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# Life Testing Experiment | ISI MStat 2017 PSB Problem 5

This is a problem from the ISI MStat 2017 Entrance Examination and tests how good are your skills in modeling a life testing experiment using an exponential distribution.

## The Problem:

The lifetime in hours of each bulb manufactured by a particular company follows an independent exponential distribution with mean $\lambda$. We need to test the null hypothesis $H_0: \lambda=1000$ against $H_1:\lambda=500$.
A statistician sets up an experiment with $50$ bulbs, with $5$ bulbs in each of $10$ different locations, to examine their lifetimes.

To get quick preliminary results,the statistician decides to stop the experiment as soon as one bulb fails at each location.Let $Y_i$ denote the lifetime of the first bulb to fail at location $i$.Obtain the most powerful test of $H_0$ against $H_1$ based on $Y_1,Y_2,…Y_{10}$ and compute its power.

## Prerequisites:

1.Properties of Exponential/Gamma distribution.

3.Order Statistics.

## Proof:

As it is clear from the arrangement of the bulbs, the first to fail(among 5 in a given location) has the smallest lifetime among the same.

That is, in more mathematical terms, for a location $i$, we can write $Y_i = \text{min}(X_{i1},X_{i2},..,X_{i5})$.

Here, $X_{ij}$ denotes the $j$ th unit in the $i th$ location where $i=1,2,..,10$ and $j=1,2,..,5$

It is given that $X_{ij} \sim \text{Exp}(\lambda)$.

Can you see that $Y_i \sim \text{Exp}(5 \lambda)$? You may try to prove this result for this:

If $X_1,..,X_n$ be a random sample from $\text{Exp}(\lambda)$ distribution,

then $X_{(1)}=\text{min}(X_1,....,X_n) \sim \text{Exp}(n \lambda)$.

So, now we have $Y_1,Y_2,..Y_{10}$ in hand each having $\text{Exp}(5 \lambda)$ distribution.

Let the joint pdf be $f(\mathbf{y} )=\frac{1}{(5 \lambda)^{10}} e^{-\frac{\sum_{i=1}^{10} y_i}{5 \lambda}}$.

For testing $H_0: \lambda=1000$ against $H_1:\lambda=500$, we use the Neyman Pearson Lemma.

We have the critical region of the most powerful test as $\frac{f_{H_1}(\mathbf{y})}{f_{H_0}(\mathbf{y})} >c$

which after simplification comes out to be $\bar{Y} > K$ where $K$ is an appropriate constant.

Also, see that $\bar{Y} \sim \text{Gamma}(10,50 \lambda)$.

Can you use this fact to find the value of $K$ using the size ($\alpha$) criterion ? (Exercise to the reader)

Also, find the power of the test.

## Challenge Problem:

The exponential distribution is used widely to model lifetime of appliances. The following scenario is based on such a model.

Suppose electric bulbs have a lifetime distribution with pdf $f(t)=\lambda e^{-\lambda t}$ where $t \in [0, \infty)$ .

These bulbs are used individually for street lighting in a large number of posts.A bulb is replaced immediately after it burns out.

Let's break down the problem in steps.

(i)Starting from time $t=0$ , the process is observed till $t=T$.Can you calculate the expected number of replacements in a post during the interval $(0,T)$ ?

(ii) Hence,deduce $g(t) \text{dt}$ ,the probability of a bulb being replaced in $(t,t+ \text{dt})$ for $t < T$,irrespective of when the bulb was put in.

(iii)Next,suppose that at the end of the first interval of time $T$,all bulbs which were put in the posts before time $X < T$ and have not burned out are replaced by new ones,but the bulbs replaced after ttime $X$ continue to be used,provided,of course,that they have not burned out.

Prove that with such a mixture of old and new bulbs, the probability of a bulb having an expected lifetime > $\tau$ in the second interval of length $T$ is given by

$S_2(\tau)=\frac{1}{2}e^{-\lambda \tau}(1+ e^{-\lambda X})$

Also, try proving the general case where the lifetimes of the bulbs follow the pdf $f(t)$ . Here, $f(t)$ need not be the pdf of an exponential distribution .

You should be getting: $S_2(\tau)=(1-p)S_1(\tau) + \int_{0}^{x} g(T-x)S_1(x)S_1(\tau +x) \text{dx}$ ; where $\tau<T$

where, $p$ is the proportion of bulbs not replaced at time $t=T$ and $S_1(t)$ is the probability that a bulb has lifetime > $t$.

This is a problem from the ISI MStat 2017 Entrance Examination and tests how good are your skills in modeling a life testing experiment using an exponential distribution.

## The Problem:

The lifetime in hours of each bulb manufactured by a particular company follows an independent exponential distribution with mean $\lambda$. We need to test the null hypothesis $H_0: \lambda=1000$ against $H_1:\lambda=500$.
A statistician sets up an experiment with $50$ bulbs, with $5$ bulbs in each of $10$ different locations, to examine their lifetimes.

To get quick preliminary results,the statistician decides to stop the experiment as soon as one bulb fails at each location.Let $Y_i$ denote the lifetime of the first bulb to fail at location $i$.Obtain the most powerful test of $H_0$ against $H_1$ based on $Y_1,Y_2,…Y_{10}$ and compute its power.

## Prerequisites:

1.Properties of Exponential/Gamma distribution.

3.Order Statistics.

## Proof:

As it is clear from the arrangement of the bulbs, the first to fail(among 5 in a given location) has the smallest lifetime among the same.

That is, in more mathematical terms, for a location $i$, we can write $Y_i = \text{min}(X_{i1},X_{i2},..,X_{i5})$.

Here, $X_{ij}$ denotes the $j$ th unit in the $i th$ location where $i=1,2,..,10$ and $j=1,2,..,5$

It is given that $X_{ij} \sim \text{Exp}(\lambda)$.

Can you see that $Y_i \sim \text{Exp}(5 \lambda)$? You may try to prove this result for this:

If $X_1,..,X_n$ be a random sample from $\text{Exp}(\lambda)$ distribution,

then $X_{(1)}=\text{min}(X_1,....,X_n) \sim \text{Exp}(n \lambda)$.

So, now we have $Y_1,Y_2,..Y_{10}$ in hand each having $\text{Exp}(5 \lambda)$ distribution.

Let the joint pdf be $f(\mathbf{y} )=\frac{1}{(5 \lambda)^{10}} e^{-\frac{\sum_{i=1}^{10} y_i}{5 \lambda}}$.

For testing $H_0: \lambda=1000$ against $H_1:\lambda=500$, we use the Neyman Pearson Lemma.

We have the critical region of the most powerful test as $\frac{f_{H_1}(\mathbf{y})}{f_{H_0}(\mathbf{y})} >c$

which after simplification comes out to be $\bar{Y} > K$ where $K$ is an appropriate constant.

Also, see that $\bar{Y} \sim \text{Gamma}(10,50 \lambda)$.

Can you use this fact to find the value of $K$ using the size ($\alpha$) criterion ? (Exercise to the reader)

Also, find the power of the test.

## Challenge Problem:

The exponential distribution is used widely to model lifetime of appliances. The following scenario is based on such a model.

Suppose electric bulbs have a lifetime distribution with pdf $f(t)=\lambda e^{-\lambda t}$ where $t \in [0, \infty)$ .

These bulbs are used individually for street lighting in a large number of posts.A bulb is replaced immediately after it burns out.

Let's break down the problem in steps.

(i)Starting from time $t=0$ , the process is observed till $t=T$.Can you calculate the expected number of replacements in a post during the interval $(0,T)$ ?

(ii) Hence,deduce $g(t) \text{dt}$ ,the probability of a bulb being replaced in $(t,t+ \text{dt})$ for $t < T$,irrespective of when the bulb was put in.

(iii)Next,suppose that at the end of the first interval of time $T$,all bulbs which were put in the posts before time $X < T$ and have not burned out are replaced by new ones,but the bulbs replaced after ttime $X$ continue to be used,provided,of course,that they have not burned out.

Prove that with such a mixture of old and new bulbs, the probability of a bulb having an expected lifetime > $\tau$ in the second interval of length $T$ is given by

$S_2(\tau)=\frac{1}{2}e^{-\lambda \tau}(1+ e^{-\lambda X})$

Also, try proving the general case where the lifetimes of the bulbs follow the pdf $f(t)$ . Here, $f(t)$ need not be the pdf of an exponential distribution .

You should be getting: $S_2(\tau)=(1-p)S_1(\tau) + \int_{0}^{x} g(T-x)S_1(x)S_1(\tau +x) \text{dx}$ ; where $\tau<T$

where, $p$ is the proportion of bulbs not replaced at time $t=T$ and $S_1(t)$ is the probability that a bulb has lifetime > $t$.

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### One comment on “Life Testing Experiment | ISI MStat 2017 PSB Problem 5”

1. Rhritajit Sen says:

Here we have given that Xij follows exponential with mean lambda that is exponential distribution with parameter 1/ lambda. But it is written here that Xij follows exponential with parameter lambda. Is it not a mistake made in this solution?