expectation calculation problem

I got the answers for this and i know its 1.05 but the way it explains is very difficult to understand so im seeking for some help here.

A system made up of 7 components with independent, identically distributed lifetimes will operate until any of 1 of the system's components fails. If the life time X of each component has density function
$f(x) = \begin{cases} 3/x^4, & \text{for 1<x}\\ 0, & \text{otherwise} \end{cases}$

what is the expected lifetime until failure of the system?

I tried to find the intersect of 7 components by integrating and power it by 7 but it doesnt give me anything useful...

• $P(min > t) = P(X > t)^7,$ so you need to find the CDF of $X$, and then $P(X > t)$. Then use the CDF of the minimum to find the mean. – BruceET Apr 19 '15 at 4:04

Since the system fails as soon as the first component fails, we are looking for the minimum of the lifetimes of the seven components, call it $Y=\min(X_1,X_2\ldots,X_7)$. The cumulative distribution $G$ for $Y$ is $$G(y)=\Pr(Y\le y)$$$$=1-\Pr(Y>y)$$$$=1-\Pr(X_1>y,X_2>y,\ldots,X_7>y)$$ $$=1-[\Pr(X_1>y)\Pr(X_2>y)\ldots\Pr(X_7>y)]$$ (Because of independence, we could split the probabilities) $$=1-[(1-\Pr(X_1\le y))(1-\Pr(X_2\le y))\ldots(1-\Pr(X_7\le y))]$$ $$=1-[(1-F(y))(1-F(y))\ldots(1-F(y))]$$ $$=1-[1-F(y)]^7,$$ where $F$ is your cumulative distribution function for any $X$. Thus your density function $g$ will be the derivative, $$g(y) = 7[1-F(y)]^6f(y)$$ Can you go from there?
• THANKS! once i saw the notation $min{{X_1,X_2,.....}}$ i know what i need to do – natsu Apr 19 '15 at 4:00
\begin{align} \mathsf E[X_{(7)}] & = \binom{7}{1}\int_1^\infty x\cdot f_{X}(x)\cdot (1-F_X(x))^6 \operatorname d x \\ & = 7\int_1^\infty x \cdot\frac {3}{x^4}\cdot \left(\int_x^\infty \frac {3}{y^4}\operatorname d y\right)^6\operatorname d x \end{align}