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I read that in survival analysis, when the hazard function $$\lambda(t)=\underset { dt\rightarrow 0 }{ lim } \frac { Pr(t\le T<t+dt|T\ge t) }{ dt } =c$$ then the event occurrence is random and independent (memoryless).

I can understand the concept and implications of monotonically increasing hazard function. However is there any theory or proof that when the $\lambda(t)=c$ then the event occurrence is random and independent?

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  • $\begingroup$ What is the meaning that you are ascribing to the word "random" when you say "the event occurrence is random"? Similarly, what do you mean when you say "...and independent." What is independent? Independence requires at least two events or random variables or.... and is a mutual property even though we colloquially say "A is independent of B" instead of the more formal "A and B are mutually independent". $\endgroup$ – Dilip Sarwate Mar 8 '13 at 2:46
  • $\begingroup$ @DilipSarwate: The words "random" and "independent" are words I found in textbooks I am reading at the time. Unfortunately they are not being explained in the context of this subject. Thank you for your explanation. $\endgroup$ – nostock Mar 8 '13 at 7:14
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If the positive random variable $T$ denotes the time of failure of a system with hazard rate $\lambda(t)$, then it is straightforward to show that the cumulative probability distribution function of $T$ is given by $$F_T(t) = 1 - \exp\left(-\int_0^t \lambda(\tau)\,\mathrm d\tau\right), ~ t \geq 0.$$ Thus, if the hazard rate $\lambda(t)$ equals a constant $c$ for all $t \geq 0$, we have that $$\begin{align} F_T(t) &= 1 - \exp(-ct), & t \geq 0,\\ P\{T > t\} &= \exp(-ct), & t \geq 0,\\ f_t(t) = \frac{\mathrm d}{\mathrm dt}F_T(t) &= c\exp(-ct), & t \geq 0, \end{align}$$ showing that $T$ is an exponential random variable with parameter $c$. As is well-known, exponential random variables are memoryless: $$\begin{align} P\{T > s+t \mid T > t\} &= \frac{P\{(T > s+t)\cap (T > t)\}}{P\{T > t\}}\\ &= \frac{P\{T > s+t\}}{P\{T > t\}}\\ &= \frac{\exp(-c(t+s))}{\exp(-ct)}\\ &= \exp(-cs)\\ &= P\{T > s\}\\ \end{align}$$ which is better understood as saying that a system that has survived till time $t$ forgets how old it is, and so the probability that it lasts for an additional time $s$ is the same as the probability of a brand new system surviving till time $s$. Systems with linearly monotonically increasing hazard rates that the OP refers to very definitely do not have this memoryless property.

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  • $\begingroup$ Great explanation. Thank you. BTW I am not referring to linearly increasing hazard rates rates. $\endgroup$ – nostock Mar 8 '13 at 7:14
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Maybe I misunderstand your question, but it might help to just think of the graph of the hazard function, that is, it's a line with no slope. When the hazard is constant, there is no dependence on time and the instantaneous failure rate is the same at say, $t = 2$ as $t = 200$. The exponential distribution has constant hazard, which makes sense when you think about the memoryless property of that distribution.

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  • $\begingroup$ The OP requests a theoretical proof, not a thought experiment. $\endgroup$ – ECII Mar 7 '13 at 21:29

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