Factorization of Expectations Under Independence

When Does an Expectation Factor?

If \(X\) and \(Y\) are independent, information about one does not change the law of the other. One important consequence is that, for suitable measurable functions \(f\) and \(g\), the expectation of \(f(X)g(Y)\) is the product of the separate expectations. The hypotheses matter: independence is essential, and the expectations must be defined. We will first derive the result for nonnegative functions, then use positive and negative parts to handle integrable real-valued functions.

Throughout, \((\Omega,\mathcal{F},\mathbb{P})\) is a probability space, hence a finite measure space; it is not assumed complete. The random variables \(X,Y:\Omega\to\mathbb{R}\) are measurable, and functions of real variables below are Borel measurable and real-valued. Expectations are Lebesgue integrals with respect to \(\mathbb{P}\). The variables \(X\) and \(Y\) are independent if the sigma-algebras \(\sigma(X)\) and \(\sigma(Y)\) are independent: for every \(A\in\sigma(X)\) and \(B\in\sigma(Y)\), \(\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)\).

For Borel sets \(C,D\subseteq\mathbb{R}\), independence gives \(\mu_{X,Y}(C\times D)=\mu_X(C)\mu_Y(D)\). The product probability measure \(\mu_X\times\mu_Y\) has the same value on every such rectangle. Rectangles form a pi-system generating \(\mathcal{B}_{\mathbb{R}^2}\); by the standard uniqueness theorem for probability measures on a generating pi-system, the two measures agree on all Borel sets. Thus \(\mu_{X,Y}=\mu_X\times\mu_Y\). This identifies the joint law; the factorization of expectations follows from integrating functions against it.

Factorization for Nonnegative Functions

Proof. Write \(\mu_X,\mu_Y\) for the marginal laws. First suppose \(u\) and \(v\) are nonnegative simple functions, with disjoint level sets \(A_i\) and \(B_j\), so \(u=\sum_i a_i\chi_{A_i}\) and \(v=\sum_j b_j\chi_{B_j}\). Using \(\mu_{X,Y}=\mu_X\times\mu_Y\), the integral of \(u(x)v(y)\) against the joint law is

Each sum in parentheses is the integral of the corresponding simple function against its marginal law. This proves the identity for simple \(u,v\).

For general nonnegative Borel \(u,v\), choose nonnegative simple Borel functions \(u_n\uparrow u\) and \(v_n\uparrow v\). Such sequences are obtained by rounding values down to dyadic levels and truncating at increasing finite bounds. Since both sequences are nondecreasing, \(u_nv_n\uparrow uv\) pointwise. The Monotone Convergence Theorem applies to each sequence of integrands on the respective probability spaces, including the joint-law space. Therefore, $$ \int_{\mathbb{R}^2}u(x)v(y)\,d(\mu_X\times\mu_Y) =\lim_n\int_{\mathbb{R}^2}u_n(x)v_n(y)\,d(\mu_X\times\mu_Y) =\left(\int u\,d\mu_X\right)\left(\int v\,d\mu_Y\right). $$ The last equality follows by taking limits of the simple-function identity; both marginal integrals are finite by hypothesis. The left side is consequently finite. By the definition of the laws and integration of nonnegative measurable functions, this is exactly the asserted expectation and integrability. \(\square\)

Proof. Apply the proposition to \(u=|f|\) and \(v=|g|\). Its hypotheses hold by the stated integrability assumptions, and it gives $$ \mathbb{E}[|f(X)g(Y)|] =\mathbb{E}[|f(X)|]\mathbb{E}[|g(Y)|]<\infty. $$ Thus the product is integrable. Define \(f^+=\max(f,0)\), \(f^-=\max(-f,0)\), and likewise \(g^+,g^-\). All four functions are nonnegative and have finite expectation after composition with the appropriate random variable, since \(f^+,f^-\le |f|\) and \(g^+,g^-\le |g|\). Applying the proposition to each of the four pairs and using linearity for integrable random variables gives $$ \begin{aligned} \mathbb{E}[f(X)g(Y)] &=\mathbb{E}[f^+(X)g^+(Y)]-\mathbb{E}[f^+(X)g^-(Y)]\\ &\quad-\mathbb{E}[f^-(X)g^+(Y)]+\mathbb{E}[f^-(X)g^-(Y)]\\ &=(\mathbb{E}[f^+(X)]-\mathbb{E}[f^-(X)]) (\mathbb{E}[g^+(Y)]-\mathbb{E}[g^-(Y)])\\ &=\mathbb{E}[f(X)]\mathbb{E}[g(Y)]. \end{aligned} $$ The decomposition and linearity are valid because the absolute expectation of the product is finite, as shown above. \(\square\)

Computations from Discrete Laws and Densities

Why Independence Cannot Be Omitted

A common mistake is to factor an expectation simply because the integrand is a product. The theorem requires independence of the random variables, not merely integrability of each factor. For a concrete counterexample, let \(X=Y\) be a single random variable with \(\mathbb{P}(X=1)=\mathbb{P}(X=-1)=1/2\). Then both variables are integrable, but they are not independent: for example, \(\mathbb{P}(X=1,Y=1)=1/2\), whereas \(\mathbb{P}(X=1)\mathbb{P}(Y=1)=1/4\).

The incorrect calculation would be $$ \mathbb{E}[XY]=\mathbb{E}[X]\mathbb{E}[Y]=0\cdot0=0. $$ But \(XY=X^2=1\) at both possible outcomes, so the correct value is \(\mathbb{E}[XY]=1\). This example also checks that all expectations exist; the failure is specifically the missing independence hypothesis.

There is a second issue to check before applying the theorem: finiteness. An expectation of a signed random variable is defined as a finite Lebesgue integral only when its absolute value has finite expectation. The theorem's separate assumptions \(\mathbb{E}|f(X)|<\infty\) and \(\mathbb{E}|g(Y)|<\infty\) guarantee both that the product is integrable and that the displayed factorization is an equality of finite real numbers. For nonnegative functions, the proposition likewise assumes both separate expectations finite, avoiding any ambiguous product involving zero and infinity.