# Expected lifetime: Maximum of $3$ exponential random variables

$A, B$ and $C$ are $3$ independent components of a device with lifetime of density function $$f(x)={1\over200}{e^{-x/200}}$$

1. What is the expected lifetime of $A$?

I think is is $\int_0^\infty {xf(x)dx}$ but I am not sure.

2. Suppose system works as long as at least one component works. What is the expected time until the system fails?

Is it $3$ times of expected lifetime of $A$?

• The distribution of max is less pleasant than the distribution of min. There is a cute way of using information about min to show that expectation of max is $200+\frac{200}{2}+\frac{200}{3}$. – André Nicolas Jan 7 '16 at 11:48

## 3 Answers

If $X_1$, $X_2$, and $X_3$ are the lifetimes of the components then the life time of the system is $M=\max(X_1,X_2,X_3)$. The cumulative distribution function of $M$ is $F_M(x)=P(M<x)=P(X_1<x, X_2<x, X_3<x)=P(X_1<x)P(X_2<x)P(X_3<x)=F_X(x)^3$ $F_X(x)=1-\exp(-x/200)$. $F_M(x)=(1-\exp(-x/200))^3$. The density of $M$ is $f_M(x)=F_M'(x)=3(1-\exp(-x/200))^2\frac{\exp(-x/200)}{200}$. The expected time is $\int_0^\infty xf_M(x)dx.$

The distribution given by $f(x)$ is $\exp(λ=1/200)$. So,

1. Its expected value is equal to $$E[A]=\frac1{1/λ}=200$$ see here. If you want to calculate it directly without using this, then yes your formula $$E[A]=\int_{0}^{+\infty}xf(x)dx$$ is correct.
2. Now, for your second question, denote with $X_1$ the minimum, with $X_2$ the middle and with $X_3$ the maximum lifetime of the three components. As is already mentioned you need to calculate $E[X_3]$. We will use that the minimum of $n$ iid $\exp(λ)$ random variables has again the exponential distribution with parameter $nλ$, see here. Write $$E[X_3]=E[X_1+(X_2-X_1)+(X_3-X_2)]=E[X_1]+E[X_2-X_1]+E[X_3-X_2]$$ by linearity of expectation. Now,

1. $X_1$ is the minimum of $3$ iid $\exp(λ)$ hence $X_1\sim\exp(3λ)$ and

2. $X_2-X_1$ is distributed as (by the memoryless property of the exponential) as the minimum of $2$ iid $\exp(λ)$, hence $X_2-X_1\sim \exp(2λ)$. Why is that? Because after $X_1$ is realized, you have two more components running. By the memoryless property of the exponential distribution, the remaining lifetime of each of the components is again exponentially distributed. So, $X_2-X_1$ is just the time that the minimum of $2$ iid exponentials will be realised.

3. By the same argument $X_3-X_2$ is distributed as $X_3-X_2\sim\exp(λ)$.

So, $$E[X_3]=\frac1{3λ}+\frac1{2λ}+\frac1{λ}=\frac{200}3+\frac{200}2+200$$

If $X$ is exponentially distributed with parameter $\lambda$, i.e. $f(x)=\lambda e ^{- \lambda x}, x \geq 0$, then $E(X)= 1 / \lambda.$

In the question; since $\lambda=1/200$, the expected lifetime is $E(X)= 200$.

• Thanks. $\int_0^\infty {xf(x)dx} = 200$ already. So I think I am right. Do you have any thoughts about the second question? – Vitiello Jan 7 '16 at 10:58
• The second question: The system failure time is the maximum lifetime among the 3 components. You first calculate the CDF, and using the CDF/Survival function to compute the expectation. – BGM Jan 7 '16 at 11:30