Factorization of Expectations Under Independence

Why Expectations Factor

Independence is a statement about the joint distribution of random variables, but it also has a useful consequence for calculations: expectations of products of separate functions can be computed from the marginal distributions alone. The qualification “separate” matters. In an expression such as $g(X)h(Y)$, one factor depends only on $X$ and the other only on $Y$. Under independence, the expectation of this product factors into two expectations, provided the expectations are well-defined.

The integrability conditions are not a technical afterthought. An expectation of a signed random variable is finite in the usual sense when its absolute expectation is finite. The result below states conditions that guarantee both sides exist and that the product itself is integrable.

Definition: A real-valued random variable $Z$ is integrable if $E[|Z|]<\infty$. A Borel measurable function is a function for which the inverse image of every Borel set is Borel measurable. This condition ensures that applying the function to a random variable produces a random variable.

The Product-Expectation Identity

Let $X$ and $Y$ be independent real-valued random variables, with marginal distributions $\mu$ and $\nu$. Independence means that the joint distribution of $(X,Y)$ is the product measure $\mu\otimes\nu$. Thus expectations of functions of $(X,Y)$ can be integrated with respect to this product measure.

Theorem: Suppose $X$ and $Y$ are independent, and $g$ and $h$ are Borel measurable real-valued functions such that $E[|g(X)|]<\infty$ and $E[|h(Y)|]<\infty$. Then $g(X)h(Y)$ is integrable and $$ E[g(X)h(Y)]=E[g(X)]E[h(Y)]. $$

The proof uses Tonelli’s theorem for nonnegative functions and then handles signed functions by separating their positive and negative parts. This also verifies that the product expectation exists rather than assuming it does.

Use the joint law.
Because $X$ and $Y$ are independent, the joint law is $\mu\otimes\nu$, so the expectation of a measurable function of $(X,Y)$ is an integral with respect to this product measure.

Check absolute integrability.
Apply Tonelli’s theorem to the nonnegative function $|g(x)h(y)|=|g(x)||h(y)|$. Its integral factors into the product of the two finite absolute expectations.

Separate positive and negative parts.
Expand $g$ and $h$ into their positive and negative parts. Tonelli’s factorization applies to each nonnegative term, and the finite results can be combined to obtain the signed identity.

Here are the details. Write $g^+=\max(g,0)$ and $g^-=\max(-g,0)$, and define $h^+$ and $h^-$ in the same way. Tonelli’s theorem gives

\begin{aligned} E[|g(X)h(Y)|] &=\int_{\mathbb R}\int_{\mathbb R}|g(x)||h(y)|\,\nu(dy)\,\mu(dx)\\ &=\left(\int_{\mathbb R}|g(x)|\,\mu(dx)\right) \left(\int_{\mathbb R}|h(y)|\,\nu(dy)\right)\\ &=E[|g(X)|]E[|h(Y)|]<\infty. \end{aligned}

The iterated-integral equality follows because the inner integral of $|h(y)|$ is a finite constant, independent of $x$. In particular, $g(X)h(Y)$ is integrable. Applying the same Tonelli argument to each pair of nonnegative parts gives, for $a,b\in\{+,-\}$, $E[g^a(X)h^b(Y)]=E[g^a(X)]E[h^b(Y)]$. All four terms are finite, since each part is bounded above by the corresponding absolute value. Therefore linearity of expectation yields

\begin{aligned} E[g(X)h(Y)] &=E[g^+(X)h^+(Y)]-E[g^+(X)h^-(Y)]\\ &\quad-E[g^-(X)h^+(Y)]+E[g^-(X)h^-(Y)]\\ &=(E[g^+(X)]-E[g^-(X)]) (E[h^+(Y)]-E[h^-(Y)])\\ &=E[g(X)]E[h(Y)]. \end{aligned}

This proves the theorem. $\square$

Worked Examples

Worked Example: Powers of Independent Discrete Variables

Let $X\sim\operatorname{Bernoulli}(0.4)$, and let $Y$ be independent of $X$, with $P(Y=-1)=0.25$ and $P(Y=2)=0.75$. Find $E[X^2Y]$, and verify the result by summing over the joint outcomes.

Since $X$ takes only the values 0 and 1, $X^2=X$, so $E[X^2]=0.4$. For $Y$,

$$ E[Y]=(-1)(0.25)+(2)(0.75)=-0.25+1.5=1.25. $$

Both variables have finite moments, and independence allows us to use the theorem with $g(x)=x^2$ and $h(y)=y$. Consequently,

$$ E[X^2Y]=E[X^2]E[Y]=(0.4)(1.25)=0.5. $$

For a direct check, $X^2Y$ is zero whenever $X=0$. When $X=1$, it equals $Y$. Independence gives

$$ E[X^2Y]=(0.4)(0.25)(-1)+(0.4)(0.75)(2) =-0.1+0.6=0.5. $$

The direct joint-outcome calculation agrees with the factorized calculation.

Worked Example: A Product of Bounded Functions

Let $X\sim N(0,1)$ and $Y\sim N(0,1)$ be independent. Find $E[\mathbf{1}_{\{X>0\}}\mathbf{1}_{\{Y\leq 1\}}]$, where $\mathbf{1}_A$ is 1 on event $A$ and 0 otherwise.

The indicator functions are measurable and bounded by 1, so their absolute expectations are finite. By symmetry of the standard normal distribution, $P(X>0)=1/2$. Write $\Phi$ for the standard normal CDF; then $P(Y\leq1)=\Phi(1)$. Factorization gives

\begin{aligned} E[\mathbf{1}_{\{X>0\}}\mathbf{1}_{\{Y\leq1\}}] &=E[\mathbf{1}_{\{X>0\}}]E[\mathbf{1}_{\{Y\leq1\}}]\\ &=P(X>0)P(Y\leq1)\\ &=\frac12\Phi(1)\approx 0.42067. \end{aligned}

The expectation of the product of the indicators is the probability that both events occur. Independence makes that joint probability the product of the two marginal probabilities.

Worked Example: A Poisson Second Moment and a Product

Let $X\sim\operatorname{Poisson}(2)$ and $Y\sim\operatorname{Poisson}(3)$ be independent. Compute $E[X^2Y]$.

First calculate $E[X^2]$. For a Poisson random variable with parameter 2, $P(X=k)=e^{-2}2^k/k!$ for $k=0,1,2,\ldots$. Since $k^2=k(k-1)+k$,

\begin{aligned} E[X^2] &=e^{-2}\sum_{k=0}^{\infty}k^2\frac{2^k}{k!}\\ &=e^{-2}\left( \sum_{k=0}^{\infty}k(k-1)\frac{2^k}{k!} +\sum_{k=0}^{\infty}k\frac{2^k}{k!}\right)\\ &=e^{-2}\left( 4\sum_{j=0}^{\infty}\frac{2^j}{j!} +2\sum_{j=0}^{\infty}\frac{2^j}{j!}\right)\\ &=e^{-2}(4e^2+2e^2)=6. \end{aligned}

In the first sum on the third line, the terms for $k=0,1$ are zero and setting $j=k-2$ gives the displayed series. In the second, the $k=0$ term is zero and setting $j=k-1$ gives its displayed series. Also $E[Y]=3$. The Poisson variables have finite absolute moments of the required orders, so

$$ E[X^2Y]=E[X^2]E[Y]=(6)(3)=18. $$

This illustrates how a product involving a higher power can be evaluated once the relevant marginal moments are known.

Powers, Bounded Functions, and the Scope of the Rule

Taking $g(x)=x^r$ and $h(y)=y^s$ gives a useful moment identity. For nonnegative integers $r$ and $s$, if $E[|X|^r]<\infty$ and $E[|Y|^s]<\infty$, then independence implies

$$ E[X^rY^s]=E[X^r]E[Y^s]. $$

The absolute-moment conditions ensure that the expectations are finite even when an odd power can be negative. For $r=0$ or $s=0$, the corresponding power is the constant 1, so the identity reduces to the expectation of the other power. The general theorem also applies to non-integer powers when the functions are measurable on the variables’ ranges and the required absolute expectations are finite.

Bounded measurable functions require no separate moment calculation: if $|g(x)|\leq M$ and $|h(y)|\leq N$ for finite constants $M,N$, then $E[|g(X)|]\leq M$ and $E[|h(Y)|]\leq N$. Thus the theorem applies automatically. Indicators are one important case, but bounded functions need not be indicators.

More generally, the theorem concerns functions of separate independent variables, not just powers. If $g(X)$ and $h(Y)$ are integrable, their product expectation factors. It does not say that the expectation of an arbitrary function of both variables factors, nor does it make $g(X)$ and $h(Y)$ independent unless the conditions establishing independence are met.

Common Mistake: Applying the Identity Without Independence

The product-expectation identity is not a rule for arbitrary pairs of random variables. For example, let $X$ take the values $-1$ and $1$, each with probability $1/2$, and set $Y=X$. Then $E[X]=E[Y]=0$, but

E[XY]=E[X^2]=1\neq 0=E[X]E[Y].

Here the two variables are dependent: once $X$ is known, $Y$ is determined. The incorrect reasoning would be to see two factors in $XY$ and multiply their expectations. The correction is to check independence first, then check that the functions have finite absolute expectations. If either condition is missing, the theorem does not justify the factorization.

A related pitfall is to assume that finiteness of a signed expectation is automatic. The stated theorem avoids ambiguity by requiring $E[|g(X)|]<\infty$ and $E[|h(Y)|]<\infty$. Under these assumptions the proof also establishes $E[|g(X)h(Y)|]<\infty$, so every expectation in the identity is finite.

Key takeaway: For independent $X$ and $Y$, separate measurable functions factor in expectation when each function of its variable is integrable: $E[g(X)h(Y)]=E[g(X)]E[h(Y)]$. Powers and bounded functions are common applications, but independence and the relevant finiteness conditions must still be checked.

Check Your Understanding

For each question, identify the independence and integrability conditions relevant to the calculation.

If independent $X$ and $Y$ satisfy $E[X^2]<\infty$ and $E[|Y|]<\infty$, what identity follows for $E[X^2Y]$?
Why does boundedness of $g$ and $h$ guarantee the integrability conditions in the theorem?
Suppose $X$ and $Y$ are independent and $E[|X|^3]<\infty$, $E[|Y|^2]<\infty$. State the factorization for $E[X^3Y^2]$.
In the dependent example with $Y=X$ and $X$ equally likely to be $-1$ or $1$, calculate $E[X]$, $E[Y]$, and $E[XY]$. Which hypothesis of the theorem fails?
For independent variables, what does Tonelli’s theorem show about $E[|g(X)h(Y)|]$ when both separate absolute expectations are finite?