Hazard measures

Gérard Letac^{∗}

3rd October 2001

Abstract Given a probability measure p on IR with unbounded support on the right, the measure q_{p}(dx) = (p([x, ∞))^{−1}p(dx) is called the hazard measure associated to p. We caracterize the set Q of possible hazard measure, we show that given q ∈ Q there exists a unique p such that q = q_{p} and we compute it.

Theorem A: Let P be the space of probability measures p on IR such that for all x ∈ IR one has r_{p}(x) = p([x, ∞)) > 0. For p ∈ P, define the positive measure q_{p}(dx) = (r_{p}(x))^{−1}p(dx). Let now q be a positive measure on IR, with q = q^{(c)} + q^{(a)} where q^{(c)} and q^{(a)} are the continuous and atomic parts of q. Write

q^{(a)} = ∑_{b∈D} m_{b}δ_{b},

where D ⊂ IR is the finite or countable set of the atoms of q. Then

1. There exists p in P such that q = q_{p} if and only if m_{b} < 1 for all b ∈ D, q((−∞, x]) < ∞ for all x ∈ IR, and q is an unbounded measure.

2. Under these circumstances, such a p is unique and

r_{p}(x) = e^{−q^{(c)}((−∞,x])} ∏_{b<x}(1 − m_{b}).  (0.1)

Proof: Part 1 ⇒ being easy, we show simultaneously Parts 2 and 1 ⇐ . Denote by r(x) the right hand side of (0.1). Clearly r is decreasing, left continuous, strictly positive with limits 1 and 0 at end points of IR. Thus p ∈ P such that r_{p} = r does exist, with p(dx) = −dr(x). Now we denote r^{(c)}(x) = e^{−q^{(c)}((−∞,x])}, which satisfies

−dr^{(c)}(x) = r^{(c)}(x)q^{(c)}(dx).  (0.2)

For b ∈ D, we denote r^{(b)}(x) = 1 for x ≤ b and r^{(b)}(x) = 1 − m_{b} for x > b, which satisfies

−dr^{(b)}(x) = m_{b}δ_{b}(dx) = r^{(b)}(x)m_{b}δ_{b}(dx)  (0.3)

∗Laboratoire de Statistique et Probabilité, Université Paul Sabatier, 31062 Toulouse.

(the last equality uses r^{(b)}(b) = 1). Since r = r^{(c)} ∏_{b∈D} r^{(b)} we get from (0.2) and (0.3)

p(dx) = −dr(x) = r(x) (\frac{−dr^{(c)(x)}}{r^{(c)(x)}} + ∑_{b∈D} \frac{−dr^{(b)(x)}}{r^{(b)}(x)} ) = r(x)(q^{(c)}(dx)+q^{(a)}(dx)) = r(x)q(dx),

and the proof of 1⇐ is complete. To see that p is unique, suppose that there exists an other solution p_{1} ∈ P such that q = q_{p_{1}}. Denote r_{1} = r_{p_{1}} for simplicity. Clearly q, p and p_{1} are absolutely continuous with respect to each other. Therefore we can write r_{1} = e^{f}r, where f is some unknown function on IR with locally bounded variation. Now:

r_{1}(x)q(dx) = −dr_{1}(x) = e^{f(x)}(−dr(x)) + r_{1}(x)df(x) = r_{1}(x)q(dx) + r_{1}(x)df(x).

Thus the signed measure df(x) is 0 and f is a constant, which is easily shown to be 0. Thus uniqueness of the solution p of the equation q = q_{p} is proved and the proof of Theorem A is complete.