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# ISI MStat PSB 2014 Problem 4 | The Machine's Failure

This is a very simple sample problem from ISI MStat PSB 2014 Problem 4. It is based on order statistics, but generally due to one's ignorance towards order statistics, one misses the subtleties . Be Careful !

## Problem- ISI MStat PSB 2014 Problem 4

Consider a machine with three components whose times to failure are independently distributed as exponential random variables with mean $\lambda$. the machine continue to work as long as at least two components work. Find the expected time to failure of the machine.

### Prerequisites

Exponential Distribution

Order statistics

Basic counting

## Solution :

In the problem as it is said, let the 3 component part of the machine be A,B and C respectively, where $X_A, X_B$ and $X_C$ are the survival time of the respective parts. Now it is also told that $X_A, X_B$ and $X_C$ follows $exponential(\lambda)$, and clearly these random variables are also i.id.

Now, here comes the trick ! It is told that the machine stops when two or all parts of the machine stop working. Here, we sometimes gets confused and start thinking combinatorially. But the we forget the basic counting of combinatorics lies in ordering ! Suppose we start ordering the life time of the individual components .i.e. among $X_A, X_B$ and $X_C$, there exists a ordering and say if we write it in order, we have $X_{(1)} \le X_{(2)} \le X_{(3)}$.

Now observe that, after $X_{(2)}$ units of time, the machine will stop !! (Are you sure ?? think it over ).

So, expected time till the machine stops , is just $E(X_{(2)})$, but to find this we need to know the distribution of $X_{(2)}$.

We have the pdf of $X_{(2)}$ as, $f_{(2)}(x)= \frac{3!}{(2-1)!(3-2)!} [P(X \le x)]^{2-1}[P(X>x)]^{3-2}f_X(x)$.

Where $f_X(x)$ is the pdf of exponentional with mean $\lambda$.

So, $E(X(2))= \int^{\infty}_0 xf_{(2)}(x)dx$. which will turn out to be $\frac{5\lambda}{6}$, which I leave on the readers to verify , hence concluding my solution.

## Food For Thought

Now, suppose, you want install an alarm system, which will notify you some times before the machine wears our!! So, what do you think your strategy should be ? Given that you have a strategy, you now replace the weared out part of the machine within the time period between the alarm rings and the machine stops working, to continue uninterrupted working.What is the expected time within which you must act ?

Keep the machine running !!

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This is a very simple sample problem from ISI MStat PSB 2014 Problem 4. It is based on order statistics, but generally due to one's ignorance towards order statistics, one misses the subtleties . Be Careful !

## Problem- ISI MStat PSB 2014 Problem 4

Consider a machine with three components whose times to failure are independently distributed as exponential random variables with mean $\lambda$. the machine continue to work as long as at least two components work. Find the expected time to failure of the machine.

### Prerequisites

Exponential Distribution

Order statistics

Basic counting

## Solution :

In the problem as it is said, let the 3 component part of the machine be A,B and C respectively, where $X_A, X_B$ and $X_C$ are the survival time of the respective parts. Now it is also told that $X_A, X_B$ and $X_C$ follows $exponential(\lambda)$, and clearly these random variables are also i.id.

Now, here comes the trick ! It is told that the machine stops when two or all parts of the machine stop working. Here, we sometimes gets confused and start thinking combinatorially. But the we forget the basic counting of combinatorics lies in ordering ! Suppose we start ordering the life time of the individual components .i.e. among $X_A, X_B$ and $X_C$, there exists a ordering and say if we write it in order, we have $X_{(1)} \le X_{(2)} \le X_{(3)}$.

Now observe that, after $X_{(2)}$ units of time, the machine will stop !! (Are you sure ?? think it over ).

So, expected time till the machine stops , is just $E(X_{(2)})$, but to find this we need to know the distribution of $X_{(2)}$.

We have the pdf of $X_{(2)}$ as, $f_{(2)}(x)= \frac{3!}{(2-1)!(3-2)!} [P(X \le x)]^{2-1}[P(X>x)]^{3-2}f_X(x)$.

Where $f_X(x)$ is the pdf of exponentional with mean $\lambda$.

So, $E(X(2))= \int^{\infty}_0 xf_{(2)}(x)dx$. which will turn out to be $\frac{5\lambda}{6}$, which I leave on the readers to verify , hence concluding my solution.

## Food For Thought

Now, suppose, you want install an alarm system, which will notify you some times before the machine wears our!! So, what do you think your strategy should be ? Given that you have a strategy, you now replace the weared out part of the machine within the time period between the alarm rings and the machine stops working, to continue uninterrupted working.What is the expected time within which you must act ?

Keep the machine running !!

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