Why Supports Matter
A density ratio is an appealing way to compare two distributions: if \(f_Y(x)/f_X(x)\) increases with \(x\), the distribution of \(Y\) places relatively more weight at larger values. But the quotient is defined only where its denominator is positive. If one distribution assigns positive density where the other assigns zero, a ratio argument on the region where both are positive can miss exactly the pairs of points that determine whether the order holds.
The reliable starting point is therefore a cross-product inequality. It does not divide by a density, so it remains interpretable at zeros and on unequal supports. The ratio criterion is useful when available, but support compatibility must be checked as well.
\(f_X(x)f_Y(y)\ge f_Y(x)f_X(y)\)
for \(\mu\otimes\mu\)-almost every pair \(x<y\). For continuous distributions, \(\mu\) may be Lebesgue measure; for discrete distributions, it may be counting measure on the union of the supports. The almost-everywhere qualification makes the definition independent of changes to density values on null sets.
When \(f_X(x)>0\) and \(f_X(y)>0\), division by \(f_X(x)f_X(y)\) shows that the inequality is equivalent to \(f_Y(x)/f_X(x)\le f_Y(y)/f_X(y)\). Thus the usual “increasing density ratio” criterion is a consequence of the cross-product definition on the region where the denominator is positive at both points. It does not, by itself, check pairs involving points where \(f_X=0\).
A Support-Aware Criterion
Call a point common-support if both densities are positive there, X-only if \(f_X>0\) and \(f_Y=0\), and Y-only if \(f_Y>0\) and \(f_X=0\). These descriptions concern the chosen density or mass functions, and are understood up to null sets. They need not coincide pointwise with the topological support, especially at endpoints.
- On pairs \(x<y\) for which \(f_X(x)>0\) and \(f_X(y)>0\), the ratio \(f_Y/f_X\) is nondecreasing, almost everywhere.
- There is no positive-measure set of pairs \(x<y\) for which \(f_X(x)=0\), \(f_Y(x)>0\), and \(f_X(y)>0\).
Proof. Suppose first that the cross-product inequality holds. On pairs where \(f_X(x)f_X(y)>0\), divide it by that product to obtain \(f_Y(x)/f_X(x)\le f_Y(y)/f_X(y)\). If instead \(f_X(x)=0\), \(f_Y(x)>0\), and \(f_X(y)>0\) for \(x<y\), then the left side of the cross-product inequality is zero while the right side is \(f_Y(x)f_X(y)>0\). That contradicts the inequality, so such pairs cannot occur except on a null set.
Conversely, assume the two requirements. Consider an ordered pair \(x<y\), outside the exceptional null sets. If both \(f_X(x)\) and \(f_X(y)\) are positive, requirement 1 gives the cross-product inequality after multiplication by \(f_X(x)f_X(y)>0\). If \(f_X(x)>0\) and \(f_X(y)=0\), the right-hand product \(f_Y(x)f_X(y)\) is zero, while the left-hand product is nonnegative. If \(f_X(x)=0\) and \(f_X(y)>0\), requirement 2 forces \(f_Y(x)=0\), so both sides are zero. Finally, if both \(f_X(x)=0\) and \(f_X(y)=0\), both products are zero. These cases exhaust all possibilities, proving the cross-product inequality. \(\square\)
The cases are worth reading literally. In particular, if \(x\) is common-support and \(y\) is Y-only, then \(f_X(y)=0\), so the right-hand product \(f_Y(x)f_X(y)\) is zero and the inequality holds. If \(x\) is X-only and \(y\) is Y-only, then \(f_Y(x)=0\), so the right-hand product is again zero. These pairs are not support obstructions. The obstruction is the reverse arrangement: a Y-only point followed by a point where \(X\) has positive density or mass.
Worked Example: Equal Supports and an Increasing Ratio
Let \(X\sim\operatorname{Exp}(2)\) and \(Y\sim\operatorname{Exp}(1)\), where the parameter is the rate. Their densities are \(f_X(x)=2e^{-2x}\) and \(f_Y(x)=e^{-x}\) for \(x\ge0\), and both are zero for \(x<0\). Determine their likelihood-ratio order.
On their common positive support \(x\ge0\), the density ratio is
Its derivative is \(\tfrac12 e^x>0\), so the ratio is increasing on the common support. To verify the cross-product condition, take \(0\le x<y\). Then
because \(x-y<0\), hence \(e^{x-y}<1\). If \(x<0\le y\), both densities at \(x\) are zero, and both products are zero. If \(x<y<0\), all four density values are zero. Therefore the cross-product inequality holds for every ordered pair, and \(X\le_{\mathrm{lr}}Y\).
Unequal Supports: What to Check
For distributions with interval supports, a useful pattern is that X-only mass may lie to the left of the common region and Y-only mass may lie to its right. This arrangement can be compatible with likelihood-ratio order. But a Y-only region to the left of a later X-positive region creates an immediate violation: choose \(x\) in the former and \(y\) in the latter. The left-hand product is zero, while the right-hand product is positive.
Worked Example: Ordered Nonidentical Supports
Let \(X\) be uniform on \((0,2)\) and \(Y\) uniform on \((1,3)\). Their densities are \(f_X(x)=1/2\) on \((0,2)\) and zero elsewhere, and \(f_Y(x)=1/2\) on \((1,3)\) and zero elsewhere. Decide whether \(X\le_{\mathrm{lr}}Y\).
On the common interval \((1,2)\), the ratio \(f_Y/f_X\) is identically \(1\), so it is nondecreasing. The support regions also have the compatible arrangement: the X-only interval is \((0,1)\), and the Y-only interval is \((2,3)\). There is no Y-only point to the left of a later point where \(f_X\) is positive.
For a direct check, divide ordered pairs into regions. If both points lie in \((1,2)\), the two products are both \(1/4\). If \(x\in(0,1)\), then \(f_Y(x)=0\), so the right-hand product is zero for every \(y>x\); the left-hand product is nonnegative. If \(y\in(2,3)\) and \(x\in(1,2)\), then \(f_X(y)=0\), so again the right-hand product is zero. If \(x\in(0,1)\) and \(y\in(2,3)\), both products are zero: \(f_Y(x)=0\) and \(f_X(y)=0\). Pairs outside the union of the supports contribute zero unless already covered. Endpoints do not affect the conclusion because they have Lebesgue measure zero. Hence \(X\le_{\mathrm{lr}}Y\).
Worked Example: A Common-Support Ratio That Misleads
Now let \(X\) be uniform on \((1,3)\) and \(Y\) uniform on \((0,2)\). On the common interval \((1,2)\), both densities equal \(1/2\), so the ratio \(f_Y/f_X\) is again the constant \(1\). That calculation alone does not establish likelihood-ratio order.
Take \(x=1/2\) and \(y=3/2\). At \(x\), \(Y\) is positive and \(X\) is zero: \(f_Y(x)=1/2\), \(f_X(x)=0\). At \(y\), both densities are \(1/2\). Consequently,
The required inequality fails. The ratio on the common interval missed the Y-only region \((0,1)\), which lies to the left of a later X-positive point. Thus \(X\not\le_{\mathrm{lr}}Y\). Notice that \(f_Y/f_X\) is not defined on \((0,1)\); assigning it an informal value such as “infinity” does not repair the failed cross-product inequality.
Zeros, Boundaries, and Discrete Laws
A zero denominator is not an invitation to divide. At a point where \(f_X(x)=0\), the expression \(f_Y(x)/f_X(x)\) is undefined, whether its numerator is zero or positive. If both densities vanish, the ratio is still \(0/0\), and it supplies no comparison. If \(f_Y(x)>0\) while \(f_X(x)=0\), one may informally think of the ratio as infinite, but that convention alone does not define a monotone function or prove the cross-product inequalities with later points.
Boundary values require similar care. For a continuous distribution, changing a density at one endpoint does not change the distribution or an almost-everywhere order. For a discrete distribution, however, an endpoint may be an atom with positive probability and cannot be ignored. The cross-product definition handles both settings: use Lebesgue measure for ordinary continuous densities or counting measure for probability masses, and retain any points that have positive mass.
Worked Example: Checking Atoms at the Boundary
Let \(X\) and \(Y\) be Bernoulli random variables with \(P(X=1)=0.3\) and \(P(Y=1)=0.7\). Using counting measure on \(\{0,1\}\), their mass functions are \(f_X(0)=0.7,\ f_X(1)=0.3,\ f_Y(0)=0.3,\ f_Y(1)=0.7\). There is only one ordered pair of distinct support points, \(0<1\). Its cross-products are
Since \(0.49\ge0.09\), the inequality holds. Equivalently, the ratios at the two atoms are \(f_Y(0)/f_X(0)=3/7\) and \(f_Y(1)/f_X(1)=7/3\), which increase. Both atoms matter: if one were a zero-probability boundary point instead, it would not contribute under counting measure, but a positive-probability atom must be checked.
Common Pitfalls and a Reliable Check
- Checking the ratio only where both densities are positive. This tests common-support pairs but can miss a Y-only point that precedes an X-positive point. Check support compatibility as well.
- Dividing by zero. The ratio is undefined wherever \(f_X=0\); an informal infinite value is not a substitute for the cross-product test.
- Assuming every support mismatch is fatal. An X-only region to the left and a Y-only region to the right can be compatible with the order, as the shifted uniform example shows.
- Discarding atoms as if they were endpoints of a continuous interval. A point with positive probability must be included in the discrete cross-product check.
- Treating a density value at a null point as decisive. For continuous distributions the order is an almost-everywhere statement, so changing a density at a single endpoint cannot change the distributional comparison.
Check Your Understanding
For each question, attend to both the ratio and the locations where its denominator vanishes.
- For \(x<y\), if \(x\) is common-support and \(y\) is Y-only, which cross-product is zero, and why does the inequality hold?
- State the support arrangement that causes a cross-product violation when a Y-only point precedes an X-positive point.
- Two densities have a constant ratio on their common support. Why is that fact alone insufficient to establish likelihood-ratio order?
- Why can changing a continuous density at one endpoint be irrelevant, while omitting an endpoint atom in a discrete comparison is not?
- For \(X\sim\operatorname{Bernoulli}(0.4)\) and \(Y\sim\operatorname{Bernoulli}(0.6)\), compute both cross-products for the pair \(0<1\) and decide whether the likelihood-ratio inequality holds.