Q. 8.5 - A First Course in Probability

Question

8.5 The amount of time that a certain type of component functions before failing is a random variable with probability density function

$f\left(x\right)=2x 0<x< 1$ 

Once the component fails, it is immediately replaced by
another one of the same type. If we let denote the life-time of the $i$th component to be put in use, then $S_n=\sum _{i=1}^n{X}_i$represents the time of the $n$th failure. The long-term rate at which failures occur, call it $r$, is defined by
$r=\underset{n\to \infty }{\lim }\frac{n}{S_n}$

Assuming that the random variables $X i, i \ge 1,$are independent, determine $r$.

Step-by-Step Solution

Answer

The r is $\frac{3}{2}$.

Step 1: Given information

A random variable with probability density function is $f\left(x\right)=2x 0<x< 1.$Let $X_i$ denote the life-time of the $i$th component , then $S_n=\sum _{i=1}^n{X}_i$ represents the time of the $n$th failure. The long-term rate failures occur,  is defined by $r=\underset{n\to \infty }{\lim }\frac{n}{S_n}$.

Step 2: Explanation

Let $X_i$ denote the $i$th component and understand that $X_1,X_2,...$, is a sequence of independent and similarly distributed random variables, with finite mean.
$\mu =E\left[X_i\right]={\int }_0^{1}xf\left(x\right)dx={\int }_0^{1}2x^{2}dx={\left.2\frac{x^3}{3}\right|}_0^1=\frac{2}{3}$
Then, $S_n=\sum _{i=1}^n{X}_i$ represents the time of $n$th failure, with probability $1$,
$\frac{S_n}{n}\to \mu ,$as $n\to \infty$

Then,

$r=\underset{n\to \infty }{\lim }\frac{n}{S_n}=\underset{n\to \infty }{\lim }\frac{1}{\frac{S_n}{n}}=\frac{1}{\mu }=\frac{3}{2}$

Hence, $r=\frac{3}{2}$