Distribution-Order Comparisons With Mixed Supports

Why Mixed Supports Need Care

Distributional comparisons often involve laws that do not share the same kind of support. One law may put positive probability at a single value and spread the rest continuously; another may have atoms at different points, or continuous mass on a different interval. In these settings, comparing only ordinary Lebesgue densities can miss probability mass, and checking a density ratio only where its denominator is positive can give a false likelihood-ratio conclusion.

The remedy is to formulate comparisons at the level of the probability measures, or to use densities with respect to a common measure that records both atoms and continuous mass. The order being used also matters: stochastic order compares tails, likelihood-ratio order imposes a stronger monotone cross-product condition, and convex order compares expectations of convex functions subject to an equal-mean requirement.

Three Orders for General Distributions

Definition: Let \(X\) and \(Y\) be real-valued random variables. Usual stochastic order, written \(X\le_{\mathrm{st}}Y\), means \(P(X>t)\le P(Y>t)\) for every real \(t\). Equivalently, \(F_X(t)\ge F_Y(t)\) for every \(t\), where \(F_X(t)=P(X\le t)\). Likelihood-ratio order, written \(X\le_{\mathrm{lr}}Y\), means that for a common dominating measure \(\mu\), with densities \(p=dP_X/d\mu\) and \(q=dP_Y/d\mu\), the inequality \(p(x)q(y)\ge q(x)p(y)\) holds for \(\mu\otimes\mu\)-almost every pair \(x<y\). Convex order, written \(X\le_{\mathrm{cx}}Y\), means that \(E[\phi(X)]\le E[\phi(Y)]\) for every convex function \(\phi\) for which the expectations exist. In particular, convex order is used for integrable variables and requires \(E[X]=E[Y]\).

The common-measure definition applies to atoms and continuous components together. For example, if both laws have an atom at 0 and otherwise have densities with respect to Lebesgue measure, one can take \(\mu\) to be Lebesgue measure plus a unit point mass at 0. The Radon–Nikodym densities then encode the atom as well as the continuous part. More generally, \(\mu=P_X+P_Y\) is always a finite common dominating measure.

The cross-product condition is the safe formulation of likelihood-ratio order when supports differ. Where \(p\) and \(q\) are both positive, it says \(q(y)/p(y)\ge q(x)/p(x)\) for \(x<y\). But the ratio statement alone does not check pairs where one density vanishes. The cross-product inequality checks those pairs too.

Worked Example: A Stochastic Comparison With an Atom

Let \(X\) have probability \(1/2\) at 0 and, with probability \(1/2\), be uniform on \((0,1)\). Let \(Y\) be uniform on \((0,2)\). Compare their distribution functions.

For \(t<0\), both distribution functions are 0. At \(t=0\), \(F_X(0)=1/2\) because of the atom, whereas \(F_Y(0)=0\). For \(0\le t\le1\), \(F_X(t)=1/2+t/2\) and \(F_Y(t)=t/2\), so \(F_X(t)\ge F_Y(t)\). For \(1<t<2\), \(F_X(t)=1\) and \(F_Y(t)=t/2<1\). For \(t\ge2\), both are 1.

Thus \(F_X(t)\ge F_Y(t)\) at every \(t\), including at the jump, and \(X\le_{\mathrm{st}}Y\). The atom is not a technical exception: it contributes directly to the CDF and must be included in the comparison.

Likelihood-Ratio Order and Its Support Condition

A common informal shortcut is to form \(q/p\) and check whether it increases on the region where \(p>0\). This is not enough if the supports differ. The pairwise definition requires the inequality across the entire ordered support, including points where one of the densities is zero.

Worked Example: A Valid Likelihood-Ratio Comparison With Mixed Support

Let \(X\) put mass \(0.6\) at 0 and have density \(0.4\) on \((0,1)\). Let \(Y\) put mass \(0.3\) at 0 and have density \(0.7\) on \((0,1)\). Take \(\mu=\delta_0+\lambda\) on \([0,1]\), where \(\delta_0\) is unit mass at 0 and \(\lambda\) is Lebesgue measure. The densities with respect to \(\mu\) are \(p(0)=0.6,\ q(0)=0.3\), and \(p(x)=0.4,\ q(x)=0.7\) for \(0<x<1\).

For two points in \((0,1)\), the cross-products are equal: \(p(x)q(y)=q(x)p(y)=0.28\). For the ordered pair \(x=0<y\) with \(0<y<1\), \(p(0)q(y)=0.6(0.7)=0.42\), while \(q(0)p(y)=0.3(0.4)=0.12\). Hence the required inequality holds for every ordered pair in the support, so \(X\le_{\mathrm{lr}}Y\). The atom and the continuous part are both included in the check.

Here \(E[X]=0.4(1/2)=0.2\) and \(E[Y]=0.7(1/2)=0.35\). The larger mean is consistent with \(Y\) being shifted upward in the usual stochastic order, but a mean comparison alone would not establish any of these orders.

Theorem: If \(X\le_{\mathrm{lr}}Y\) under the common-measure cross-product definition, then \(X\le_{\mathrm{st}}Y\).

Proof. Fix \(t\in\mathbb R\), and write \(p=dP_X/d\mu\), \(q=dP_Y/d\mu\). Integrating the nonnegative cross-product difference over \(x\le t<y\) gives

\begin{aligned} 0 &\le \int_{\{x\le t\}}\int_{\{y>t\}} \bigl[p(x)q(y)-q(x)p(y)\bigr]\,d\mu(y)\,d\mu(x)\\ &=F_X(t)P(Y>t)-F_Y(t)P(X>t)\\ &=F_X(t)\{1-F_Y(t)\}-F_Y(t)\{1-F_X(t)\}\\ &=F_X(t)-F_Y(t). \end{aligned}

The factorization is valid because the two integrands are products of nonnegative integrable densities and \(\mu\) is finite. Thus \(F_X(t)\ge F_Y(t)\) for every \(t\), which is exactly \(X\le_{\mathrm{st}}Y\). \(\square\)

The support requirement can be seen in a simple failed comparison. Let \(X\) be uniform on \((1,2)\) and \(Y\) uniform on \((0,1)\), with Lebesgue densities \(p\) and \(q\). On the region where \(p>0\), the ratio \(q/p\) is constantly 0, which is nondecreasing. But for \(x\in(0,1)\) and \(y\in(1,2)\), \(x<y\) and \(p(x)q(y)=0<1=q(x)p(y)\). The cross-product condition fails; indeed \(X\le_{\mathrm{lr}}Y\) is false. A ratio checked only where its denominator is positive has ignored the support ordering.

Convex Order: Equal Means and Stop-Loss Checks

Convex order asks whether one distribution is less spread out than another in the sense of expectations of convex functions. Equal means are essential: since both \(x\) and \(-x\) are convex, \(X\le_{\mathrm{cx}}Y\) forces \(E[X]=E[Y]\), provided the variables are integrable. A useful test, including for mixed supports, is the stop-loss characterization.

Theorem: For integrable \(X\) and \(Y\), \(X\le_{\mathrm{cx}}Y\) if and only if \(E[X]=E[Y]\) and \(E[(X-t)_+]\le E[(Y-t)_+]\) for every real \(t\), where \((u)_+=\max(u,0)\). Necessity follows because \(u\mapsto(u-t)_+\) is convex. The converse is the stop-loss characterization: with equal means, the hinge functions \(u\mapsto(u-t)_+\), together with affine functions, generate the convex-function comparisons.

Atoms must be counted explicitly in a stop-loss expectation. If a law has atoms \(a_j\) with masses \(r_j\) and an absolutely continuous component with density \(g\), with no singular-continuous component, then

E[(X-t)_+]=\sum_j r_j(a_j-t)_+ + \int_{\mathbb R}(x-t)_+g(x)\,dx.

The formula separates the point masses from the continuous contribution; omitting the sum can change the comparison, particularly at thresholds that coincide with atoms.

Worked Example: Convex Order Without Stochastic Order

Let \(X=1/2\) with probability 1. Let \(Y=0\) with probability \(1/2\), and with probability \(1/2\) let \(Y\) be uniform on \((0,2)\). Both variables have mean \(1/2\): \(E[X]=1/2\), and \(E[Y]=(1/2)0+(1/2)(1)=1/2\).

For \(t<0\), \(E[(X-t)_+]=1/2-t\). The uniform component of \(Y\) has \(E[(U-t)_+]=1-t\) for \(t<0\), so \(E[(Y-t)_+]=(1/2)(1-t)+(1/2)(-t)=1/2-t\). For \(0\le t\le1/2\), the stop-loss values are \(E[(X-t)_+]=1/2-t\) and \(E[(Y-t)_+]=(1/2)\int_t^2(x-t)\,dx/2=(2-t)^2/8\). Their difference is \((2-t)^2/8-(1/2-t)=t^2/8+t/2\ge0\).

For \(1/2<t<2\), \(E[(X-t)_+]=0\), while \(E[(Y-t)_+]=(2-t)^2/8>0\). For \(t\ge2\), both are 0. The stop-loss inequalities hold at every threshold, and the means agree, so \(X\le_{\mathrm{cx}}Y\).

Neither stochastic ordering holds. At \(t=1/4\), \(P(X>t)=1\) but \(P(Y>t)=(1/2)(1-1/8)=7/16\), so \(X\le_{\mathrm{st}}Y\) fails. At \(t=1\), \(P(X>t)=0\) but \(P(Y>t)=1/4\), so \(Y\le_{\mathrm{st}}X\) fails. The continuous component and atom in \(Y\) together make the stop-loss comparison possible without ordering all tails in one direction.

A Practical Comparison Sequence

For mixed-support laws, a disciplined check prevents support and atom errors. First identify which order matches the question: tails for usual stochastic order, ordered density ratios for likelihood-ratio order, or convex-function expectations for convex order. Then represent every component of each law, including point masses, before comparing.

Write the full laws.
Record each atom and its probability, along with every density or other continuous component. Check that the total probability is one.

Use the criterion for the chosen order.
For stochastic order, compare CDFs or tails at every threshold. For likelihood-ratio order, use a common dominating measure and check ordered pairs, including across support gaps. For convex order, check equal finite means and all stop-loss thresholds.

Audit boundaries and zeros.
Check CDF jumps at atoms, stop-loss values at atom locations, and cross-products involving points where one density vanishes. A calculation on open intervals alone may miss the decisive case.

Common Mistakes

Checking a likelihood ratio only where its denominator is positive. This can miss support violations. Use the cross-product condition over the common measure, or state and verify explicit support conditions alongside a ratio argument.
Treating an atom as if it were part of a Lebesgue density. A point mass has zero Lebesgue measure and is not represented by an ordinary density value. Include it through a common dominating measure or a separate probability term.
Using a mean comparison as evidence of stochastic order. The mean summarizes a distribution but does not establish tail inequalities at every threshold.
Claiming convex order without matching means. Equal means are necessary. Even with equal means, verify the convex or stop-loss criterion rather than inferring the order from a variance comparison alone.

Key takeaway: Mixed supports do not require a new order definition, but they do require complete accounting. Include atoms, choose a common measure for likelihood-ratio comparisons, check support across ordered pairs, and enforce equal means for convex order.

Check Your Understanding

For each question, keep atoms and support boundaries explicit.

What does a common dominating measure accomplish when one law has an atom and another has a density?
Why does monotonicity of \(q/p\) only where \(p>0\) fail to establish likelihood-ratio order in general?
State the cross-product inequality that defines \(X\le_{\mathrm{lr}}Y\), and identify which ordered pairs it checks.
Which two conditions does the stop-loss characterization require for \(X\le_{\mathrm{cx}}Y\)?
In the convex-order example, why is the stop-loss comparison at \(t=1\) enough to rule out \(Y\le_{\mathrm{st}}X\)?