# How to extract rates and expectations from a poisson distribution.

I'm reading this wikipedia article to try to get an understanding of the Poisson distribution (again). From what I'm reading, it seems like the Poisson distribution is useful for answering the question: "Suppose I have a computer failure rate of $1$ per month. What is the probability of $X$ computer failures per month?" My intuition tells me that I could rewrite the $1$ per month in different units (such as per day) and get the probability of a computer failing in a day or year, or whatever units I'd like.

However, I'm interested in being able to manipulate this distribution to answer other questions. For example, suppose that I have $10$ (or $m$) computers in a network, and I require that any of the computers fail in order for the network to fail. Intuition says (assumption 1) that I could take the Exponential distribution to find the time to failure, $1/\lambda = 1/1$. Since I now have 10 computers, but only one of them needs to fail, that this means, on average, I will have $10$ (or $m\lambda$) failures per month, or that the expected time for a network failure would be , $\frac{1}{m\lambda} = 1/10$.

Finally, suppose that I add some redundancy to my network, so that now I need 2 (or $n$) computers to fail instead of 1. Assuming the above is correct, my intuition is (assumption 2) to look for 2 (or $n$) failures, which is a sum of exponentials, and thus has expectation arrival time $\frac{n}{m\lambda} = 2/10$. However, I don't like where this intuition takes me. To me, this implies that if I have 10 computers with perfect redundancy, such that all 10 ($n$) need to fail for the network to fail, then the expected time to failure of my perfectly redundant network is exactly equal to the expected time to failure of a single computer, because for $n=m$ we have expected time to failure $1/\lambda$. This strikes me as obviously incorrect.

Therefore, one of my two assumptions above must be wrong. Can someone point out exactly what the problem is?

Minimum of exponential lifetimes. Your first assumption is true. Let each of the computers have lifetime $T \sim Exp(\lambda),$ where $\lambda$ is its failure rate and $E(T) = 1/\lambda.$ Let $V$ be the lifetime of a network of $m$ computers that survives only until the first computer fails. Then $$P(V > t) = P(T_1 > t, T_2 > t, \dots, T_m > t) = \prod_{i=1}^m P(T_i > t) = \prod_{i=1}^m e^{-\lambda t} = e^{-m\lambda t},$$ and so $V \sim Exp(m\lambda)$ and $E(V) = 1/m\lambda,$ as you suppose.

Maximum of exponential lifetimes. However, the mechanism of redundancy is not quite so intuitive, and your second assumption is incorrect. Let $W$ be the lifetime of your second system of $n$ computers that continues working until the last one fails. Then

$$P(W \le t) = P(T_1 \le t, \dots, T_n \le t) = \prod_{i=1}^n P(T_i \le t) = (1 - e^{-\lambda t})^n,$$ and so $W$ does not have an exponential distribution.

Intuitively, $W$ cannot be exponential. An exponential random variable $X$ has the no-memory property, sometimes described in reliability terms as "used is as good as new": $P(X > t+s|X > t) = P(X > s).$ However a redundant system cannot have a constant failure rate. A system that has been working until all but one component has failed is more vulnerable than it was at the start when back-ups were available. (You might want to read further on the Internet by searching for 'parallel redundancy reliability' and similar phrases.)

Simulation and computation of simple examples. The simulation below using R statistical software illustrates the difference between the distributions of the minimum and maximum of two independent exponential lifetimes, each with rate $\lambda = 1.$ From the formulas above, you should be able to compute some of the exact values, using $m = n = 2$, $\lambda = 1.$ An intuitive explanation of $E(W)$ is to wait average time 0.5 for the first failure (min) and then additional average time 1 for the second (max).

 x1 = rexp(10^4, 1);  x2 = rexp(10^4, 1)
v = pmin(x1, x2);  w = pmax(x1, x2)
mean(v);  mean(w);  mean(v > 1);  mean(w > 1)
## 0.4998082  # Simulates E(V) = .5
## 1.496407   # Simulates E(W) = 1.5
## 0.1349     # Simulates P(V > 1) = .1353
## 0.5963     # Simulates P(W > 1) = .6004


You are correct that the time to the first computer failure out of $$m$$ has an exponential distribution with rate $$m \lambda$$ and so an expected time to this failure of $$\frac{1}{m \lambda}$$. You can show this exponential distribution and rate by considering the probability that none have failed by time $$t$$, which is $$\left(e^{-\lambda t}\right)^m=e^{-(m\lambda) t}$$.

Your error is in what happens next. You are left with $$m-1$$ functioning computers, and so the time between the first failure and the second failure is also an exponential distribution but with a slower rate $$(m-1) \lambda$$ and so a higher expected additional time to this failure of $$\frac{1}{(m-1) \lambda}$$. The combined time does not have an exponential distribution but you can say the total expected time for both the first and second failure to occur is $$\frac{1}{m \lambda}+\frac{1}{(m-1) \lambda}$$, which is greater than $$\frac{2}{m \lambda}$$.

You can extend this to more failures required for total system failure in the same way, and if all $$m$$ need to fail you get a total expected time of $$\frac1\lambda(\frac1m+\frac1{m-1}+\cdots+\frac11)$$ where the sum in brackets is the $$m$$th harmonic number. For $$m=10$$ this is about $$\frac{2.929}{\lambda}$$, rather more than the $$\frac1\lambda$$ you rejected as being obviously incorrect.