The hazard function The hazard function gives the so-called “instantaneous” risk of death (or failure) at time t, assuming survival up to time t. Estimate h(t) by the quotient

The hazard function is also called the instantaneous failure rate or force of mortality or conditional mortality rate or age-specific failure rate. In a real sense it gives the risk of failure (death) per unit time over the progress of aging. We have seen f(y)=-d/dy(S(y)), and so that f, S, and h are all related and each can be obtained from the others. Hazard functions can be flat, increasing, decreasing, or more complex…

Consider a simple hazard function, the constant hazard h(y)= for all y≥0. Here we assume , where 0. We have seen that so if we evaluate this for h(y)= , we get Since f(y)=-d/dy(S(y)), we have the exponential probability density with parameter . This means the expected value is  and the variance is 2.

Definition 2.1 writes Y as Y ~ exp( with h(y)=1/ and notes that this is one of the most commonly used models for lifetime distributions. One reason is because of the “memoryless property” of the exponential distribution given on page 20: Theorem 2.1: If Y ~ exp( then for any y>0 and t>0 we have P(Y>y+t | Y>y)=P(Y>t) Note this says that given survival past y, the conditional probability of surviving an additional t is the same as the unconditional probability of surviving t. Thus there is no “aging” with an increased risk of dying… Go over Example 2.1 to see a picture of exponential data which would then have a constant hazard (of value =1/mean(Y)) Note from Theorem 2.2 that if Y ~ exp( , Y/ ~ exp(1). This tells us that if we multiply an exponential survival variable by a constant, the MTTF is correspondingly multiplied by the same constant.

How do we decided whether a set of survival data is following the exponential distribution? That the hazard is constant? Look over Example 2.1 - 200 randomly generated exponential variables with mean=100. Characteristic skewed distribution, sample mean=107.5, sample s.d.=106.1; (Recall that if Y~exp( then E(Y)=SD(Y)= . ) The sample stemplot and the sample mean and sd approximate the true shape, center and spread of the exponential. The estimated hazards (rightmost column) approx. .01 (1/100) - constant - see the formula on p.22 for getting these values… But another way to check the distribution is to compare the quantiles of the exponential distribution with the sample quantiles in a plot known as a qqplot. See R#3 for a way to compute the quantiles and do the plot… Recall that the p-th quantile of a distribution of a r.v. Y is the value Q s.t. P(Y<=Q)=p. So we must compute the quantiles of the theoretical distribution and compare them (smallest to smallest, next smallest to next smallest, etc.) to the sample quantiles.

Power Hazard: Note this is of the form (constant)yconstant and if 1 this reduces to the constant hazard we just considered. Note that and so We say in this case that Y ~ Weibull(. The mean and variance of Y are given in Theorem 2.5 in terms of the gamma function Note that Y ~ Weibull(1,exp(and if >1 the hazard function is increasing; if 1, the hazard is decreasing; if 1 then the hazard is constant (as in the exponential survival case) Note that if 2, the hazard is linear in y. Go over Example 2.2 on pages 24-25. Y ~ Weibull(2,sqrt(3)). Use R (gamma()) to compute the values of the Gamma function… Also in R, shape=alpha and scale=beta in qweibull.

To check graphically if a distribution is Weibull, do essentially a qqplot on the log-log scale (see section 4.4, p. 61-63). The key formulas are: (4.2) Now substitute the ordered data and take natural logs (4.2a) Take logs again… (4.2b) Write as a linear equation: (4.3) (4.3a) So plot the points in 4.4, look for a straight line and the slope will equal 1/ and the intercept will equal log()