Factorization of Expectations for Independent Variables

Why Factorization Matters

Independence is often introduced through the factorization of probabilities of events. A closely related and extremely useful consequence is that, under suitable integrability conditions, the expectation of a product factors into a product of expectations.

The integrability condition matters. Even if two variables are independent and each has a finite expectation, their product must still be checked for integrability before its expectation is treated as an ordinary finite number. Independence supplies the needed control when both variables are integrable.

Independence and Simple Variables

We use independence in its sigma-algebra form: the sigma-algebras \(\sigma(X)\) and \(\sigma(Y)\) are independent if, for every \(A\in\sigma(X)\) and \(B\in\sigma(Y)\), \(\mathbb P(A\cap B)=\mathbb P(A)\mathbb P(B)\). Equivalently, for independent real-valued variables, their joint law is the product of their marginal laws.

A finite-valued simple random variable is a measurable random variable taking only finitely many values. Suppose \(S\) and \(T\) are simple, with representations on finite measurable partitions:

where \(A_i\in\sigma(X)\), \(B_j\in\sigma(Y)\), and the respective collections partition the sample space. If \(\sigma(X)\) and \(\sigma(Y)\) are independent, then \(\mathbb P(A_i\cap B_j)=\mathbb P(A_i)\mathbb P(B_j)\) for every \(i,j\).

Proof. Expanding the product gives \(ST=\sum_{i=1}^{m}\sum_{j=1}^{n}a_ib_j\mathbf{1}_{A_i\cap B_j}\). Taking expectations and using event independence,

All sums are finite, so this calculation does not require a limiting argument. The values \(a_i,b_j\) may be negative; the argument still applies. \(\square\)

A Nonnegative Product Estimate

To pass from simple variables to general integrable variables, we need to control products while the approximations converge. The following nonnegative version follows from the simple-variable result and monotone convergence.

Proof. Choose nonnegative simple random variables \(U_k\) and \(V_k\), measurable with respect to \(\sigma(X)\) and \(\sigma(Y)\), respectively, such that \(U_k\uparrow U\) and \(V_k\uparrow V\). Such approximations are obtained by truncating the values and rounding down to successively finer finite grids. The simple-variable proposition gives \(\mathbb E[U_kV_k]=\mathbb E[U_k]\mathbb E[V_k]\) for every \(k\).

Since both sequences are nonnegative and increasing, \(U_kV_k\uparrow UV\). By the monotone convergence theorem, \(\mathbb E[U_kV_k]\to\mathbb E[UV]\), \(\mathbb E[U_k]\to\mathbb E[U]\), and \(\mathbb E[V_k]\to\mathbb E[V]\). The last two limits are finite by assumption, so their products converge to \(\mathbb E[U]\mathbb E[V]\). Taking limits in the simple-variable identity proves the lemma. \(\square\)

In particular, applying the lemma to \(U=|X|\) and \(V=|Y|\) shows \(\mathbb E|XY|=\mathbb E|X|\mathbb E|Y|<\infty\). Thus the product of independent integrable variables is integrable.

Factorization for Integrable Variables

Proof. First approximate each variable by finite-valued simple variables. For \(n\in\mathbb N\), clip a real number to the interval \([-n,n]\), then round down to a grid of spacing \(2^{-n}\). Applying this operation to \(X\) and \(Y\) gives simple variables \(X_n\) and \(Y_n\), measurable with respect to \(\sigma(X)\) and \(\sigma(Y)\). For each real \(x\), the approximation error is bounded by \(|x|\mathbf{1}_{\{|x|>n\}}+2^{-n}\). Consequently, dominated convergence gives \(\mathbb E|X_n-X|\to0\) and \(\mathbb E|Y_n-Y|\to0\).

By the simple-variable proposition, \(\mathbb E[X_nY_n]=\mathbb E[X_n]\mathbb E[Y_n]\). To pass to the limit in the left side, write \(X_nY_n-XY=(X_n-X)Y_n+X(Y_n-Y)\). The triangle inequality and the nonnegative-product lemma imply

The sequence \(\mathbb E|Y_n|\) is bounded, since \(\mathbb E|Y_n|\leq\mathbb E|Y|+\mathbb E|Y_n-Y|\). The first term on the final line therefore tends to zero, and the second tends to zero because \(\mathbb E|X|<\infty\) and \(\mathbb E|Y_n-Y|\to0\). Hence \(X_nY_n\to XY\) in \(L^1\), so \(\mathbb E[X_nY_n]\to\mathbb E[XY]\).

Also, \(L^1\) convergence implies \(\mathbb E[X_n]\to\mathbb E[X]\) and \(\mathbb E[Y_n]\to\mathbb E[Y]\). Passing to the limit in \(\mathbb E[X_nY_n]=\mathbb E[X_n]\mathbb E[Y_n]\) proves the stated identity. The product is integrable by the preceding nonnegative-product estimate. \(\square\)

Products of Measurable Functions

The same factorization applies when the variables are first transformed by measurable functions. Let \(f,g:\mathbb R\to\mathbb R\) be Borel measurable, and suppose \(\mathbb E|f(X)|<\infty\) and \(\mathbb E|g(Y)|<\infty\). Since the joint law of \((X,Y)\) is \(\mathcal L(X)\otimes\mathcal L(Y)\), the product-measure integration theorem and Fubini’s theorem give

The integrability needed for Fubini is ensured because \(\int_{\mathbb R^2}|f(x)g(y)|\,d(\mathcal L(X)\otimes\mathcal L(Y)) =\mathbb E|f(X)|\,\mathbb E|g(Y)|<\infty\). Thus the expectation on the left is well-defined and finite. The functions need not be bounded; their absolute expectations are the relevant condition.

Interpretation and a Common Pitfall

The factorization is a consequence of independence, not merely of the separate distributions of \(X\) and \(Y\). Knowing the marginal laws alone does not determine \(\mathbb E[XY]\): the joint law matters unless independence is given. Nor should the identity be used without checking integrability. In this theorem, finite absolute expectations ensure that the product is integrable, making the equality an equality of finite real numbers.

A useful summary is: event factorization establishes the result for simple variables; approximation and integrability extend it to general variables; and product-measure integration gives the corresponding formula for measurable functions.