{i=1}^n\mathbb P(X_i\in A_i).
$$
The requirement covers all these <events>, not only one selected combination of values. Equivalently, the generated <sigma-algebras> are <independent sigma-algebras>, or the <joint probability distribution> is the <product measure> of the <marginal distributions>. For an infinite family, the same condition must hold for every finite subfamily. For <random elements>, replace <Borel sets> by <measurable sets> in their specified target spaces.
For <discrete random variables>, checking every entry of the <joint probability mass function> is enough:
$$
p_{X_1,\ldots,X_n}(x_1,\ldots,x_n)=\prod_{i=1}^n p_{X_i}(x_i).
$$
Summing this identity over the desired sets proves the event formulation. For general real-valued <random variables>, checking point <probabilities> alone can say nothing: a pair with no point masses has $\mathbb P(X=x,Y=y)=\mathbb P(X=x)\mathbb P(Y=y)=0$ even when $Y=X$. The all-set definition avoids that trap. See https://www.math.cmu.edu/~gautam/c/2026-425/notes/independence.html[Gautam Iyer, Conditional Probability and Independence, definitions 12, 16 and 18 and proposition 19], with the continuous case treated in his https://www.math.cmu.edu/~gautam/c/2026-425/notes/cts-rvs.html[Continuous Random Variables notes]{ref}.
For a concrete comparison, let $X,Y$ be <independent> fair <bits>: each has a <Bernoulli distribution> with parameter $1/2$. Their <joint probability mass function>, with rows $X=0,1$ and columns $Y=0,1$, is
$$
J_{\mathrm{ind}}=\begin{pmatrix}1/4&1/4\\1/4&1/4\end{pmatrix}.
$$
All four entries equal the product of the corresponding row and column totals, $(1/2)(1/2)$. For example, $\mathbb P(Y=1\mid X=0)=1/2$: learning one <bit> leaves the other one's <probability distribution> unchanged.
Now keep a fair $X$ but set $Y=X$. The same <marginal distributions> give the same product table, yet the actual <joint probability mass function> is
$$
J_{\mathrm{same}}=\begin{pmatrix}1/2&0\\0&1/2\end{pmatrix}.
$$
Here $\mathbb P(X=0,Y=1)=0$ differs from $(1/2)(1/2)=1/4$, and $\mathbb P(Y=1\mid X=1)=1$. Thus matching <marginal distributions> does not establish <independence>. Grinstead and Snell formulate the discrete product criterion in https://chance.dartmouth.edu/teaching_aids/books_articles/probability_book/Chapter4.pdf[Introduction to Probability, section 4.1, definition 4.4, printed page 143].
\Image[/independent-and-dependent-probability-tables.png]
{title=Independence compares every joint probability with the product of the marginals}
{description=Each row compares an actual <joint probability mass function> with its product of <marginal distributions>. Marginal totals are shown at the right and bottom of every table; each whole table sums to one. The last example has zero <covariance> but a visibly different joint table.}
<Independence> is stronger than zero <covariance>. For <square-integrable> <independent random variables>, $\mathbb E[XY]=\mathbb E[X]\mathbb E[Y]$, so they are <uncorrelated random variables>. The converse fails: take $X$ with a <discrete uniform distribution> on $\{-1,0,1\}$ and $Y=X^2$. Direct calculation gives
$$
\mathbb E[X]=0,\qquad \mathbb E[Y]=2/3,\qquad
\mathbb E[XY]=\mathbb E[X^3]=0,\qquad
\operatorname{Cov}(X,Y)=0.
$$
Nevertheless, $Y$ is a nonconstant <function> determined by $X$. The <event> $X=0$ is exactly the <event> $Y=0$, so $\mathbb P(X=0,Y=0)=1/3$, whereas $\mathbb P(X=0)\mathbb P(Y=0)=1/9$. The positive and negative contributions to the <covariance> cancel without erasing the <statistical dependence>.
The usefulness of <independence> is that a large <joint probability distribution> can be assembled from small pieces. For $n$ fair <independent> <bits>, each of the $2^n$ patterns has <probability> $2^{-n}$. Products also simplify <expected values>, and vanishing cross <covariances> simplify the <variance of a sum>. Separate <measurable functions> of <independent random variables> remain <independent>: each output <event> is already an input <event>, so the defining product rule still applies. These are consequences of a model's <independence> assumption; whether the assumption describes actual observations is a question for <statistical inference>.
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