Hazard Rate and Survivor Function

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SUMMARY

The discussion focuses on the hazard rate function defined as h(t) = λ for 0 ≤ t ≤ 50 and h(t) = λ + α(t - 50) for t > 50. The survivor function S is derived as S = e^(-λt) for 0 ≤ t ≤ 50. The user seeks assistance in integrating to find the complete survivor function, including constants of integration. The conversation highlights the importance of correctly applying integration techniques and recognizing constants in statistical functions.

PREREQUISITES
  • Understanding of hazard rate functions in survival analysis
  • Knowledge of integration techniques in calculus
  • Familiarity with exponential functions and their properties
  • Basic concepts of statistical constants and their significance
NEXT STEPS
  • Study the derivation of the survivor function in survival analysis
  • Learn about the role of constants in integration and their impact on functions
  • Explore advanced topics in hazard functions and their applications
  • Investigate the relationship between hazard rates and survival probabilities
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Statisticians, data analysts, and researchers in survival analysis or reliability engineering who are looking to deepen their understanding of hazard rates and survivor functions.

rad0786
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I really doubt that anybody would help me out on this...because by experience, nobody ever replies to the stats stuff on this forum...

But i'll try anyway

Question


the hazard rate function is:


h(t) = lamda > 0 ...... for 0 =< t =< 50

h(t) = lamda + alpha(t - 50) ... for t> 50



Find the surviovr function S'/S = -h.


Answer


So far...all i know is that S = e^-(lamda)t for 0 =< t =< 50

I'm not sure what to do from here...

Can somebody give me a boost please?
 
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I would integrate. S'/S = ln(S)'

So calling lamda L and alpha A (I assume they're both constants, right?)
So ln(S) = -(Lt + At2/2 -50At + C) where C is a constant.

In fact, you missed for your other value of S that it's D*e-lamda*t because you didn't do what I did above, and missed the constant of integration
 

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