# The relation of mean time between failure and the exponential distribution

Suppose that we calculate the mean time between failures(mtbf) to be $m$ by simply taking $n$ samples where $n$ is sufficiently large. Now that we know this the failure probability of the system under inspection is modeled with the the exponential distribution($Exp$) parametrized with $\frac{1}{\lambda} = m$. So my question is why do we model it like this. I mean the memoryless property(I know Exp is the only continuous distribution with this property) of $Exp$ and that $E(Exp(\lambda)) = \frac{1}{\lambda} = m$ is obviously desired properties but for example being positively skewed is not necessarily desired. Suppose I want to give higher probability of failure at a given $(t_1, t_2)$ time frame and low outside that also keeping the mean at $m$. Or that's not really an interesting use-case?

## 1 Answer

Your question is why we don't use a model shaped like that and the answer is that the standard model seems to work for a majority of cases. It allows you to move forward when you don't have any better information to go on. If you do have more specific information about a particular situation, then by all means develop your own probability distribution based on observation or experience, and use that. If your observations are correct and reasonably accurate, your predictive results will be better than a generic solution. It might be worth checking some of that actual data against the standard model before you go to all that trouble, to see how much benefit you would be buying with the hassle of setting up this individualized approach.