Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A system consists of $n$ components in parallel. The lifetimes of the components are i.i.d. exp($\lambda$) random variables. The system functions as long as at least one of the $n$ components is functioning. Let $T$ be the lifetime of the system. Compute $E[T]$.


Ok, so I've done problems like these in the past and they usually involved a finite number of components with a fixed probability. That is, the components usually had a fixed probability $p$ of failure. So I would just plug that into a Binomial distribution and call it a day. This problem is a bit different because $p$ varies with time.

So I do have some ideas for this problem, but I eventually get stuck.


Because this is a Poission process, the PP is counting the number of failures up to time $T$. So because the system fails when the last component fails, we essentially want to find the expected lifetime of the last component. In math terms:

Let $N(t)$ = the number of failures by time $t$
Let $T$ = the lifetime of the system
Let $S_i$ = the lifetime of component $i$
Let $X_i$ = the interarrival time between failure i and i-1$

So, essentially, we know that $T=max\{S_i | i \epsilon 1,2,...,n\}$

We also know that $E[T] = \sum_0^n E[S_i] $

So would the answe to this really be just the sum of all $\frac{1}{\lambda}$ or am I missing something?

share|improve this question

2 Answers 2

up vote 1 down vote accepted

Due to the lack of memory property of the exponential distribution, when the $i$th failure happens, there are $n-i$ components left, whose residual lifetimes are still i.i.d. exponential of parameter $\lambda$. Hence the $(i+1)$th failure will occur at the time of the $i$th failure plus an exponential time of parameter $(n-i)\lambda$.

In other words, for every $1\leqslant i\leqslant n$, $X_i$ is exponential with parameter $(n-i+1)\lambda$ and $T=X_1+X_2+\cdots+X_n$ hence $$ E(T)=\frac1\lambda\,\sum_{i=1}^n\frac1{n-i+1}=\frac1\lambda\,\sum_{k=1}^n\frac1k=\frac1\lambda\,H_n. $$ Exercise: The naïve approach yields the alternative expression $E(T)=\frac1\lambda t_n$ where $$ t_n=\sum_{k=1}^n{n\choose k}\frac{(-1)^{k+1}}k. $$ One can show that $t_n=H_n$ thanks to a recursion on $n$ using the well known identities $$ {n+1\choose k}={n\choose k}+{n\choose k-1},\quad{n\choose k-1}\frac1k=\frac1{n+1}{n+1\choose k}, $$ and $$ \sum_{k=1}^{n+1}{n+1\choose k}(-1)^{k+1}=1. $$

share|improve this answer
    
What is $H_n$, the heavyside step function? –  audiFanatic Mar 27 at 16:55
    
@audiFanatic The $n$th harmonic number (as explicitely written in the post). –  Did Mar 27 at 17:00

We first find the distribution of the maximum life $T$. The maximum of the $S_i$ is $\le t$ if and only if all the $S_i$ are $\le t$. This has probability $(1-e^{-\lambda t})^n$. Differentiate. The density function of $T$ is $\lambda n e^{-\lambda t}(1-e^{-\lambda t})^{n-1}$ (for $t\gt 0$). Call this $f_T(t)$

We want to find $\int_0^\infty tf_T(t)\,dt$. That is in principle no problem. Do the natural integration by parts, $u=t$, $dv=$ the rest. We end up needing to integrate $(1-e^{-\lambda t})^{n}$. Expand, using the Binomial Theorem.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.