If f(t) represents the probability density of failure rate, then how it it possible that f(t) will follow exponential distribution whereas the failure rate is constant?
You have confused the distribuion of waiting time and the failure rate.
Let f(t) denote the probability density of time T to failure, i.e. the probability of failure in the interval [t, t+dt] is f(t)*dt. Let F denote the distribution function, i.e. F(t)=P(T
Dear Parag Sen,
your problem concerns the relationship between Poisson distribution and exponential distribution - namely:
If the random variable X represents the number of errors (system failures) in a given time period and has the Poisson distribution, then the intervals between every two consecutive errors have the exponential distribution.
For example, see Wikipedia:
If for every t > 0 the number of arrivals in the time interval [0, t] follows the Poisson distribution with mean value λt, then the sequence of inter-arrival times are independent and identically distributed exponential random variables having mean 1/λ.
https://en.wikipedia.org/wiki/Poisson_distribution
In general, the exponential distribution describes the distribution of time intervals between every two subsequent Poisson events.
The answer to your question can be found at the following addresses:
- 7. Reliability (see the the Pages 3 and 4) https://cosmologist.info/teaching/STAT/CHAP7.pdf
- Relation between the Poisson and exponential distributions. https://neurophysics.ucsd.edu/courses/physics_171/exponential.pdf
- Relationship of Poisson and exponential distributions http://www.csee.usf.edu/~kchriste/tools/poisson.pdf
- Exponential Distribution and Poisson Process; see Page 9 http://www.utdallas.edu/~metin/Probability/Folios/expoPois.pdf
- PROBABILITY DISTRIBUTIONS, see the Pages: 8 –20 http://176.32.89.45/~hideaki/res/lecture/ipm/1_Poisson_process.pdf
Best regards
Anatol Badach
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