Hazard-Rate Order Through Hazard and Survival Ratios

Introduction

A lifetime distribution describes how long a component, patient, or system lasts. Its survival function gives the probability of lasting beyond a specified time, while its hazard function describes the instantaneous failure rate among those still surviving. Hazard-rate order compares two lifetime distributions through how their relative survival changes over time.

The direction of the order is important. If $X$ is below $Y$ in hazard-rate order, $X$ has at least as large a hazard as $Y$ whenever both are still at risk. This is a comparison of failure rates, not a claim that one particular realized lifetime is always shorter than another.

Hazards, Survival, and the Order

Definition: Let $T$ be a nonnegative, absolutely continuous random variable with density $f_T$, distribution function $F_T$, and survival function $S_T(t)=P(T>t)=1-F_T(t)$. At times $t$ for which $S_T(t)>0$, its hazard function is $$ h_T(t)=\frac{f_T(t)}{S_T(t)}. $$ The hazard is defined up to changes on sets of Lebesgue measure zero, because a density itself is only specified almost everywhere.

The numerator $f_T(t)$ describes probability mass per unit time near $t$, and the denominator restricts attention to individuals that have survived to $t$. Consequently, at times where the local derivative exists, $h_T(t)$ is the limiting failure probability per unit time conditional on survival to $t$. The hazard has units of inverse time; it is not itself a probability.

Definition: For nonnegative, absolutely continuous lifetimes $X$ and $Y$, write $X\le_{\mathrm{hr}}Y$ if the ratio $$ R(t)=\frac{S_Y(t)}{S_X(t)} $$ is nondecreasing for $0\le t<\tau_X$, where $\tau_X=\sup\{t\ge 0:S_X(t)>0\}$. The ratio is considered only while $X$ has positive survival probability.

This convention makes the orientation explicit: the survival function of $Y$ is in the numerator. A nondecreasing ratio means that, as time passes, survival under $Y$ becomes no smaller relative to survival under $X$. If $Y$'s survival probability reaches zero while $X$'s remains positive, the ratio falls to zero and cannot be nondecreasing; that case does not satisfy $X\le_{\mathrm{hr}}Y$.

How Survival Ratios Encode Hazards

Where both survival functions are positive, they are absolutely continuous, and so is their ratio on every compact interval contained in that region. Differentiating the ratio connects its direction of change to the hazards.

Result: At almost every $t$ for which $S_X(t)>0$ and $S_Y(t)>0$, $$ R'(t)=R(t)\bigl(h_X(t)-h_Y(t)\bigr). $$

Indeed, $S_X'(t)=-f_X(t)$ and $S_Y'(t)=-f_Y(t)$ almost everywhere. The quotient rule therefore gives $$ \begin{aligned} R'(t) &=\frac{-f_Y(t)S_X(t)+f_X(t)S_Y(t)}{S_X(t)^2}\\ &=\frac{S_Y(t)}{S_X(t)} \left(\frac{f_X(t)}{S_X(t)}-\frac{f_Y(t)}{S_Y(t)}\right)\\ &=R(t)\bigl(h_X(t)-h_Y(t)\bigr). \end{aligned} $$ Since $R(t)>0$ while both survival functions are positive, the ratio increases locally exactly when $h_X(t)\ge h_Y(t)$ almost everywhere.

Theorem: Let $X$ and $Y$ be nonnegative, absolutely continuous lifetimes. Then $X\le_{\mathrm{hr}}Y$ if and only if $\tau_X\le\tau_Y$ and $$ h_X(t)\ge h_Y(t) $$ for almost every $t\in[0,\tau_X)$.

Proof. First suppose $X\le_{\mathrm{hr}}Y$. If $\tau_Y<\tau_X$, then after $Y$'s survival function reaches zero and before $X$'s does, $R(t)=0$. At earlier times when both survival functions are positive, $R(t)>0$. This would make $R$ decrease, contradicting its nondecreasing property. Thus $\tau_X\le\tau_Y$. On every compact subinterval of $[0,\tau_X)$, both survival functions are positive and $R$ is absolutely continuous. A nondecreasing absolutely continuous function has nonnegative derivative almost everywhere. The derivative identity then implies $h_X(t)-h_Y(t)\ge0$ almost everywhere there. Taking a countable increasing sequence of such compact intervals establishes the inequality almost everywhere on $[0,\tau_X)$.

Conversely, suppose $\tau_X\le\tau_Y$ and $h_X\ge h_Y$ almost everywhere on $[0,\tau_X)$. On each compact subinterval of this interval, $R$ is absolutely continuous and $$ R'(t)=R(t)\bigl(h_X(t)-h_Y(t)\bigr)\ge0 $$ almost everywhere. An absolutely continuous function whose derivative is nonnegative almost everywhere is nondecreasing, by the fundamental theorem of calculus. Hence $R$ is nondecreasing on each such compact interval, and therefore on all of $[0,\tau_X)$. This is $X\le_{\mathrm{hr}}Y$. $\square$

Worked Comparisons

Worked Example: Exponential Lifetimes

Let $X\sim\operatorname{Exponential}(2)$ and $Y\sim\operatorname{Exponential}(1)$, where the second argument is the rate. For an exponential lifetime with rate $\lambda>0$, $S(t)=e^{-\lambda t}$, $f(t)=\lambda e^{-\lambda t}$, and $h(t)=\lambda$ for $t\ge0$.

Thus $h_X(t)=2$ and $h_Y(t)=1$. The survival ratio is $$ R(t)=\frac{e^{-t}}{e^{-2t}}=e^t. $$ It is increasing, and $h_X\ge h_Y$ at every time. Therefore $X\le_{\mathrm{hr}}Y$. In this comparison, the lifetime $X$ has the greater instantaneous failure rate throughout.

Worked Example: Weibull Lifetimes With the Same Shape

A Weibull lifetime with shape $k>0$ and scale $\lambda>0$ has survival function $S(t)=\exp[-(t/\lambda)^k]$, for $t\ge0$. Compare $X$, with shape $2$ and scale $1$, to $Y$, with shape $2$ and scale $2$.

Their survival functions and hazards are $$ \begin{aligned} S_X(t)&=e^{-t^2}, & h_X(t)&=2t,\\ S_Y(t)&=e^{-t^2/4}, & h_Y(t)&=\frac{t}{2}. \end{aligned} $$ For every $t\ge0$, $2t\ge t/2$. Equivalently, the ratio is $$ R(t)=\frac{e^{-t^2/4}}{e^{-t^2}} =e^{3t^2/4}, $$ which is nondecreasing. Hence $X\le_{\mathrm{hr}}Y$. Both hazards start at zero, but the hazard of $X$ is larger for every positive time.

Worked Example: Uniform Lifetimes With Different Endpoints

Let $X$ be uniform on $[0,1]$ and $Y$ uniform on $[0,2]$. For a uniform lifetime on $[0,a]$, the density is $1/a$ on that interval and the survival function is $1-t/a$ for $0\le t<a$. Therefore, for $0\le t<1$, $$ h_X(t)=\frac{1}{1-t}, \qquad h_Y(t)=\frac{1}{2-t}. $$ Since $1-t<2-t$ and both are positive, $h_X(t)>h_Y(t)$. The survival ratio on the interval where $X$ survives is $$ R(t)=\frac{1-t/2}{1-t}. $$ Differentiating gives $$ R'(t)=\frac{1}{2(1-t)^2}>0. $$ Thus $X\le_{\mathrm{hr}}Y$. The ratio is considered only for $t<1$, the endpoint of $X$'s positive-survival interval; it grows without bound as $t$ approaches $1$.

Worked Example: Crossing Hazards Do Not Give an Order

Let $X$ be exponential with rate $1$, and let $Y$ be Weibull with shape $2$ and scale $1$. Then $$ S_X(t)=e^{-t},\quad h_X(t)=1, \qquad S_Y(t)=e^{-t^2},\quad h_Y(t)=2t. $$ The hazards are equal at $t=1/2$. For $t<1/2$, $h_X(t)>h_Y(t)$, while for $t>1/2$, $h_X(t)<h_Y(t)$. The survival ratio confirms that it changes direction: $$ R(t)=e^{-t^2+t}, \qquad R'(t)=e^{-t^2+t}(1-2t). $$ It increases before $1/2$ and decreases afterward, so it is not nondecreasing. Therefore $X\not\le_{\mathrm{hr}}Y$. The reverse order also fails: it would require $h_Y\ge h_X$ almost everywhere, which is false for $t<1/2$. The two lifetimes are incomparable in hazard-rate order.

Interpreting the Order and Avoiding Pitfalls

Hazard-rate order is useful when the question concerns how risk evolves among survivors. For example, a larger hazard for $X$ means a larger instantaneous failure rate for $X$ at times when both populations still contain survivors. The order requires this comparison to hold consistently across time, as captured by the monotonicity of the survival ratio.

Reversing the ratio reverses the direction. With $X\le_{\mathrm{hr}}Y$, the ratio is $S_Y/S_X$, not $S_X/S_Y$. In the exponential example, $S_Y/S_X=e^t$ increases; its reciprocal decreases.
A hazard is not a survival probability. Comparing $h_X$ and $h_Y$ concerns conditional instantaneous rates, not the probability that a randomly selected lifetime exceeds a fixed time.
A comparison at one time is not an order. A single inequality such as $h_X(1)\ge h_Y(1)$ does not establish the required comparison over the full interval where both can survive.
Crossing hazards matter. If one hazard is larger early and smaller later, the survival ratio may increase and then decrease. The crossing example shows why a single hazard-rate order may not exist.

Key takeaway: For absolutely continuous lifetimes, $X\le_{\mathrm{hr}}Y$ means that $S_Y(t)/S_X(t)$ is nondecreasing while $X$ can survive. Equivalently, $X$'s hazard is at least $Y$'s almost everywhere on that interval, with $X$'s positive-survival endpoint no later than $Y$'s.

Check Your Understanding

Use the survival-ratio definition and the hazard comparison carefully.

For exponential lifetimes with rates $3$ and $2$, which lifetime is below the other in hazard-rate order? Show the survival ratio.
Why must the hazard comparison be made almost everywhere rather than at every individual time?
For the uniform lifetimes on $[0,1]$ and $[0,2]$, on which time interval is the survival ratio used to define the order?
What feature of the hazard functions in the final worked example prevents either lifetime from being below the other in hazard-rate order?
If $S_Y(t)/S_X(t)$ is decreasing on some interval, what does that tell you about the sign of $h_X(t)-h_Y(t)$ almost everywhere on that interval where both survival functions are positive?