Why Hazard-Rate Order Implies Stochastic Order

From Local Hazard to Survival Probability

A hazard rate describes an instantaneous rate of failure among individuals who have survived up to a given time. A survival function, by contrast, gives the probability of lasting beyond a specified time. The key connection is that the hazard accumulates over time: integrating it gives the cumulative hazard, which determines survival through an exponential transformation. This turns a pointwise hazard comparison into a comparison of entire survival curves.

The direction of the comparison matters. A larger hazard means that failure is occurring at a higher instantaneous rate among those still at risk. Under suitable regularity conditions, this leads to a smaller survival probability at every time. The result is a useful bridge from hazard-rate order to the usual stochastic order.

Definition: Let \(X\) and \(Y\) be nonnegative, absolutely continuous lifetimes with no mass at zero and with survival functions \(S_X(t)=P(X>t)\) and \(S_Y(t)=P(Y>t)\) positive for every finite \(t\geq 0\). Write their densities as \(f_X\) and \(f_Y\), and define their hazards, almost everywhere, by \(h_X(t)=f_X(t)/S_X(t)\) and \(h_Y(t)=f_Y(t)/S_Y(t)\). We say \(X\leq_{\mathrm{hr}}Y\) if \(h_X(t)\geq h_Y(t)\) for almost every \(t\geq 0\). The usual stochastic order \(X\leq_{\mathrm{st}}Y\) means \(P(X>t)\leq P(Y>t)\) for every real \(t\).

The hazard-order convention is worth keeping in view: the hazard of \(X\) is at least the hazard of \(Y\), so \(X\) is the lifetime that is stochastically smaller. Hazards are only defined up to changes on sets of Lebesgue measure zero, which is why their comparison is stated almost everywhere. The positivity of each survival function ensures that the hazard is defined throughout every finite time interval.

Cumulative Hazard and the Survival Function

For a lifetime \(X\) satisfying the assumptions above, define its cumulative hazard by \(H_X(t)=\int_0^t h_X(u)\,du\). The integral adds the instantaneous hazard over the interval from zero to \(t\). The following identity explains why comparing hazards is enough.

Result: For every finite \(t\geq 0\), \(H_X(t)=-\log S_X(t)\), and hence \(S_X(t)=\exp\{-H_X(t)\}\).

To justify the identity, fix any finite \(T>0\). The survival function \(S_X\) is absolutely continuous on \([0,T]\), with derivative \(S_X'(t)=-f_X(t)\) almost everywhere. It is continuous and positive on this compact interval, so it has a positive minimum there. Consequently, \(\log S_X\) is absolutely continuous on \([0,T]\), and the chain rule for absolutely continuous functions gives

\frac{d}{dt}\log S_X(t) =\frac{S_X'(t)}{S_X(t)} =-\frac{f_X(t)}{S_X(t)} =-h_X(t) \quad\text{for almost every }t\in[0,T].

Integrating this equality from zero to \(t\leq T\), and using \(S_X(0)=P(X>0)=1\), yields

\log S_X(t)-\log S_X(0) =-\int_0^t h_X(u)\,du, \qquad \log S_X(t)=-H_X(t).

Exponentiating gives the survival representation. Since \(T\) was arbitrary, it holds for every finite \(t\geq 0\).

Worked Example: Comparing Exponential Lifetimes

Let \(X\sim\operatorname{Exponential}(2)\) and \(Y\sim\operatorname{Exponential}(1)\), where the parameter is the rate. For a rate \(\lambda>0\), the density is \(f(t)=\lambda e^{-\lambda t}\) and the survival function is \(S(t)=e^{-\lambda t}\) for \(t\geq0\). Thus its hazard is

h(t)=\frac{\lambda e^{-\lambda t}}{e^{-\lambda t}}=\lambda.

It follows that \(h_X(t)=2\) and \(h_Y(t)=1\) at every \(t\geq0\), so \(X\leq_{\mathrm{hr}}Y\). The cumulative hazards are \(H_X(t)=2t\) and \(H_Y(t)=t\). Therefore,

S_X(t)=e^{-2t}\leq e^{-t}=S_Y(t) \quad\text{for every }t\geq0.

For instance, at \(t=1\), the survival probabilities are \(e^{-2}\approx0.1353\) and \(e^{-1}\approx0.3679\). The hazard comparison therefore gives the claimed stochastic comparison, not merely a comparison at one selected time.

Hazard-Rate Order Implies Usual Stochastic Order

The main implication follows directly by integrating the hazard inequality. The regularity assumptions in the definition ensure that the cumulative-hazard identity applies to both lifetimes.

Theorem: Suppose \(X\) and \(Y\) are nonnegative, absolutely continuous lifetimes, have no mass at zero, and satisfy \(S_X(t)>0\) and \(S_Y(t)>0\) for every finite \(t\geq0\). If \(h_X(t)\geq h_Y(t)\) for almost every \(t\geq0\), then \(X\leq_{\mathrm{st}}Y\).

Proof. For any fixed \(t\geq0\), integrate the assumed hazard inequality over \([0,t]\). This gives

H_X(t)=\int_0^t h_X(u)\,du \geq \int_0^t h_Y(u)\,du=H_Y(t).

By the cumulative-hazard identity, \(S_X(t)=e^{-H_X(t)}\) and \(S_Y(t)=e^{-H_Y(t)}\). The exponential function is increasing, so the function \(v\mapsto e^{-v}\) is decreasing. Applying it to the cumulative-hazard inequality yields

P(X>t)=S_X(t)=e^{-H_X(t)} \leq e^{-H_Y(t)}=S_Y(t)=P(Y>t).

This holds for every \(t\geq0\). For \(t<0\), nonnegativity gives \(P(X>t)=P(Y>t)=1\). Thus the survival inequality holds for every real \(t\), which is precisely \(X\leq_{\mathrm{st}}Y\). \(\square\)

The proof also identifies when the comparison is strict. If \(h_X\geq h_Y\) almost everywhere and, for a particular \(t\), the difference \(h_X-h_Y\) has a strictly positive integral over \([0,t]\), then \(H_X(t)>H_Y(t)\), so \(S_X(t)<S_Y(t)\). In particular, if the hazard inequality is strict on a set of positive measure within \([0,t]\), then the survival inequality is strict at that \(t\). A strict hazard inequality at just one isolated time is not enough: changing a hazard at one point does not change its integral.

Worked Example: Comparing Weibull Lifetimes

Consider Weibull lifetimes with survival function \(S(t)=\exp\{-(t/\lambda)^k\}\) for \(t\geq0\), where \(k>0\) is the shape and \(\lambda>0\) is the scale. Take \(X\) to have \(k=2,\lambda=1\), and \(Y\) to have \(k=2,\lambda=2\). Differentiating the survival function gives the density \(f(t)=2t\lambda^{-2}\exp\{-(t/\lambda)^2\}\), and hence

h(t)=\frac{f(t)}{S(t)}=\frac{2t}{\lambda^2}.

Therefore \(h_X(t)=2t\), while \(h_Y(t)=t/2\). For every \(t\geq0\), \(2t\geq t/2\), so \(X\leq_{\mathrm{hr}}Y\). The cumulative hazards and survival functions are

H_X(t)=\int_0^t 2u\,du=t^2, \qquad H_Y(t)=\int_0^t \frac{u}{2}\,du=\frac{t^2}{4}, $$ $$ S_X(t)=e^{-t^2} \leq e^{-t^2/4}=S_Y(t).

At \(t=2\), the two survival probabilities are \(e^{-4}\approx0.0183\) and \(e^{-1}\approx0.3679\). They are equal at \(t=0\), as both lifetimes survive beyond zero with probability one, and the inequality is strict for every \(t>0\).

A Survival-Ratio View and a Piecewise Example

The same integration argument gives a useful ratio identity. Under the theorem's assumptions, for every finite \(t\geq0\),

\frac{S_X(t)}{S_Y(t)} =\exp\!\left\{-\int_0^t\bigl(h_X(u)-h_Y(u)\bigr)\,du\right\}.

If \(h_X\geq h_Y\) almost everywhere, the integral on the right is nondecreasing in \(t\). Its negative exponential is therefore nonincreasing, and the survival ratio starts at one. In particular, it cannot rise above one. This is another way to see why the survival curves have the required order. The identity also makes clear that the hazard comparison controls how the ratio changes over time, while the stochastic-order conclusion compares the survival probabilities themselves.

Worked Example: Piecewise-Constant Hazards

Suppose \(X\) and \(Y\) have hazards

h_X(t)= \begin{cases} 1,&0\leq t<1,\\ 2,&t\geq1, \end{cases} \qquad h_Y(t)= \begin{cases} \frac12,&0\leq t<1,\\ 1,&t\geq1. \end{cases}

Both hazards are positive, and \(h_X(t)\geq h_Y(t)\) on both intervals. Their cumulative hazards are obtained by integrating each constant segment. For \(0\leq t<1\),

H_X(t)=t,\qquad H_Y(t)=\frac{t}{2}, \qquad S_X(t)=e^{-t}\leq e^{-t/2}=S_Y(t).

For \(t\geq1\), include the contribution from the first interval and then integrate the hazard after time one:

H_X(t)=1+2(t-1)=2t-1, \qquad H_Y(t)=\frac12+(t-1)=t-\frac12.

Their difference is \(H_X(t)-H_Y(t)=t-\frac12\), which is positive for \(t\geq1\). Consequently,

S_X(t)=e^{-(2t-1)} \leq e^{-(t-1/2)}=S_Y(t) \quad (t\geq1).

At \(t=1.5\), the cumulative hazards are \(H_X(1.5)=2\) and \(H_Y(1.5)=1\), giving \(S_X(1.5)=e^{-2}\approx0.1353\) and \(S_Y(1.5)=e^{-1}\approx0.3679\). The comparison holds across the change in hazard at time one because the accumulated hazards include both time segments.

Why the Implication Matters

Usual stochastic order summarizes a global comparison of tails: at every threshold, the probability that \(X\) exceeds that threshold is no larger than the corresponding probability for \(Y\). The theorem supplies a sufficient route to that conclusion from a local comparison among survivors at each time. Once the usual stochastic order is established, its standard increasing-test-function characterization can be used when that is the relevant way to express the comparison. The hazard calculation is especially useful for lifetime models, where a model may describe the failure rate more directly than it describes tail probabilities.

The result is an implication, not an equivalence. A survival comparison alone need not give the pointwise hazard comparison: survival curves can remain ordered even when their local rates cross. The hazard-rate condition is stronger because it compares those rates throughout time, rather than comparing only the accumulated probabilities.

Common Mistakes and Assumption Checks

Reversing the sign. The convention here is \(h_X\geq h_Y\), which yields \(X\leq_{\mathrm{st}}Y\). A larger hazard accumulates a larger \(H_X\), and the negative exponential then gives a smaller \(S_X\).
Comparing hazards at one time only. A pointwise comparison at a single time does not establish an integral inequality. The hypothesis must hold almost everywhere over the time intervals being compared.
Confusing hazard order with stochastic order. The theorem proves that hazard-rate order implies usual stochastic order under the stated assumptions. It does not claim that the converse holds.
Ignoring support and positivity. The proof uses \(h=f/S\) and \(\log S\), so it requires positive survival on each finite interval. If a lifetime has a finite right endpoint, its hazard may cease to be defined after that endpoint; the theorem as stated does not cover that situation without a separate support analysis.
Treating strictness as automatic at every time. The two survival functions agree at zero. Strict ordering at a later time requires a strictly positive accumulated hazard difference by that time.

Key takeaway: For the stated class of continuous lifetimes, hazard-rate order gives \(h_X\geq h_Y\) almost everywhere. Integration yields \(H_X\geq H_Y\); applying \(S=e^{-H}\) reverses the inequality and gives \(S_X\leq S_Y\), or \(X\leq_{\mathrm{st}}Y\).

Check Your Understanding

Use the direction of the hazard comparison and the cumulative-hazard identity to answer each question.

If \(h_X(t)\geq h_Y(t)\) almost everywhere, which cumulative hazard is larger, and which survival function is smaller?
Two exponential lifetimes have rates \(3\) and \(2\), respectively. Which lifetime is smaller in hazard-rate order, and what are their survival functions?
Why does a hazard inequality that differs only at one isolated time leave the cumulative-hazard comparison unchanged?
If \(H_X(t)>H_Y(t)\) at a particular time, what strict inequality follows for the survival probabilities at that time?
Does the theorem, as stated, cover a lifetime whose survival function reaches zero at a finite time? Explain which assumption needs attention.