Products of Uniform Variables

Why the Product Has a Logarithmic Distribution

Let \(X\) and \(Y\) be independent Uniform\((0,1)\) random variables, and define their product by \(Z=XY\). Although each factor has a constant density, the product does not: many pairs in the unit square yield products near zero, while a product near one requires both factors to be near one. The resulting density contains a logarithm and becomes unbounded at zero.

The support is bounded, which makes the geometry particularly clear. The event \(XY\le z\) is a region of the unit square; its area gives the CDF. A change of variables gives the density directly. These two approaches also clarify why the density is singular at one boundary but still integrates to a finite probability.

Deriving the CDF from the Unit Square

Because \(X\) and \(Y\) are independent Uniform\((0,1)\), probabilities for their pair are areas in the unit square. Fix \(z\) with \(0<z<1\). For each fixed \(x\), the condition \(XY\le z\) is equivalent to \(Y\le z/x\) when \(x>0\). If \(0<x\le z\), then \(z/x\ge1\), so every possible value of \(Y\) satisfies the condition. If \(z<x<1\), only values \(Y\le z/x\) qualify. Thus

The first integral is the full vertical strip \(0<x\le z\); the second is the area beneath the curve \(y=z/x\) for \(z<x<1\). For \(z\le0\), \(XY\ge0\), and the probability \(P(XY\le z)\) is zero. For \(z\ge1\), \(XY\le1\), so the CDF is one. At the endpoints, \(F_Z(0)=0\) and \(F_Z(1)=1\); neither endpoint has positive probability.

Deriving the Density

The earlier material on finite-to-one transformations gives a density by summing contributions from inverse branches. Here, away from \(x=0\), the transformation can be written as \((x,y)\mapsto(x,z)=(x,xy)\), which is one-to-one. Its inverse is \(y=z/x\), with \(x\) unchanged. The absolute Jacobian determinant of the inverse transformation is \(1/x\). The line \(x=0\) has probability zero, so it does not affect the density.

The original joint density is one inside the unit square. In the new coordinates, the region \(0<x<1,\ 0<y<1\) becomes \(0<x<1,\ 0<z<x\). Therefore the transformed joint density is \(1/x\) on that region. Integrating out \(x\), for \(0<z<1\), gives

This agrees with differentiating the CDF on \(0<z<1\): $$ \frac{d}{dz}\bigl[z(1-\ln z)\bigr] =1-\ln z-1 =-\ln z. $$ The change-of-variables derivation identifies the density directly; the CDF derivative provides an independent check on the formula.

To check that the density accounts for all probability, evaluate its improper integral: $$ \int_0^1-\ln z\,dz =\lim_{\epsilon\downarrow0}\bigl[-z\ln z+z\bigr]_{\epsilon}^{1} =1. $$ Here \(\epsilon\ln\epsilon\to0\) as \(\epsilon\downarrow0\). There is no atom at zero: since \(X,Y\) are nonnegative, \(XY=0\) requires \(X=0\) or \(Y=0\), and each event has probability zero. Similarly, \(XY=1\) requires \(X=Y=1\), also an event of probability zero. Together with the density integrating to one, this shows there is no remaining probability mass at either endpoint.

Behavior at the Support Boundaries

The density increases without bound as \(z\downarrow0\), because \(-\ln z\to\infty\). This is an integrable singularity, not an atom. For \(0<\epsilon<1\), the probability of a product in the short interval \((0,\epsilon)\) is $$ P(0<Z<\epsilon)=F_Z(\epsilon) =\epsilon(1-\ln\epsilon), $$ which tends to zero as \(\epsilon\downarrow0\). The density is unbounded there, but its integral over a shrinking interval still vanishes.

At the other boundary, \(f_Z(z)\to0\) as \(z\uparrow1\), and \(F_Z(z)\to1\). The upper tail has a useful exact form: $$ P(Z>z)=1-F_Z(z)=1-z+z\ln z,\qquad 0<z<1. $$ To see the geometry of this tail near one, put \(z=1-\delta\), where \(\delta\downarrow0\). Using \(\ln(1-\delta)=-\delta-\delta^2/2+O(\delta^3)\) gives $$ P(Z>1-\delta)=\frac{\delta^2}{2}+O(\delta^3). $$ Thus a product close to one is less likely on the order of the square of the distance to the boundary. Both factors must lie close to one, so the relevant corner region has approximately quadratic area.

Moments as a Check on the Distribution

Independence also makes the moments of \(Z\) straightforward to calculate. For any real \(r>-1\), the moment \(E[X^r]\) is finite and equals \(\int_0^1x^r\,dx=1/(r+1)\). Since \(Z^r=X^rY^r\) and \(X,Y\) are independent, $$ E[Z^r]=E[X^r]E[Y^r]=\frac{1}{(r+1)^2}. $$ For \(r>-1\), the corresponding integral using the product density also equals this value: $$ \int_0^1 z^r(-\ln z)\,dz=\frac{1}{(r+1)^2}. $$ For example, this integral follows by integration by parts on \((\epsilon,1)\) and taking \(\epsilon\downarrow0\); the boundary term \(\epsilon^{r+1}\ln\epsilon\) tends to zero because \(r+1>0\). Agreement between these calculations checks both the density and the role of its logarithmic factor.

Common Mistakes and Why the Formula Matters

• Assuming the product is uniform. Uniform factors do not imply a uniform product. The CDF calculation shows that \(F_Z(z)=z(1-\ln z)\), not \(z\), and the density varies across its support.
• Treating the infinite density at zero as an atom. A density value at one point does not determine the probability of that point. Here the density diverges as \(z\downarrow0\), but \(\int_0^\epsilon-\ln z\,dz=\epsilon(1-\ln\epsilon)\to0\), and \(P(Z=0)=0\).
• Using the density outside its support. The expression \(-\ln z\) applies only for \(0<z<1\). The density is zero outside that interval; the logarithm is not a density formula for negative \(z\) or \(z>1\).
• Dropping the full strip in the CDF calculation. When \(0<x\le z\), every \(y\in(0,1)\) satisfies \(xy\le z\). Omitting this region loses the first term \(z\) and gives the wrong CDF.

The exact distribution is useful whenever a statistic is the product of two independent proportions or random fractions, and it is a clear example of how a smooth transformation can produce an unbounded density at a support boundary. The CDF is often the simplest route to threshold and interval probabilities; the density is especially useful for integration and moment calculations. In either method, keeping the support and the boundary behavior explicit prevents errors.