Product Densities for Independent Continuous Variables

From Independence to a Joint Density

For continuous random variables, a marginal density describes each variable separately. A joint density describes how probability is distributed across combinations of their values. When the variables are independent, the joint density has a particularly simple form: it is the product of the marginal densities. This formula turns probabilities of events involving several variables into integrals over regions in a multidimensional space.

The product formula is a consequence of independence, not a definition of it. To use it correctly, we must assume mutual independence and that each variable has a density. Once the joint density is identified, the integration limits must still describe the event and respect the variables' supports.

Definition: Random variables \(X_1,\ldots,X_d\) are mutually independent if, for all Borel sets \(A_1,\ldots,A_d\subseteq\mathbb{R}\), \(P(X_1\in A_1,\ldots,X_d\in A_d)=\prod_{i=1}^d P(X_i\in A_i)\). A random variable \(X_i\) is continuous with density \(f_i\) if its distribution is absolutely continuous with respect to Lebesgue measure and \(P(X_i\in A)=\int_A f_i(x)\,dx\) for every Borel set \(A\), where \(f_i\geq 0\) and \(\int_{\mathbb{R}} f_i(x)\,dx=1\).

The Product-Density Theorem

Theorem: Suppose \(X_1,\ldots,X_d\) are mutually independent random variables, and each \(X_i\) has density \(f_i\) with respect to Lebesgue measure on \(\mathbb{R}\). Then the random vector \((X_1,\ldots,X_d)\) has joint density \(f(x_1,\ldots,x_d)=\prod_{i=1}^d f_i(x_i)\). In particular, for every Borel set \(B\subseteq\mathbb{R}^d\), \(P((X_1,\ldots,X_d)\in B)=\int_B \prod_{i=1}^d f_i(x_i)\,dx_1\cdots dx_d\).

Here is a proof that establishes the formula for the whole joint distribution, not just for a particular event. Define \(h(x_1,\ldots,x_d)=\prod_{i=1}^d f_i(x_i)\). This is a nonnegative measurable function. By Tonelli's theorem, its integral over \(\mathbb{R}^d\) factors into the product of its one-dimensional integrals:

\int_{\mathbb{R}^d} h(x_1,\ldots,x_d)\,dx_1\cdots dx_d =\prod_{i=1}^d \int_{\mathbb{R}} f_i(x_i)\,dx_i =1.

Thus \(h\) defines a probability measure \(\nu\) on \(\mathbb{R}^d\) by integration. Let \(\mu\) be the actual joint distribution of \((X_1,\ldots,X_d)\). For any Borel sets \(A_1,\ldots,A_d\), Tonelli's theorem and the marginal density formulas give

\begin{aligned} \nu(A_1\times\cdots\times A_d) &=\int_{A_1\times\cdots\times A_d}\prod_{i=1}^d f_i(x_i)\,dx_1\cdots dx_d\\ &=\prod_{i=1}^d \int_{A_i} f_i(x_i)\,dx_i\\ &=\prod_{i=1}^d P(X_i\in A_i)\\ &=P(X_1\in A_1,\ldots,X_d\in A_d)\\ &=\mu(A_1\times\cdots\times A_d). \end{aligned}

The measurable rectangles \(A_1\times\cdots\times A_d\) form a collection closed under finite intersections and generate the Borel sets of \(\mathbb{R}^d\). Since \(\mu\) and \(\nu\) are probability measures that agree on these rectangles, the uniqueness theorem for probability measures implies \(\mu=\nu\) on all Borel sets. Therefore \(h\) is a joint density of the random vector, and integrating it over any Borel event gives that event's probability. \(\square\)

Rectangular Regions

A rectangle in the joint space has the form \(A_1\times\cdots\times A_d\). Integrating the product density over it factors into separate one-dimensional integrals. This agrees with the defining probability property of independence.

Rectangular probability formula: For Borel sets \(A_i\subseteq\mathbb{R}\), \(P(X_1\in A_1,\ldots,X_d\in A_d)=\prod_{i=1}^d\int_{A_i}f_i(x_i)\,dx_i\). For intervals, each factor is the probability that one variable falls in its specified interval.

Worked Example: A Rectangular Event for Two Uniform Variables

Let \(X\sim U(-1,1)\) and \(Y\sim U(0,2)\) be independent. Find \(P(-1/2<X<1/2,\ 1/2<Y<3/2)\).

The marginal densities are \(f_X(x)=1/2\) for \(-1<x<1\), and zero otherwise, and \(f_Y(y)=1/2\) for \(0<y<2\), and zero otherwise. Independence gives the joint density \(f(x,y)=1/4\) on \((-1,1)\times(0,2)\). The event is a rectangle of width \(1\) in the \(x\)-direction and width \(1\) in the \(y\)-direction, so

\begin{aligned} P(-1/2<X<1/2,\ 1/2<Y<3/2) &=\int_{-1/2}^{1/2}\int_{1/2}^{3/2}\frac14\,dy\,dx\\ &=\int_{-1/2}^{1/2}\frac14\left(\frac32-\frac12\right)\,dx\\ &=\frac14\left(\frac12-(-\frac12)\right)=\frac14. \end{aligned}

Equivalently, each marginal interval has probability \(1/2\), and independence gives \((1/2)(1/2)=1/4\), confirming the integral.

Nonrectangular Regions

An event need not specify separate intervals for each variable. For example, it may compare the variables or impose a curved boundary. The product-density theorem still applies: describe the event as a region \(B\), then integrate the joint density over \(B\). In two dimensions, a nonrectangular region is often handled by describing one coordinate's limits in terms of the other.

For nonnegative densities, Tonelli's theorem permits changing the order of integration, even before finiteness of an iterated integral is established. A useful practical check is to sketch or otherwise describe the region and verify that the chosen limits cover it exactly once. Endpoints of integration do not affect probabilities when they differ only by a Lebesgue-null boundary.

Worked Example: A Triangular Region for Independent Uniform Variables

Let \(X\) and \(Y\) be independent \(U(0,1)\) random variables. Find \(P(X<Y)\).

The joint density is \(f(x,y)=1\) on the unit square \(0<x<1,\ 0<y<1\), and zero elsewhere. The event \(X<Y\) is the part of the square above the diagonal \(y=x\). For each \(x\) between 0 and 1, \(y\) ranges from \(x\) to 1. Therefore,

\begin{aligned} P(X<Y) &=\int_0^1\int_x^1 1\,dy\,dx\\ &=\int_0^1(1-x)\,dx\\ &=\left[x-\frac{x^2}{2}\right]_0^1\\ &=1-\frac12=\frac12. \end{aligned}

The diagonal \(X=Y\) has two-dimensional Lebesgue measure zero, so it has probability zero under this joint density. By symmetry, the two events \(X<Y\) and \(Y<X\) consequently have equal probability, consistent with the calculation.

Worked Example: Comparing Independent Exponential Variables

Let \(X\sim\operatorname{Exponential}(2)\) and \(Y\sim\operatorname{Exponential}(3)\) be independent, where the numbers denote the rates. Find \(P(X<Y)\).

Their densities are \(f_X(x)=2e^{-2x}\) for \(x>0\) and \(f_Y(y)=3e^{-3y}\) for \(y>0\), each zero outside its stated support. The joint density on the positive quadrant is \(6e^{-2x-3y}\). The event \(X<Y\) has limits \(x>0\) and \(y>x\), so

\begin{aligned} P(X<Y) &=\int_0^\infty\int_x^\infty 6e^{-2x-3y}\,dy\,dx\\ &=\int_0^\infty 6e^{-2x}\left[\frac{e^{-3x}}{3}\right]\,dx\\ &=\int_0^\infty 2e^{-5x}\,dx\\ &=2\left[-\frac{e^{-5x}}{5}\right]_0^\infty\\ &=\frac25. \end{aligned}

The inner integral uses the \(y\)-support from \(x\) to infinity; the remaining integral is finite because the integrand decays as \(e^{-5x}\). Thus the probability is \(2/5\).

Support, Independence, and Common Pitfalls

A density may be written over all of \(\mathbb{R}\), taking value zero outside its support, or only on the region where it is positive. Either convention is valid if the integration respects it. For instance, the exponential calculation uses only \(x>0\) and \(y>0\); integrating over negative values as though the densities were positive there would describe a different function and produce an invalid probability.

Multiplying densities without independence. Marginal densities alone do not determine the joint density. The product formula is justified here by mutual independence; dependent variables generally have a different joint density.
Assuming pairwise independence is enough. For three or more variables, independence of every pair does not in general imply mutual independence. The theorem requires the joint factorization in the definition.
Integrating over the wrong region. For \(X<Y\), the limits \(0<y<x<1\) describe \(Y<X\), not \(X<Y\). Write the inequalities first, then translate them into limits.
Treating a density value as a probability. A density at a point is not the probability of that point. Probabilities come from integrating density over sets.

The product-density result is useful whenever an event involves several independent continuous quantities: it supplies a joint integrand, while the event's geometry supplies the domain of integration. Rectangles often factor into marginal probabilities; curved or comparative events require limits that follow their boundaries. Keeping those two tasks separate—construct the density from independence, then describe the region—makes the calculation easier to check.

Check Your Understanding

Use the product-density theorem and careful integration limits to answer these questions.

State the hypotheses needed for the joint density of \(X_1,\ldots,X_d\) to be the product of their marginal densities.
If independent variables \(X\) and \(Y\) have densities \(f_X\) and \(f_Y\), write the integral for \(P(a<X<b,\ c<Y<d)\).
For independent \(U(0,1)\) variables \(X,Y\), describe the region corresponding to \(X+Y<1\) and set up an integral for its probability.
Why does the product-density theorem not follow from knowing only the marginal densities?
For independent exponential variables \(X\sim\operatorname{Exponential}(\alpha)\) and \(Y\sim\operatorname{Exponential}(\beta)\), with positive rates, set up an integral for \(P(Y<X)\) without evaluating it.