Usual Stochastic Order from CDF and Survival Functions

Comparing Distributions Through Their Tails

Two random variables can have very different distributions yet still be ordered in a useful sense: one may be less likely than the other to exceed every threshold. This comparison is called the usual stochastic order. It appears in reliability, risk analysis, and comparisons of outcomes, where the question is often not whether two distributions have the same shape, but whether one consistently places more probability on larger values.

The direction can initially seem counterintuitive. If \(X\) is stochastically smaller than \(Y\), then \(X\) has a larger CDF at each point and a smaller upper tail. The CDF accumulates probability from the left; the survival function measures probability still lying to the right.

Definition: Let \(X\) and \(Y\) be real-valued random variables, not necessarily defined on the same probability space. We say that \(X\) is less than or equal to \(Y\) in the usual stochastic order, and write \(X\le_{\mathrm{st}}Y\), if \(F_X(x)\ge F_Y(x)\) for every \(x\in\mathbb R\), where \(F_X(x)=P(X\le x)\) and \(F_Y(x)=P(Y\le x)\). Equivalently, using the strict survival functions \(S_X(x)=P(X>x)\) and \(S_Y(x)=P(Y>x)\), the condition is \(S_X(x)\le S_Y(x)\) for every real \(x\).

Why the CDF and Survival Conditions Agree

For any real-valued random variable and any real \(x\), the events \(\{X\le x\}\) and \(\{X>x\}\) are complements. Therefore, \(S_X(x)=1-F_X(x)\), and the same identity holds for \(Y\). Subtracting a CDF inequality from one reverses its direction. This gives the basic equivalence, with no assumption that either distribution is continuous.

Theorem: For real-valued \(X\) and \(Y\), \(X\le_{\mathrm{st}}Y\) if and only if \(P(X>x)\le P(Y>x)\) for every \(x\in\mathbb R\).

Start with the CDF condition.
Assume \(F_X(x)\ge F_Y(x)\) for every real \(x\). Since \(P(X>x)=1-F_X(x)\), subtracting gives \(P(X>x)\le P(Y>x)\) for every \(x\).

Reverse the argument.
If \(P(X>x)\le P(Y>x)\) for every real \(x\), then \(1-F_X(x)\le 1-F_Y(x)\). Rearranging gives \(F_X(x)\ge F_Y(x)\) everywhere, which is exactly \(X\le_{\mathrm{st}}Y\). \(\square\)

The word “every” is essential. A comparison at a single threshold, or even at several selected thresholds, does not establish the usual stochastic order. The criterion compares the entire upper-tail profiles.

Tail Sums for Integer-Valued Distributions

Suppose \(X\) and \(Y\) take values in the integers, with probability mass functions \(p_X(j)=P(X=j)\) and \(p_Y(j)=P(Y=j)\). At an integer threshold \(k\), the event \(X>k\) consists of the disjoint possibilities \(X=k+1,k+2,\ldots\). Thus its probability is a sum of the corresponding masses.

Tail-sum criterion: If \(X\) and \(Y\) are integer-valued, then \(X\le_{\mathrm{st}}Y\) if and only if \(\sum_{j=k+1}^{\infty}p_X(j)\le\sum_{j=k+1}^{\infty}p_Y(j)\) for every integer \(k\). Equivalently, \(\sum_{j=-\infty}^{k}p_X(j)\ge\sum_{j=-\infty}^{k}p_Y(j)\) for every integer \(k\).

To verify that checking integer thresholds is enough, take any real \(x\) and let \(k=\lfloor x\rfloor\), the greatest integer not exceeding \(x\). For integer-valued variables, \(X>x\) holds exactly when \(X>k\), and likewise for \(Y\). Therefore, the integer-threshold tail comparisons imply the comparisons at every real threshold. Conversely, if the comparisons hold for every real threshold, they hold in particular at each integer \(k\). The cumulative-sum expression follows because \(P(X\le k)=\sum_{j=-\infty}^{k}p_X(j)\), and the CDF condition is the reverse inequality to the tail condition.

Worked Example: Ordering Two Finite Discrete Laws

Let \(X\) have probabilities \(P(X=0)=0.5\), \(P(X=1)=0.3\), and \(P(X=2)=0.2\). Let \(Y\) have probabilities \(P(Y=1)=0.2\), \(P(Y=2)=0.5\), and \(P(Y=3)=0.3\). Determine whether \(X\le_{\mathrm{st}}Y\).

For integer thresholds \(k=0,1,2\), calculate the strict upper tails:

\begin{aligned} P(X>0)&=0.3+0.2=0.5, & P(Y>0)&=0.2+0.5+0.3=1,\\ P(X>1)&=0.2, & P(Y>1)&=0.5+0.3=0.8,\\ P(X>2)&=0, & P(Y>2)&=0.3. \end{aligned}

Each tail for \(X\) is no larger than the corresponding tail for \(Y\). For \(k<0\), both variables exceed \(k\) with probability one; for \(k\ge3\), both tails are zero. The criterion therefore holds at every integer threshold, so \(X\le_{\mathrm{st}}Y\).

For a CDF check at \(x=1\), \(F_X(1)=0.5+0.3=0.8\), while \(F_Y(1)=0.2\). The larger CDF value for \(X\) agrees with the stochastic-order direction.

Tail Integrals for Absolutely Continuous Distributions

If \(X\) has a probability density function \(f_X\), then its probability above \(x\) is the integral of the density from \(x\) to infinity. This turns the survival-function comparison into a direct comparison of tail integrals. The same applies to \(Y\) when it also has a density.

Tail-integral criterion: If \(X\) and \(Y\) are absolutely continuous with densities \(f_X\) and \(f_Y\), then \(X\le_{\mathrm{st}}Y\) if and only if \(\int_x^\infty f_X(t)\,dt\le\int_x^\infty f_Y(t)\,dt\) for every \(x\in\mathbb R\).

Indeed, absolute continuity gives \(P(X>x)=\int_x^\infty f_X(t)\,dt\); a single point has probability zero, so using \(>\) rather than \(\ge\) makes no difference for these laws. Substituting this identity for each variable into the survival-function criterion proves both directions. The integral criterion does not require the density functions themselves to be ordered pointwise. It is the integrated tails that must be ordered. For distributions without densities, the probability-tail criterion remains valid even though a density integral may not be available.

Worked Example: Comparing Two Exponential Lifetimes

Let \(X\sim\operatorname{Exp}(2)\) and \(Y\sim\operatorname{Exp}(1)\), where the rate parameter is the number following \(\operatorname{Exp}\). Their densities are \(f_X(t)=2e^{-2t}\) and \(f_Y(t)=e^{-t}\) for \(t\ge0\), and zero for \(t<0\). Decide whether \(X\le_{\mathrm{st}}Y\).

For \(x<0\), both variables are nonnegative, so \(P(X>x)=P(Y>x)=1\). For \(x\ge0\), integrate each density:

P(X>x)=\int_x^\infty 2e^{-2t}\,dt=e^{-2x}, \qquad P(Y>x)=\int_x^\infty e^{-t}\,dt=e^{-x}.

Since \(x\ge0\), \(e^{-2x}\le e^{-x}\): equivalently, \(e^{-2x}=e^{-x}e^{-x}\) and \(e^{-x}\le1\). Thus \(P(X>x)\le P(Y>x)\) for all real \(x\), and \(X\le_{\mathrm{st}}Y\). At \(x=1\), the two tails are \(e^{-2}\approx0.1353\) and \(e^{-1}\approx0.3679\). The faster-decaying lifetime has the smaller chance of exceeding the threshold.

Worked Example: Bernoulli Variables and the Order Direction

Let \(X\sim\operatorname{Bernoulli}(0.35)\) and \(Y\sim\operatorname{Bernoulli}(0.60)\), so each variable is zero or one and its parameter is the probability of one. Check the order directly.

If \(x<0\), both strict tails equal one. If \(0\le x<1\), exceeding \(x\) means taking the value one, so \(P(X>x)=0.35\) and \(P(Y>x)=0.60\). If \(x\ge1\), both tails are zero. Hence \(P(X>x)\le P(Y>x)\) at every real \(x\), and \(X\le_{\mathrm{st}}Y\).

The CDFs give the same conclusion: on \(0\le x<1\), \(F_X(x)=0.65\) and \(F_Y(x)=0.40\), so \(F_X(x)\ge F_Y(x)\). This example also shows why atoms require care. At \(x=0\), \(P(X>0)=0.35\), whereas \(P(X\ge0)=1\); these are different events.

Inclusive Tails, Atoms, and Common Mistakes

Some settings define a tail using \(P(X\ge x)\) rather than \(P(X>x)\). These conventions must not be mixed at an atom, where \(P(X=x)\) can be positive. Nevertheless, the inclusive-tail comparison is also equivalent when imposed at every threshold: \(X\le_{\mathrm{st}}Y\) if and only if \(P(X\ge t)\le P(Y\ge t)\) for all real \(t\). To see the forward implication, take \(x<t\). Then \(\{X\ge t\}\subseteq\{X>x\}\), so \(P(X\ge t)\le P(X>x)\le P(Y>x)\). Let \(x\) increase to \(t\); the events \(\{Y>x\}\) decrease to \(\{Y\ge t\}\), and continuity of probability for decreasing events gives the desired inequality. The reverse implication follows by taking \(t\) down to \(x\): the events \(\{X\ge t\}\) increase to \(\{X>x\}\), and likewise for \(Y\). The strict-tail criterion then applies.

Worked Example: Equal Means Do Not Establish the Order

Let \(X\) take values 0 and 2, each with probability \(1/2\), and let \(Y=1\) with probability one. Both means equal one: \(E[X]=(0)(1/2)+(2)(1/2)=1\) and \(E[Y]=1\). Does either stochastic order hold?

At threshold \(x=1\), \(P(X>1)=1/2\) but \(P(Y>1)=0\), so the condition for \(X\le_{\mathrm{st}}Y\) fails. For the reverse order, check \(x=0\): \(P(Y>0)=1\), whereas \(P(X>0)=1/2\). The condition \(P(Y>x)\le P(X>x)\) fails. Thus neither order holds, despite the equal means.

Two errors are especially common:

Reversing the CDF inequality. For \(X\le_{\mathrm{st}}Y\), the correct directions are \(F_X(x)\ge F_Y(x)\) and \(P(X>x)\le P(Y>x)\). A smaller variable accumulates probability sooner in the CDF and has less probability beyond each threshold.
Checking only one tail or comparing pointwise densities. The order requires a comparison at every threshold. For density-based laws, comparing \(f_X(x)\) and \(f_Y(x)\) pointwise is not the criterion; compare the integrals from each threshold to infinity.

The tail formulation makes the meaning of the order concrete: at no threshold does \(X\) have a greater chance than \(Y\) of lying above it. For integer-valued laws, this becomes a comparison of sums of masses; for absolutely continuous laws, it becomes a comparison of integrated densities. In both cases, the same all-threshold requirement and the same direction of inequality govern the conclusion.

Check Your Understanding

Use the CDF or tail criterion, paying attention to whether thresholds include atoms.

State the CDF and strict-survival inequalities that define \(X\le_{\mathrm{st}}Y\).
For integer-valued variables, write \(P(X>k)\) as a sum of mass-function values. Why is it enough to check integer \(k\)?
What tail-integral comparison characterizes the order when both laws have densities?
If \(X\) and \(Y\) are integer-valued and \(P(X>k)\le P(Y>k)\) for every integer \(k\), what can be concluded about their CDFs?
Why can replacing \(P(X>x)\) with \(P(X\ge x)\) change a probability at an atom?