Order statistics: What's the probability that all three components will fail within 2 years of each other? Suppose an instrument has three independent parts, all of whose lifetimes (in years) are modeled by an exponential pdf which is
$f_Y(y)=e^{-y}, y>0. $ What's the probability that all three parts will fail within two years of one another?
I understand how to find the cdf (it would be $ F_Y(y)=1-e^{-y} $). I also know the formula to find the pdf of the $i^{th}$ order statistic is as follows: $f^{'}_{Y_i}(y)=\frac{n!}{(i-1)!(n-i)!}[F_Y(y)]^{i-1}[1-F_Y(y)]^{n-i}f_Y(y)  $ for $ 1 \le i \le n.$
However I'm having trouble understanding how to incorporate the fact that they would all fail within two years of each other $(0 \le y \le 2?)$; how can this incorporated?
Thank you!
 A: My earlier comments gave the memoryless property of the exponential distribution as a short cut.
If you want the long and more general way of doing this for $n$ i.i.d. continuous non-negative random variables each with density $f(x)$ and with cumulative distribution function $F(x)$, you could try the following steps:


*

*The probability none have failed by time $t\gt 0$ is $(1-F(t))^n$

*So the probability at least one has failed by time $t$ is $G(t) = 1-(1-F(t))^n$

*So the density function for the time of the first failure is $g(x) = G'(t) = n f(t) (1 - F(t))^{n-1} $

*Given that the first to fail has failed at time $t$, the probability a particular other one has failed by time $s+t$ with $s\gt 0$ is $\frac{F(s+t)-F(t)}{1-F(t)}$

*So given that the first to fail has failed at time $t$, the probability the other $n-1$ have failed by time $s+t$ with $s\gt 0$ is $\left(\frac{F(s+t)-F(t)}{1-F(t)}\right)^{n-1}$

*So the overall probability that all three fail within time $s$ of each other is $$\int\limits_{t=0}^\infty g(t) \left(\frac{F(s+t)-F(t)}{1-F(t)}\right)^{n-1} \, dt = \int\limits_{t=0}^\infty n\, f(t)\, \left({F(s+t)-F(t)}\right)^{n-1} \, dt$$ 

*Here $s=2$ and $n=3$
