On such occasions, a reversed hazard rate is more appropriate than a hazard rate to analyze lifetime data due to the fact that estimators of hazard rates are unstable when data are left censored.

The reversed hazard rate of T given in (1) can be written as

where [[lambda].sub.0](f) is the baseline reversed hazard rate, g(*) is a nonnegative function of X and [beta],p x 1 vector of regression parameters, and []lambda(t | x) is the reversed hazard rate of T given the covariate X.

The baseline reversed hazard rate of T is then obtained as

Note that the baseline reversed hazard rate is decreasing as t increases.

Proportional reversed hazard rate models are more suitable for modeling the left censored lifetime data.

Finkelstein, "On the reversed hazard rate," Reliability Engineering and System Safety, vol.

Xu, "Reversed hazard rate order of equilibrium distributions and a related aging notion," Statistical Papers, vol.

It should be noted that, for two subjects, the ratio of reversed hazard rates is independent of the time t.

Let X and Y be two nonnegative discrete (or continuous) r.v.'s with respective distribution functions F(t) = P(X [less than or equal to] t) and G(t) = P(Y [less than or equal to] t), probability density functions f(t) = P(X = t) and g(t) = P(Y = t), survival functions [bar.F](t) and [bar.G](t), hazard (failure) rate functions [r.sub.X](t) = f(t)/P(X [greater than or equal to] t) and [r.sub.Y](t) = g(t)/P(Y [greater than or equal to] t) and reversed hazard rate functions [[bar.r].sub.X] (t) = f(t)/P(X [less than or equal to] t) and [[bar.r].sub.Y](t) = g(t)/ P(Y [less than or equal to] t).

Similarly, X is said to be smaller than Y in reversed hazard rate ordering, denoted by X [[less than or equal to].sub.rh] Y, if [[bar.r].sub.X] [less than or equal to] [[bar.r].sub.Y] (t) for all t [greater than or equal to] 0.

By Theorem 3.1, X is smaller than or equal to Y in likelihood ratio (reversed hazard rate) ordering sense if, and only if, 0 < [lambda] [less than or equal to] [[lambda].sub.0] = E([[LAMBDA].sup.2]/([e.sup.[LAMBDA]] - 1))/E([LAMBDA]/([e.sup.[LAMBDA]] - 1)).