Why mixture hazards can cross after averaging

Averaging two exponential rates can order the hazards of a two-component mixture. Add the same slower component to both mixtures, and the hazard curves can cross. The unchanged component alters the weights among the survivors, which changes the comparison.

The example below proves this with exact fractions. It also explains why a proof about a sum cannot automatically be reused for a ratio of sums.

A mixture's hazard changes with its survivors

An exponential component with rate $\lambda$ has survival probability $e^{-\lambda t}$ . A population with component weights $p_i$ has survival function

S(t)=\sum_i p_i e^{-\lambda_i t},\qquad \sum_i p_i=1.

Its hazard is the instantaneous failure rate among the units still present:

h(t)=-\frac{S'(t)}{S(t)} =\frac{\sum_i p_i\lambda_i e^{-\lambda_i t}} {\sum_i p_i e^{-\lambda_i t}}.

The original weights are fixed. The weights among survivors are proportional to $p_i e^{-\lambda_i t}$ , so faster components lose influence over time. Even a component that is identical in both populations affects the denominator of each hazard.

Two components, then three

Begin with equal-weight mixtures having rates $(3,5)$ and $(4,4)$ . The second is obtained by averaging the two rates. Its hazard is constantly four. The first has hazard

h_{(3,5)}(t)=\frac{3+5e^{-2t}}{1+e^{-2t}}\le4,

with strict inequality for positive time. These two hazards are ordered.

Now add a rate-one component to each mixture and give every component equal weight. Write $X$ for rates $(1,3,5)$ and $Y$ for rates $(1,4,4)$ . The same averaging operation still relates their rate vectors, and both mean rates are three. Their hazards are

h_X(t)=\frac{1+3q^2+5q^4}{1+q^2+q^4},\qquad h_Y(t)=\frac{1+8q^3}{1+2q^3},\qquad q=e^{-t}.

Two evaluations settle the comparison:

Time	$h_X(t)$	$h_Y(t)$	Difference
$\log 2$	$11/7$	$8/5$	$-1/35$
$\log 4$	$103/91$	$12/11$	$41/1001$

The first difference is negative and the second positive. Neither hazard stays above the other for all time. The extra component has destroyed the two-component hazard ordering.

This does not destroy every ordering. The survival probabilities obey the exact identity

S_X(t)-S_Y(t)=\frac{q^3(1-q)^2}{3}\ge0.

Thus $X$ is larger in the usual stochastic order: its survival probability is at least that of $Y$ at every time. Ordered survival curves can coexist with crossing hazard curves.

There is exactly one crossing

Subtracting the hazards and factoring gives

h_X-h_Y= \frac{2q^2(1-q)(1-2q-q^3-q^4)} {(1+q^2+q^4)(1+2q^3)}.

For positive time, $0<q<1$ . Every factor outside the last polynomial in the numerator is positive. Let

B(q)=1-2q-q^3-q^4.

The derivative $B'(q)=-2-3q^2-4q^3$ is negative, while $B(0)>0$ and $B(1)<0$ . Therefore $B$ has exactly one root in the interval. As time increases, $q$ decreases, so $h_X-h_Y$ starts negative, crosses once and becomes positive. Both hazards eventually approach the common slowest rate.

The two exact fractions already disprove a hazard ordering. The factorization gives the stronger result: there is one crossing, with a determined direction on each side.

The same failure occurs with any number of components

Let $m$ be a positive integer. Compare equal-weight mixtures with rates

\lambda=(\underbrace{1,\ldots,1}_{m},3,5),\qquad \gamma=(\underbrace{1,\ldots,1}_{m},4,4).

Only the last two coordinates change. The hazard difference is

h_\lambda-h_\gamma= \frac{2q^2(1-q)\{m(1-2q)-q^3(1+q)\}} {(m+q^2+q^4)(m+2q^3)}.

The expression in braces has derivative $-2m-3q^2-4q^3<0$ . Its values at the endpoints are $m$ and $-m-2$ , so it has exactly one root. The hazard curves cross for every $m\ge1$ , giving a counterexample in every dimension $n\ge3$ .

In the language of majorization, $\lambda$ majorizes $\gamma$ : the latter is obtained by averaging a pair of coordinates while preserving their sum. The survival difference remains

S_\lambda-S_\gamma=\frac{q^3(1-q)^2}{m+2}\ge0.

Majorization of these rate vectors therefore gives the expected survival comparison while failing to order the hazards.

Where a two-component proof needs another argument

For an additive quantity, unchanged terms cancel. If two terms are transformed and every other summand stays fixed, the difference of the full sums equals the difference of those two terms.

For a ratio, write the transformed contributions as $(A_1,B_1)$ and $(A_2,B_2)$ , and the unchanged contributions as $(C,D)$ . Assuming positive denominators,

\frac{A_1+C}{B_1+D}-\frac{A_2+C}{B_2+D} =\frac{A_1B_2-A_2B_1+D(A_1-A_2)+C(B_2-B_1)} {(B_1+D)(B_2+D)}.

The two extra terms involving $C$ and $D$ have no counterpart in the two-component comparison. A proof that adds common components must control their signs as well. The explicit family above shows that this additional requirement is substantive.

Reproduce the calculation

The exact verification script derives the survival and hazard identities, checks the rational witnesses, and certifies the crossing polynomial. It uses SymPy rational arithmetic and exact polynomial root counts. The manuscript and companion checks develop the mechanism further.

This article presents the mathematical result. The separate literature audit distinguishes statements checked against full texts from conditional reconstructions and retains its author-response process before release of paper-by-paper findings. The counterexample above can be checked without accepting any verdict about a particular publication.