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

Thanks for your answer. Actually, the question should be does failure function represents the probability density function of failure rate? I cannot edit the question right now.

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

Thank you very much, Dr. Anatol Badach, that is the answer I was looking for.

Dear All P.T. Followers,

A lot of misunderstanding has been caused by using the not commonly known notion of failure function. In some old works I have seen this as the function calculating the number of failures till instant t within a population of elements mounted in different regimes: without replacing old parts by new, with replacement after some a priori established time or juat after the failure of some, say 3, of them etc. In every case the observed failure function assumes natural values and is increasing: it simply counts the instants of failures till time t. Let it be N(t).

From the formulation of the question one can guess, that this time it is the number of failures observed in a population, with element started all to be used simultaneously. Then the number N(t) divided by the initial number, say N, is the observed (=empirical) probability that the life time of the element is less than t, for any t>0. Theoretical conterpart of it is the cpdf F (cumulative probability distribution function, briefly: cpf=cumulative probability function). If this the right guess, then

1 - N(t)/N is called the empirical(=observed) reliability function, its counterpart R(t)=1-F(t)

[R(t1)-R(t2)]/[R(t1)(t2-t1)] is called empirical (observed) failure rate within the time interval [t1,t2]; its counterpart is the theoretical failure rate is defined by the same formula with the use of the theoretical reliability function, of course.

For theoretcal distributions with density, as in the answer by Peter Harremoës , i.e, when F(t)=\int[0,t] f(u) du, t>0, the limit of the theoretical failure rate as t2 approaches t1, is called simply theoretical failure rate (in full content with adjective: instantaneous), and equals f(t)/R(t) = f(t)/[1-F(t)].

According to the above, there is no chance to bind the failure function (which assumes only natural values, and is a step function) with the theoretical failure rate (usually a theoretical smooth function). From the failure function, one can evaluate the empirical (observed) failure rate if additionally the initial number of elements is known. But this is practically not in use.

On the other hand the notion of failure function is more often used in maintanance theories, as the mean value of the random processes counting events of particular type. In such cases, if N(t) is the random process, then the failure function would be H(t)=E{N(t)}. In the simplest case of renewal processes, when there system is instantly renewed at any instant of failure, the expectation is simply related to the cpf of the life time between renewals (I am omitting it here). In the most simple case only, when the time between renewals is exponential say with density f(t)=a*exp(-a*t), then the counting process is a Poisson process, and H(t)=t*a. Therefore the indicated by Anatol Badach relation is actually only one of possible cases. However it nicely explains the name of failure rate (obviously equal a), since it is not just f'(t)/[1-R(t)], but as in this example really the rate of which the elements of the renewal system should be replaced, since a=E{N(t)}/t.

Best regards, Joachim