Products of Independent Positive Variables

Why Products Need a Joint Transformation

When two positive quantities combine multiplicatively, their product may have a distribution quite different from either input. For example, a hypothetical model might represent total output as the product of two independent positive factors. To find the distribution of that output, it is not enough to multiply the two densities or to apply the usual rule for sums. Instead, we transform the pair of input variables and account for every pair of inputs that can produce the same product.

Let $X$ and $Y$ be independent, continuous random variables with $P(X>0)=P(Y>0)=1$. Write their densities as $f_X$ and $f_Y$, and define $Z=XY$. Independence gives the joint density $f_{X,Y}(x,y)=f_X(x)f_Y(y)$. The positivity assumption means that a possible value $z$ of the product is positive, and for each compatible $x$, the corresponding value of $y$ is $z/x$.

Theorem: If $X$ and $Y$ are independent, positive, continuous random variables and $Z=XY$, then for $z>0$, $$ f_Z(z)=\int_{\{x>0:\,z/x\text{ is in the support of }Y\}} f_X(x)f_Y(z/x)\frac{1}{x}\,dx. $$ The integrand is understood to be zero whenever $x$ is outside the support of $X$. Equivalently, integrate over all positive $x$ using the densities extended by zero outside their supports. For $z\leq 0$, $f_Z(z)=0$.

Deriving the Product Density

The transformation keeps $X$ as the first coordinate and replaces $Y$ by the product: $$ U=X,\qquad Z=XY. $$ Because $X>0$, the inverse is $x=u,\ y=z/u$, with $u>0$. The absolute Jacobian determinant of this inverse transformation is $1/u$. Thus the joint density of $(U,Z)$ is $f_X(u)f_Y(z/u)/u$. Integrating over $u$ removes the coordinate we do not need and gives the density of $Z$.

\left|\frac{\partial(x,y)}{\partial(u,z)}\right| = \left| \begin{matrix} 1&0\\ -z/u^2&1/u \end{matrix} \right| =\frac{1}{u}, \qquad f_{U,Z}(u,z)=f_X(u)f_Y(z/u)\frac{1}{u}.

This also explains why the factor $1/x$ cannot be omitted. For a fixed product $z$, a small change in $z$ corresponds to a change in $y$ of size approximately $dz/x$; the scaling depends on the value of $x$. The Jacobian records that scaling.

Set the product.
Define $Z=XY$ and retain one input coordinate, for example $U=X$.

Invert the transformation.
Write $x=u$ and $y=z/u$, requiring $u>0$.

Apply the Jacobian.
Multiply the joint density by the absolute inverse Jacobian, $1/u$.

Restrict and integrate.
Keep only values of $u$ for which both $u$ and $z/u$ lie in their respective supports, then integrate over $u$.

The support restriction is central to using the formula correctly. If $X$ has support $(a,b)$ and $Y$ has support $(c,d)$, where all four endpoints are positive, then for a given $z$, the compatible $x$-values must satisfy $a<x<b$ and $c<z/x<d$. Since $x>0$, the latter condition is equivalent to $z/d<x<z/c$. Therefore, the integration limits are $$ \max(a,z/d)<x<\min(b,z/c), $$ provided the lower limit is less than the upper limit. If it is not, that $z$ is outside the product support and the density is zero. Endpoint inclusion does not affect a continuous density.

Worked Examples

Worked Example: A Product of Two Uniform Factors

A hypothetical model uses independent factors $X\sim\operatorname{Uniform}(1,2)$ and $Y\sim\operatorname{Uniform}(1,2)$, with $Z=XY$. Find the density of $Z$, then calculate $P(Z<2)$. The method is the product transformation with support limits; it avoids treating the product like a sum.

Each input density is 1 on $(1,2)$ and 0 elsewhere. The product can range from 1 to 4. For a fixed $z$, compatibility requires $1<x<2$ and $1<z/x<2$, or $z/2<x<z$. Combining these inequalities gives different limits in the two parts of the support.

For $1<z<2$, the limits are $1<x<z$, so $$ f_Z(z)=\int_1^z \frac{1}{x}\,dx=\ln z. $$ For $2<z<4$, they are $z/2<x<2$, so $$ f_Z(z)=\int_{z/2}^2 \frac{1}{x}\,dx =\ln 2-\ln(z/2)=\ln(4/z). $$ The density is zero outside $1<z<4$. Thus the full result is $$ f_Z(z)= \begin{cases} \ln z,&1<z<2,\\ \ln(4/z),&2<z<4,\\ 0,&\text{otherwise}. \end{cases} $$

To find $P(Z<2)$, integrate the first part of the density: $$ P(Z<2)=\int_1^2\ln z\,dz =[z\ln z-z]_1^2 =(2\ln2-2)-(0-1) =2\ln2-1 \approx0.3863. $$ The answer is a valid probability between 0 and 1. As an additional density check, the mass below 2 is $2\ln2-1$, and the mass from 2 to 4 is $1-(2\ln2-1)$; together they equal 1. The pointwise value at $z=1.5$, for example, is $f_Z(1.5)=\ln(1.5)\approx0.4055$.

Worked Example: Unequal Uniform Supports

Suppose independent positive variables have $X\sim\operatorname{Uniform}(1,3)$ and $Y\sim\operatorname{Uniform}(2,4)$. Find $f_Z(6)$ for $Z=XY$. This hypothetical calculation tests whether both support restrictions are applied before integrating.

The densities are $f_X(x)=1/2$ for $1<x<3$ and $f_Y(y)=1/2$ for $2<y<4$. At $z=6$, the condition $2<6/x<4$ gives $6/4<x<6/2$, or $1.5<x<3$. Intersecting this interval with the support $1<x<3$ of $X$ leaves $1.5<x<3$.

The density formula therefore gives $$ f_Z(6)=\int_{1.5}^3 \frac12\frac12\frac{1}{x}\,dx =\frac14[\ln x]_{1.5}^3 =\frac14\ln\left(\frac{3}{1.5}\right) =\frac14\ln2 \approx0.1733. $$ The product support is between $1\cdot2=2$ and $3\cdot4=12$, so 6 is in its interior. The integration interval is nonempty, and the calculated density is positive, as required.

Worked Example: A Product with an Integrable Density Singularity

Let $X$ and $Y$ be independent $\operatorname{Uniform}(0,1)$ variables and let $Z=XY$. Find the density of $Z$, then calculate $P(Z\leq0.25)$. Although the inputs are positive with probability 1, their supports extend arbitrarily close to zero, so the support calculation must retain that boundary.

For $0<z<1$, the input conditions are $0<x<1$ and $0<z/x<1$. The second condition requires $x>z$. Since both densities equal 1 on their supports, $$ f_Z(z)=\int_z^1\frac{1}{x}\,dx=-\ln z,\qquad 0<z<1. $$ Outside this interval, the product density is zero. The density grows without bound as $z$ approaches zero, but the singularity is integrable.

The requested probability is $$ P(Z\leq0.25)=\int_0^{0.25}-\ln z\,dz =[z(1-\ln z)]_0^{0.25} =0.25(1-\ln0.25) =0.25(1+\ln4) \approx0.5966. $$ At the lower endpoint, $z(1-\ln z)$ tends to zero as $z$ decreases to zero. To check that the derived density is normalized, $$ \int_0^1-\ln z\,dz=[z(1-\ln z)]_0^1=1. $$ Thus the unbounded value near zero does not imply an atom at zero or an invalid density.

Checks, Assumptions, and Common Traps

Independence is required for the factorization $f_{X,Y}(x,y)=f_X(x)f_Y(y)$. If $X$ and $Y$ are dependent, their joint density cannot in general be replaced by the product of their marginal densities; the same change of variables can be used, but it must start with the actual joint density. Positivity is also important for the displayed one-branch inverse $y=z/x$ with $x>0$. If inputs can be negative, the support and possible inverse branches must be handled differently.

A common exam error is to omit the Jacobian. For the first worked example at $z=1.5$, the incorrect calculation would be $$ \int_1^{1.5}1\cdot1\,dx=0.5. $$ This omits the factor $1/x$, so it is not the density of the product. The correct calculation is $$ f_Z(1.5)=\int_1^{1.5}\frac{1}{x}\,dx =\ln(1.5)\approx0.4055. $$ Another frequent error is to use the full support of $X$ as the integration interval without checking whether $z/x$ lies in the support of $Y$. The limits must satisfy both input constraints simultaneously.

The product-density formula also gives useful checks. A density must be nonnegative and integrate to 1 over its support. For independent positive variables with finite means, independence further implies $E[Z]=E[XY]=E[X]E[Y]$; this can provide a separate moment check when the relevant integrals are manageable. The density formula itself remains applicable whether or not those means are finite.

Key takeaway: For independent positive continuous $X$ and $Y$, the density of $Z=XY$ is obtained by integrating $f_X(x)f_Y(z/x)/x$ over precisely those positive $x$-values for which both inputs are supported. The factor $1/x$ and the support intersection are essential.

Check Your Understanding

For each problem, determine the compatible input range before evaluating the product density or probability.

Independent $X,Y\sim\operatorname{Uniform}(1,2)$ are used as two positive factors. Find $f_{XY}(1.6)$. (Answer: $\ln(1.6)\approx0.4700$)
Independent $X\sim\operatorname{Uniform}(1,3)$ and $Y\sim\operatorname{Uniform}(2,4)$ define $Z=XY$. Find $f_Z(6)$, showing how the support restrictions set the integration limits. (Answer: $\frac14\ln2\approx0.1733$)
Independent $X,Y\sim\operatorname{Uniform}(0,1)$ define $Z=XY$. Find $P(Z\leq0.2)$, using the product density. (Answer: $0.2(1+\ln5)\approx0.5219$)
Independent $X\sim\operatorname{Uniform}(1,2)$ and $Y\sim\operatorname{Uniform}(1,3)$ define $Z=XY$. Find $f_Z(2.5)$, including the intersection of the allowable $x$-range with the support of $X$. (Answer: $\frac12\ln2\approx0.3466$)
Independent $X,Y\sim\operatorname{Uniform}(1,2)$ define $Z=XY$. Find $P(1<Z<2)$ by integrating the density over the appropriate part of the support. (Answer: $2\ln2-1\approx0.3863$)

What You'll Learn