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The rate is the expected number of events per some unit (time, or spatial). $\text{Rate}=\frac{\text{numbers of events}}{\text{the length of time interval}}.\ldots (1)$

(Reference: https://stats.stackexchange.com/users/11887/kjetil-b-halvorsen).

The definition of hazard function (or, hazard rate) is "the instantaneous rate of death at time $t$, given that the individual survives up to $t$".

The hazard function $h(t)$ can be defined as

$$h(t)=\lim_{\Delta t\to 0} \frac{Pr(t\le T<t+\Delta t|T\ge t)}{\Delta t}.\ldots (2)$$

(Reference: Lawless, J. F. (2011). Statistical models and methods for lifetime data (Vol. 362). John Wiley & Sons.).

My question is why $h(t)$ is not defined as $\frac{\text{NUMBER of death at time t, given that the individuals survive up to t}}{\Delta t}$?

More Specifically, why is the numerator of equation (2) probability? Why is it not number of death?

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    $\begingroup$ In nature, $h(t)$ is a parameter (non-random) but not a statistic (random). $\endgroup$ – Zhanxiong Oct 30 '17 at 1:58
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    $\begingroup$ You can only die once. $\endgroup$ – StijnDeVuyst Oct 30 '17 at 13:19
  • $\begingroup$ @StijnDeVuyst But, as the James. Bond franchise showed You Only Live Twice ! $\endgroup$ – Dilip Sarwate Oct 30 '17 at 16:03
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You are confusing two very different things. The definition of $h(t)$, which is just a deterministic function $h:[0,+\infty]\to[0,+\infty]$, and the estimator $\hat{h}(t)$ of $h(t)$, which is a statistic, i.e., a function of your random sample from the population, and thus a random variable itself1.

The definition of hazard rate is what you report (not sure why the dots, I just removed them):

$$h(t)=\lim_{\Delta t\to 0} \frac{Pr(t\le T<t+\Delta t|T\ge t)}{\Delta t}$$

Now, suppose that for a given problem you don't know the expression of $Pr(t\le T<t+\Delta t|T\ge t)$, so you can't compute the above limit. If you can sample from the population, you can estimate it. Now, we need to show that

$$\hat{h}(t) = \frac{\text{number of deaths at time } t}{\text{number of individuals at risk at time } t}$$

is a reasonable estimator of $h(t)$.

Basically, the problem here is that $\text{number of individuals at risk at time } t$ is a really confusing terminology and the authors shouldn't have used it. Anyway, since, given a fixed time $t$, no individual dies exactly at $t$, the numerator must be intended as $\text{number of deaths between time } t \text{ and time } t+\delta t$, where $\delta t$ is suitably small (it's an estimate, after all). Of course, only people who were still alive at time $t$ could die in $[t, t + \delta t]$. Now, the ratio

$$\hat{Pr}(t\le T<t+\delta t|T\ge t) = \frac{\text{number of deaths between time } t \text{ and time } t+\delta t}{\text{number of individuals alive at time } t}$$

is a frequentist estimate of $Pr(t\le T<t+\delta t|T\ge t)$: this seems intuitive to me, but I add a detailed explanation as requested by the OP. Feel free to skip this part.


Let $T$ be the random variable expressing the time of death of an individual. We want to estimate $Pr(t\le T<t+\delta t)$. We could then draw a random sample $D=\{T_1,\ldots,T_N\}$ of iid random variables, which in practice means following a group of similar individuals until they die. Of course, since you can only die once, we cannot follow the same individual across multiple lives, as opposed to when we throw the same coin multiple times to estimate $Pr(\text{Heads})$. We need to follow a group of individuals, instead. Now, as explained in introductory probability courses, an estimate of $Pr(t\le T<t+\delta t)$ based on $D$ is just

$$\hat{Pr}(t\le T<t+\delta t) = \frac{\text{number of deaths between time } t \text{ and time } t+\delta t}{N}$$

However, we really want to estimate

$$Pr(t\le T<t+\delta t|T\ge t) = \frac{Pr((t\le T<t+\delta t) \cap (T\ge t))}{Pr(T\ge t)} $$

The estimate for the denominator is obvious: someone will die at a time $T\ge t$ iff he's alive at time $t$, thus

$$\hat{Pr}(T \ge t) = \frac{\text{number of individuals alive at time } t}{N}$$

For the numerator, since the event $t\le T<t+\delta t$ is included in the event $T \ge t$, $(t\le T<t+\delta t) \cap (T\ge t) = t\le T<t+\delta t$. Thus

$$\hat{Pr}(t\le T<t+\delta t|T\ge t) = \frac{\text{number of deaths between time } t \text{ and time } t+\delta t}{\text{number of individuals alive at time } t} $$


Thus, $\hat{h}(t)$ defined as

$$\hat{h}(t)=\frac{\hat{Pr}(t\le T<t+\delta t|T\ge t)}{\delta t} = \frac{\text{number of deaths between time } t \text{ and time } t+\delta t}{(\text{number of individuals alive at time } t) \times \delta t}$$

is a reasonable estimator of $h(t)$. I guess that the definition used by your book is trying to get at the same estimator, in a very convoluted way,. Probably, what the authors mean is that not all individuals alive at time $t$ are likely to die in a subsequent short time interval. Intuitively, the longer the time interval $\delta t$, the larger is the number of "at risk" individuals, among all individuals alive at time $t$. If we suppose that the number of individuals that, at time $t$, are at risk of death in $[t,t+\delta t]$, is equal to $(\text{number of individuals alive at time } t) \times \delta t$, then we get

$$\hat{h}(t) = \frac{\text{number of deaths between time } t \text{ and time } t+\delta t}{\text{number of individuals at risk at time } t}$$

which is the definition reported by your book.


1 to be precise, $\hat{h}(t)$ is a random process, not a random variable: however, for each fixed $\bar{t}\in[0,+\infty]$, $\hat{h}(\bar{t})$ is a random variable.

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  • $\begingroup$ If anyone stops deriving the estimate of $h(t)$ in this equation, $\hat{h}(t)=\frac{\hat{Pr}_{[t,t+\delta t]}(t)}{\delta t} = \frac{\text{number of deaths between time } t \text{ and time } t+\delta t}{(\text{number of individuals alive at time } t) \times \delta t}$, doesn't it suffice? $\endgroup$ – ABC Nov 3 '17 at 0:35
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    $\begingroup$ Not sure what you mean with " anyone stops deriving". If you're asking me "is $\hat{h}(t)=\frac{\hat{Pr}_{[t,t+\delta t]}(t)}{\delta t} = \frac{\text{number of deaths between time } t \text{ and time } t+\delta t}{(\text{number of individuals alive at time } t) \times \delta t}$ a valid estimator for $h(t)?" then my answer is "yes". But your book is citing a slightly different formula, so I kept going on to show you how you can get to your book's formula. $\endgroup$ – DeltaIV Nov 3 '17 at 0:54
  • $\begingroup$ It also seems to me $\hat{Pr}_{[t,t+\delta t]}(t)$ is an estimate of $Pr(t\le T<t+\delta t|T\ge t)$. But I can't provide any explanation. Could you please explain it? $\endgroup$ – ABC Nov 3 '17 at 4:20
  • $\begingroup$ @ABC yep, sure, but I'm a bit busy this weekend, so I'm not sure if I'll be able to write it before Monday. However, rest assured I'll do it :) in meantime, you may want to read a bit about conditional probability. Are you familiar with the concept that conditional probability $Pr(A\vert B)$ is a valid probability measure on the sample space $B$? $\endgroup$ – DeltaIV Nov 3 '17 at 8:03
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    $\begingroup$ @ABC thanks to you for suggesting it. It's all too easy, after a few years of dealing with conditional probabilities, how unintuitive some results are the first time you meet them. What's intuitive for me now, definitely wasn't when I started. It's not always easy to write answers on a site where the public goes from real giants in this field, to passionate learners who maybe started their journey in Statistics just a couple years ago :) $\endgroup$ – DeltaIV Nov 7 '17 at 9:21
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By definition, the hazard rate is the rate of death of a population of arbitrary size. The hazard rate you propose - let us call it $\widetilde{h(t)}$ - would be defining what you are interested in implicitly as a function of the population size $S$. In particular, the relationship between the original $h(t)$ and your $\widetilde{h(t)}$ is given by $S \times h(t) = \widetilde{h(t)}$. Naturally, you are more interested in a population-size independent function.

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  • $\begingroup$ From the reference "Lawless, J. F. (2011). Statistical models and methods for lifetime data (Vol. 362). John Wiley & Sons", they told estimate of hazard function $\widetilde {h(t)}=\frac{\text{number of death at time t}}{\text{number of individuals at risk at time t}}$. But in my proposed $\widetilde {h(t)}$, the denominator comes a time interval. Could you please explain it further? $\endgroup$ – ABC Oct 31 '17 at 0:51
  • $\begingroup$ Well, when you divide number of deaths by number of individuals at risk at time $t$, all you do is constructing $\widetilde{h(t)} = \dfrac{S(t) \times \mathbb{P}( t\leq T \leq t + \delta t|T\geq t)}{S(t) \times \delta t}$ where $S(t)$ denotes the population size at time $t$. So it is different from what you proposed to do up in the question, and should be correct (because $S(t)$ cancels). $\endgroup$ – Jeremias K Nov 1 '17 at 10:59

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