Product Distributions on the Positive Half-Line

Why Products Need Careful Support Accounting

When two independent positive quantities are multiplied, the product density is not obtained by multiplying their densities at the same argument. Instead, each possible value of the product can arise from many pairs of factor values. The density calculation must sum the contributions from all such pairs, while restricting the integral to values allowed by both factors’ supports.

The central formula is compact, but its limits are not automatic. For a fixed product value $z$, if $X=x$, then $Y=z/x$. The integral must include precisely those $x$ for which both $x$ and $z/x$ are in their respective supports.

Product-Density Formula

Theorem: Let $X$ and $Y$ be independent random variables with densities $f_X$ and $f_Y$, and suppose $P(X>0,Y>0)=1$. Then $Z=XY$ has density $$ f_Z(z)=\int_0^\infty f_X(x)f_Y(z/x)\,\frac{dx}{x}, \qquad z>0. $$ The density is zero for $z\leq 0$. The integral is understood as a nonnegative Lebesgue integral and may be infinite at exceptional points; as a density it is finite almost everywhere.

The factor $1/x$ is the absolute Jacobian determinant for solving $z=xy$ for $y$, with $x$ held fixed. Independence supplies the joint density $f_X(x)f_Y(y)$.

Conditioning form: For $z>0$, the same formula can be written as $$ f_Z(z)=\int_0^\infty f_X(x)\frac{1}{x}f_Y(z/x)\,dx. $$ For a fixed $x>0$, the conditional transformation $Y\mapsto xY$ scales the density by $1/x$. Averaging that conditional density over $X$ gives the displayed integral.

Here is a change-of-variables proof that also explains why the formula is valid. The map $(x,y)\mapsto(x,z)=(x,xy)$ is a differentiable bijection from $(0,\infty)^2$ to itself, with inverse $(x,z)\mapsto(x,z/x)$. The absolute determinant of the inverse Jacobian is $1/x$. Thus the joint density of $(X,Z)$ is $f_X(x)f_Y(z/x)/x$. Integrating out $x$ gives the stated marginal density. The change-of-variables theorem applies to nonnegative measurable densities; Tonelli’s theorem permits the marginal integration without requiring prior integrability of intermediate expressions. Since the original joint density integrates to one, the resulting marginal does also. $\square$

Limits Imposed by the Supports

Suppose $X$ has density supported on an interval $(a,b)$ and $Y$ on $(c,d)$, where $0\leq a<b\leq\infty$ and $0\leq c<d\leq\infty$. In the product formula, the integrand can be nonzero only when $a<x<b$ and $c<z/x<d$. For $z>0$, the second pair of inequalities is equivalent to $z/d<x<z/c$, with $z/\infty=0$ and $z/0=\infty$. Therefore the effective limits are $$ \max(a,z/d)<x<\min(b,z/c). $$ If the lower limit is at least the upper limit, the density is zero at that $z$. Endpoint choices do not affect a density integral.

If both densities are positive almost everywhere on their support intervals, the possible product values range between the endpoint products $ac$ and $bd$, with the usual interpretation when an endpoint is zero or infinite. But knowing this overall range does not determine the integral limits: those depend on $z$, and can change as $z$ passes particular endpoint products.

Worked Example: A Product on the Unit Interval

Let $X$ and $Y$ be independent with densities $f_X(x)=2x$ for $0<x<1$ and $f_Y(y)=3y^2$ for $0<y<1$, each zero elsewhere. Find the density of $Z=XY$.

For both densities to be nonzero, we need $0<x<1$ and $0<z/x<1$. Since $z>0$, the latter requires $x>z$. A nonempty integration interval therefore exists only for $0<z<1$, and its limits are $z<x<1$. The product formula gives $$ f_Z(z)=\int_z^1 (2x)\left(3\left(\frac{z}{x}\right)^2\right)\frac{dx}{x} =6z^2\int_z^1\frac{dx}{x^2}. $$

Since $\int_z^1 x^{-2}\,dx=[-1/x]_z^1=1/z-1$, $$ f_Z(z)=6z^2\left(\frac{1}{z}-1\right)=6z(1-z),\qquad 0<z<1. $$ Outside this interval the density is zero. As a check, its integral is $$ \int_0^1 6z(1-z)\,dz =6\left(\frac{1}{2}-\frac{1}{3}\right)=1. $$ Also, $E[X]=\int_0^1 2x^2\,dx=2/3$ and $E[Y]=\int_0^1 3y^3\,dy=3/4$, so independence gives $E[Z]=E[X]E[Y]=1/2$. The derived density gives the same result: $$ \int_0^1 z\,6z(1-z)\,dz =6\left(\frac{1}{3}-\frac{1}{4}\right)=\frac{1}{2}. $$

Piecewise Limits from Bounded Supports

When the supports do not start at zero, the product density may have several different integral limits as $z$ varies. The support restriction can be handled first, before any integration: intersect the interval for $x$ with the interval required by $z/x$.

Worked Example: Triangular Densities on Two Intervals

Let $X$ and $Y$ be independent with $$ f_X(x)=2(x-1),\quad 1<x<2, \qquad f_Y(y)=\frac{y-1}{2},\quad 1<y<3, $$ and densities zero elsewhere. Both are normalized: $\int_1^2 2(x-1)\,dx=1$, and $\int_1^3 (y-1)/2\,dy=1$. Find the product density.

For a fixed $z$, the restrictions are $1<x<2$ and $1<z/x<3$, or $z/3<x<z$. Thus the limits are $\max(1,z/3)$ and $\min(2,z)$. A nonempty interval occurs only for $1<z<6$. The limits change at $z=2$ and $z=3$, giving three cases.

The integrand simplifies to $$ f_X(x)f_Y(z/x)\frac{1}{x} =(x-1)\left(\frac{z}{x}-1\right)\frac{1}{x} =\frac{z}{x}-\frac{z}{x^2}-1+\frac{1}{x}. $$ An antiderivative with respect to $x$ is $A_z(x)=(z+1)\log x+z/x-x$. Therefore $f_Z(z)=A_z(\text{upper limit})-A_z(\text{lower limit})$.

For $1<z<2$, the limits are $1$ and $z$, so $$ f_Z(z)=A_z(z)-A_z(1)=(z+1)\log z+2-2z. $$ For $2\leq z<3$, the limits are $1$ and $2$, giving $$ f_Z(z)=A_z(2)-A_z(1)=(z+1)\log 2-\frac{z}{2}-1. $$ For $3\leq z<6$, the limits are $z/3$ and $2$, giving $$ f_Z(z)=A_z(2)-A_z(z/3)=(z+1)\log(6/z)+\frac{5z}{6}-5. $$ The density is zero outside $1<z<6$. At the breakpoints either adjacent formula gives the same value, so the choice of which formula to use at a single endpoint is immaterial. Each piece is nonnegative because it was obtained by integrating a nonnegative density over its stated interval. Normalization follows from the change-of-variables derivation: integrating the product density over $z$ recovers the integral of the joint density of $(X,Y)$, which is $1$.

A Product of Lognormal Variables

A product density can also simplify to a familiar distribution when both factors have unbounded positive support. This example illustrates the product formula without requiring a case-by-case support split.

Worked Example: Multiplying Independent Lognormal Variables

Suppose $X$ and $Y$ are independent, with $\log X\sim N(\mu_1,\sigma_1^2)$ and $\log Y\sim N(\mu_2,\sigma_2^2)$, where $\sigma_1,\sigma_2>0$. Their densities, for positive arguments, are $$ f_X(x)=\frac{1}{x\sigma_1\sqrt{2\pi}} \exp\left(-\frac{(\log x-\mu_1)^2}{2\sigma_1^2}\right), \quad f_Y(y)=\frac{1}{y\sigma_2\sqrt{2\pi}} \exp\left(-\frac{(\log y-\mu_2)^2}{2\sigma_2^2}\right). $$

Use the product formula and put $t=\log z$ and $u=\log x$. After substituting $dx=e^u\,du$, the factors $1/x$, $1/(z/x)$, and $1/x$ in the formula combine with $dx$ to leave $1/z$. Hence $$ f_Z(z)=\frac{1}{z}\int_{-\infty}^{\infty} \frac{\exp[-(u-\mu_1)^2/(2\sigma_1^2)]}{\sigma_1\sqrt{2\pi}} \frac{\exp[-(t-u-\mu_2)^2/(2\sigma_2^2)]}{\sigma_2\sqrt{2\pi}}\,du. $$

Set $s^2=\sigma_1^2+\sigma_2^2$ and $m=t-\mu_1-\mu_2$. Completing the square gives $$ \frac{(u-\mu_1)^2}{\sigma_1^2} +\frac{(t-u-\mu_2)^2}{\sigma_2^2} = \frac{(u-u_0)^2}{\tau^2}+\frac{m^2}{s^2}, \qquad \tau^2=\frac{\sigma_1^2\sigma_2^2}{s^2}, $$ where $u_0=\mu_1+\tau^2m/\sigma_2^2$. Integrating the Gaussian factor in $u$ gives $\sqrt{2\pi}\tau$. Since $\tau/(\sigma_1\sigma_2)=1/s$, the result is $$ f_Z(z)=\frac{1}{z s\sqrt{2\pi}} \exp\left[-\frac{(\log z-\mu_1-\mu_2)^2}{2s^2}\right], \qquad z>0. $$ Thus $Z$ is lognormal with log-scale mean $\mu_1+\mu_2$ and log-scale variance $\sigma_1^2+\sigma_2^2$.

Common Pitfall and a Reliable Check

A frequent error is to integrate over all $x>0$ without enforcing that $z/x$ lies in the support of $Y$. For example, in the first worked example, using limits $0<x<1$ would include values $x<z$, for which $z/x>1$ and $f_Y(z/x)=0$. Writing the full formula with $f_Y$ extended by zero is valid, but when evaluating an integral explicitly, the support restrictions must determine its effective limits.

A second error is to omit the Jacobian factor $1/x$. The conditional density of $xY$, given $X=x$, is $f_Y(z/x)/x$, not $f_Y(z/x)$. A useful check is to verify that the resulting density integrates to one; when moments are finite, comparing its mean with $E[XY]=E[X]E[Y]$ is another check under independence.

Key takeaway: For a product $Z=XY$ of independent positive continuous variables, use $f_Z(z)=\int f_X(x)f_Y(z/x)\,dx/x$. Then restrict $x$ to values for which both factors are in their supports; the resulting limits can depend on $z$ and change piecewise.

Check Your Understanding

Work through these questions by identifying the support restrictions before attempting any integration.

For independent positive variables with densities supported on $(2,5)$ and $(1,4)$, write the effective integration limits for the product density at $z>0$.
Why does the product-density formula contain $1/x$?
If $X$ is supported on $(1,2)$ and $Y$ on $(3,5)$, what is the range of possible product values?
For the densities in the first worked example, verify $E[Z]$ directly from the product density.
Suppose a proposed product density integrates to $0.8$. What does this indicate about the calculation, and what support-related mistake might have caused it?