Why a Density Comparison Controls a Tail Comparison
A density ratio compares the relative probability assigned by two distributions at each point. Hazard-rate order, by contrast, compares their remaining survival probabilities above each threshold. The connection is that a survival ratio can be written as an average of the density ratio over the corresponding upper tail. If the density ratio increases with the observation, moving the threshold to the right removes relatively smaller ratio values and raises that average.
We make this argument under common-support conditions so the density ratio is meaningful throughout the region being compared. The result gives a direct route from likelihood-ratio order to hazard-rate order; it does not require computing either hazard function explicitly.
Definitions and Common-Support Conditions
The phrase “has a nondecreasing version” allows densities, and therefore their ratio, to be specified only up to changes on sets of Lebesgue measure zero. The ratio is considered where \(f_X>0\); under the stated assumptions this is almost everywhere on \(I\). Since \(f_Y=r f_X\) almost everywhere and both densities integrate to one, \(E[r(X)]=1\).
For \(t\in I\), the common-support assumptions give \(S_X(t)>0\) and \(S_Y(t)>0\). For \(t\le a\), if \(a\) is finite, both survivals equal one. Thresholds at or beyond a finite upper endpoint have zero survival under both distributions, so the survival ratio there is not defined. The hazard-rate comparison is understood on the region where the survivals are positive.
The Survival Ratio Is a Tail Average
Indeed, \(f_Y(x)=r(x)f_X(x)\) almost everywhere on \(I\), so integrating each density over \((t,b)\) gives the identity. The denominator is positive, which makes the conditional expectation well defined. The expression says that the survival ratio at threshold \(t\) is the average density ratio among \(X\)-values that exceed \(t\).
Worked Example: Exponential Lifetimes
Let \(X\sim\operatorname{Exp}(2)\) and \(Y\sim\operatorname{Exp}(1)\), where an exponential distribution with rate \(\lambda\) has density \(f(x)=\lambda e^{-\lambda x}\) for \(x>0\). Both distributions have support \((0,\infty)\). Their density ratio is $$ r(x)=\frac{f_Y(x)}{f_X(x)} =\frac{e^{-x}}{2e^{-2x}} =\frac{1}{2}e^x,\qquad x>0. $$ Since \(r'(x)=\tfrac12 e^x>0\), the ratio is increasing. Directly, their survival functions are \(S_X(t)=e^{-2t}\) and \(S_Y(t)=e^{-t}\) for \(t\ge0\), so $$ \frac{S_Y(t)}{S_X(t)} =\frac{e^{-t}}{e^{-2t}} =e^t. $$ This ratio is increasing, establishing \(X\le_{\mathrm{hr}}Y\). The corresponding hazards are \(h_X(t)=2\) and \(h_Y(t)=1\), consistent with the direction: \(X\) has the greater instantaneous failure rate.
Why the Tail Average Increases
Here is a proof using integrals over two adjacent parts of the support. It does not require differentiating a density or interchanging a derivative and an integral.
Fix \(s,t\in I\) with \(s<t\). Write \(C=\int_s^t f_X(x)\,dx\) and \(D=\int_t^b f_X(x)\,dx\). Both are positive because \(f_X>0\) almost everywhere on \(I\).
Let \(c=C^{-1}\int_s^t r(x)f_X(x)\,dx\) and \(d=D^{-1}\int_t^b r(y)f_X(y)\,dy\). Monotonicity of \(r\) gives \(r(x)\le r(y)\) for almost every \(x\in(s,t)\) and \(y\in(t,b)\). Multiplying by \(f_X(x)f_X(y)\ge0\) and integrating yields \(CDc\le CDd\), hence \(c\le d\).
The average over \((s,b)\) is the weighted average \((Cc+Dd)/(C+D)\). Since \(c\le d\), this is at most \(d\), the average over \((t,b)\). By the tail-average identity, these two averages are \(S_Y(s)/S_X(s)\) and \(S_Y(t)/S_X(t)\), respectively. Thus the survival ratio is nondecreasing, which is precisely \(X\le_{\mathrm{hr}}Y\). \(\square\)
The comparison has a simple interpretation: the average over the tail above \(s\) combines values from \((s,t)\) and \((t,b)\), while every density-ratio value in the first region is no larger than those in the second. Removing the first region cannot lower the average.
Connection With Hazard Functions
For an absolutely continuous distribution with \(S_X(t)>0\), its hazard function is \(h_X(t)=f_X(t)/S_X(t)\), defined almost everywhere. The survival-ratio condition is equivalent, under these regularity conditions, to the hazard comparison \(h_X(t)\ge h_Y(t)\) almost everywhere. To verify the connection, write \(R(t)=S_Y(t)/S_X(t)\). On any compact subinterval where \(S_X\) and \(S_Y\) are positive, the survival functions are absolutely continuous and \(R\) is absolutely continuous. The quotient rule therefore gives, almost everywhere, $$ R'(t) =\frac{-f_Y(t)S_X(t)+f_X(t)S_Y(t)}{S_X(t)^2} =R(t)\bigl(h_X(t)-h_Y(t)\bigr). $$ Since \(R(t)>0\), the derivative is nonnegative exactly when \(h_X(t)\ge h_Y(t)\). An absolutely continuous function is nondecreasing if and only if its derivative is nonnegative almost everywhere, so the two criteria agree. In particular, the theorem proves hazard-rate order through the survival-ratio definition and also yields the hazard inequality almost everywhere.
Worked Example: Uniform Versus Triangular Density
Let \(X\sim\operatorname{Uniform}(0,1)\), so \(f_X(x)=1\) on \((0,1)\), and let \(Y\) have density \(f_Y(y)=2y\) on \((0,1)\). This is a density because \(\int_0^1 2y\,dy=1\). The density ratio is $$ r(x)=\frac{2x}{1}=2x, $$ which is increasing on the common support. For \(0\le t<1\), $$ S_X(t)=1-t,\qquad S_Y(t)=\int_t^1 2y\,dy=1-t^2. $$ Consequently, $$ \frac{S_Y(t)}{S_X(t)} =\frac{(1-t)(1+t)}{1-t} =1+t. $$ The ratio increases from \(1\) at \(t=0\), so \(X\le_{\mathrm{hr}}Y\). The hazards provide a check: \(h_X(t)=1/(1-t)\) and \(h_Y(t)=2t/(1-t^2)=2t/[(1-t)(1+t)]\). Their difference is $$ h_X(t)-h_Y(t) =\frac{1+t-2t}{(1-t)(1+t)} =\frac{1}{1+t}>0,\qquad 0\le t<1. $$
Worked Example: Shifted Normal Distributions
Let \(X\sim N(0,1)\) and \(Y\sim N(1,1)\). Both densities are positive on the whole real line, so they have common support. Using the normal density formula, $$ \frac{f_Y(x)}{f_X(x)} =\frac{\exp\{-(x-1)^2/2\}}{\exp\{-x^2/2\}} =\exp\left(x-\frac12\right). $$ The ratio is increasing because its derivative is \(\exp(x-\tfrac12)>0\). If \(\Phi\) denotes the standard normal distribution function, then $$ S_X(t)=1-\Phi(t),\qquad S_Y(t)=1-\Phi(t-1). $$ The theorem establishes that \([1-\Phi(t-1)]/[1-\Phi(t)]\) increases for every real \(t\), and hence \(X\le_{\mathrm{hr}}Y\). For a numerical check at \(t=0\), \(S_X(0)=0.5\) and \(S_Y(0)=\Phi(1)\approx0.8413\), giving a ratio of about \(1.6826\). This value alone does not prove monotonicity; the increasing density ratio and theorem establish it at all thresholds.
Common Mistake and Practical Meaning
A frequent error is to reverse the density ratio partway through the argument. The theorem uses \(f_Y/f_X\) increasing to conclude that \(S_Y/S_X\) is increasing, which means \(X\le_{\mathrm{hr}}Y\). If instead \(f_X/f_Y\) is increasing, the corresponding conclusion has the distributions reversed, provided the same support conditions hold.
The common-support condition is substantive, not cosmetic: the ratio \(f_Y/f_X\) must be comparable across the region in which the tail averages are taken. If one distribution has regions of support absent from the other, the ratio may not be defined there and the integral comparison above may fail. Support complications require separate care; the proof here applies to the stated common-support setting.
Check Your Understanding
Use the tail-average argument and keep the ratio orientation explicit.
- Why can \(S_Y(t)/S_X(t)\) be written as \(E[r(X)\mid X>t]\) when \(r=f_Y/f_X\)?
- In the theorem’s proof, why is the average of \(r\) on \((t,b)\) at least its average on \((s,t)\) when \(s<t\)?
- If \(f_X/f_Y\) is increasing on a common support, which distribution is smaller in hazard-rate order?
- For \(X\sim\operatorname{Uniform}(0,1)\) and \(f_Y(y)=2y\), calculate \(S_Y(t)/S_X(t)\) for \(0\le t<1\).
- Why does knowing the survival ratio at one threshold, even if calculated exactly, not establish hazard-rate order?