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I'm using the exponential distribution to calculate a probability of an event to occur. $\lambda = 0.007$ failures/year. I am to calculate an expected time between 2 failures.

My approach was to calculate the probability of 1 failure to occur within a year and then somehow use the memorylessness property of exponential distributions. However, I'm absolutely stuck now since I don't understand how to get time out of this probability. Is there some specific formula I should turn to ? Any hints are welcome!

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    $\begingroup$ Please add [self-study] tag to your question since it seems to be homework or self-study problem. $\endgroup$ – Tim Jul 2 '15 at 10:49
  • $\begingroup$ Are you asking for the expected time between two successive failures or for two failures to occur? $\endgroup$ – whuber Jul 2 '15 at 13:13
  • $\begingroup$ @whuber I would say that they mean the same thing? $\endgroup$ – ashishv Jul 3 '15 at 12:51
  • $\begingroup$ The latter will have twice the expectation of the former, because the time elapsed between two successive failures is the time needed to wait for just one more failure after the first occurs, whereas the time elapsed before seeing two failures requires first one failure to occur and then for another waiting period to happen until the second failure occurs. $\endgroup$ – whuber Jul 3 '15 at 13:13
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    $\begingroup$ @whuber I understand now what you mean. I meant, the time after the first failure has occurred until the second failure to occur. $\endgroup$ – ashishv Jul 3 '15 at 13:32
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Let $T$ be the time between failures. Then $T\sim Exponential(\lambda)$ with $\lambda = 0.007$ failures per year.

$$\begin{align}\text{E}[\text{Time Between 2 Failures}] &= \text{E}[T + T] \\ &=\text{E}[2T] \\ &=2\,\text{E}[T] \\ &=\frac{2}{\lambda} \;\text{ years} \end{align}$$

An alternate approach would be to recognize that the sum of $n$ identical Exponential distributions with rate $\lambda$ is an Erlang distribution. In other words, $\sum_{i=1}^n T_i \sim Erlang(n,\lambda)$ which has a mean of $\frac{n}{\lambda}$.

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