/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 A flashlight needs two batteries... [FREE SOLUTION] | 91Ó°ÊÓ

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A flashlight needs two batteries to be operational. Consider such a flashlight along with a set of \(n\) functional batteries-battery 1 , battery \(2, \ldots\), battery \(n .\) Initially, battery 1 and 2 are installed. Whenever a battery fails, it is immediately replaced by the lowest numbered functional battery that has not yet been put in use. Suppose that the lifetimes of the different batteries are independent exponential random variables each having rate \(\mu .\) At a random time, call it \(T\), a battery will fail and our stockpile will be empty. At that moment exactly one of the batteries-which we call battery \(X\) -will not yet have failed. (a) What is \(P[X=n\\}\) ? (b) What is \(P[X=1\\} ?\) (c) What is \(P[X=i\\} ?\) (d) Find \(E[T]\). (e) What is the distribution of \(T ?\)

Short Answer

Expert verified
(a) \(P[X=n] = \frac{1}{n}\) (b) \(P[X=1] = \frac{1}{2^{n-1}}\) (c) \(P[X=i] = \binom{n-1}{i-1} \frac{1}{2^{n-1}}\) (d) \(E[T] = \mu \sum_{i=1}^{n} \frac{1}{n-i+1}\) (e) The distribution of \(T\) can be represented by the sum of \(n\) independent Erlang distributions with scale parameter \(\mu\).

Step by step solution

01

(a) Finding \(P[X=n]\)#

\ Given that the lifetimes of the batteries are independent, we can use the memoryless property of exponential distribution. In this case, we want to find the probability that battery \(n\) is the last one to fail. For this to happen, the first \(n-1\) batteries must fail before battery \(n\). This will happen if each of the batteries fails faster than the remaining time of battery \(n\). When a battery fails, we can ignore it. In general, the probability that exponential random variable \(X\) with rate \(\mu_{X}\) is smaller than another exponential random variable \(Y\) with rate \(\mu_{Y}\) equals to the ratio of their rates: \(P[X<Y] = \frac{\mu_{X}}{\mu_{X}+\mu_{Y}}\) In our case, since all batteries have the same rate \(\mu\), we have: \(P[X=n] = \frac{1}{2} \cdot \frac{2}{3} \cdots \frac{n-1}{n} = \frac{1}{n}\)
02

(b) Finding \(P[X=1]\)#

\ To find the probability that battery \(1\) never fails, we need to consider that all other batteries must fail before battery \(1\). Again, this will occur when the remaining battery lifetimes are smaller than the remaining time of battery \(1\). Hence, the probability is: \(P[X=1] = P[1\,>\,2] P[1\,>\,3|1\,>\,2] \cdots P[1\,>\,n|1\,>\,2,\ldots,1\,>\,n-1]\) Using the formula for the exponential random variables, we get: \(P[X=1] = \frac{1}{1+1} \cdot \frac{1}{1+1} \cdots \frac{1}{1+1} = \frac{1}{2^{n-1}}\)
03

(c) Finding \(P[X=i]\)#

\ For the general case, when battery \(i\) is the last one to fail, all batteries \(j \neq i\) should fail before it. We can find this probability by considering each battery independently: \(P[X=i] = P[i\,>\,1,\ldots,i\,>\,i-1,i\,>\,i+1,\ldots,i\,>\,n]\) Matching the probabilities as we did before: \(P[X=i] = \frac{(i-1)! (n-i)!}{(n-1)!} = \binom{n-1}{i-1} \frac{1}{2^{n-1}}\)
04

(d) Finding \(E[T]\)#

\ In order to find the expected time until all the batteries fail, we can compute the sum of the expected lifetimes divided by the number of batteries left. Recall that the expected value of an exponential random variable with rate \(\mu\) is \(\frac{1}{\mu}\): \(E[T] = E[T_1] + E[T_2] + \cdots + E[T_n]\) Here, \(T_i\) is the time it takes for the \(i^{th}\) battery to fail given that there are \(n-i+1\) batteries left. Therefore, the expected time until the \(i^{th}\) battery fails is given by: \(E[T_i] = \frac{1}{\frac{n-i+1}{\mu}} = \frac{\mu}{n-i+1}\) Summing these up to find the total expected time: \(E[T] = \sum_{i=1}^{n} E[T_i] = \mu \sum_{i=1}^{n} \frac{1}{n-i+1}\)
05

(e) Identifying the distribution of \(T\)#

\ To find the distribution of \(T\), we need to analyze the sum of the independent exponential random variables. Since these random variables are not identically distributed, \(T\) itself will not follow an exponential distribution. Instead, we can look at the sum of the minimum lifetimes of these random variables and identify their distribution. The minimum of the \(n-i+1\) exponential random variables, which represent the time it takes for the \(i^{th}\) battery to fail, follows an Erlang distribution with parameters \(k = 1\) (the first order) and \(\lambda = \frac{n-i+1}{\mu}\) (the rate). Therefore, the distribution of \(T\) can be represented by the sum of \(n\) independent Erlang distributions with scale parameter \(\mu\). In conclusion, the time to failure \(T\) is not distributed exponentially, but it follows the sum of several Erlang distributions with scale parameter \(\mu\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Random Variables
Exponential random variables are a core concept in probability and statistics, frequently used to model the time until an event occurs. This type of random variable is suitable for modeling events in time that occur continuously and independently at a constant average rate. In mathematical terms, if a random variable \(X\) follows an exponential distribution with rate \( \mu \), then the probability density function is given by:\[ f(x) = \mu e^{-\mu x}\, \text{for} \, x \geq 0 \]This distribution is defined by one parameter, \( \mu \), which is the rate or inverse of the mean (average time). The mean or expected value \( E[X] \) of an exponential random variable is \( \frac{1}{\mu} \). This property makes it particularly handy in reliability theory and queuing models to predict the expected "wait time" until the next event occurs. Furthermore, exponential distributions are memoryless. This means that the probability of an event occurring in the next time period is independent of how much time has already passed.
Erlang Distribution
The Erlang distribution is a special case of the gamma distribution, often used in the context of waiting times where multiple events occur sequentially. It is especially useful in scenarios where you want to model the time until the first of several events has occurred. The Erlang distribution is described by two parameters: the shape \( k \), which must be a positive integer, and the rate \( \lambda \). The probability density function of the Erlang distribution is:\[ f(x; k, \lambda) = \frac{\lambda^k x^{k-1} e^{-\lambda x}}{(k-1)!}\, \text{for} \, x \geq 0 \]In scenarios like the one in the original exercise, the time until failure of a set of batteries, when each follows an exponential distribution and events occur independently, the sum of such exponential variables leads to an Erlang distribution. This is because the aggregated process can be viewed as waiting for \( k \) independent exponential events to occur, where \( k \) is the number of stages or failures required.
Independent Events
Independence is a key property in probability theory, particularly relevant when dealing with multiple random variables or events. Two events are independent if the occurrence of one event does not affect the probability of the other occurring. Mathematically, this is defined as:\[ P(A \cap B) = P(A) \times P(B) \]In the context of the given exercise, the lifetimes of batteries are stated to be independent exponential random variables. This independence implies that the lifetime of one battery does not impact or interfere with the lifetime of another. Hence, the failure or survival of one battery is entirely unrelated to that of another. This allows us to analyze each battery's failure probability in isolation, and significantly simplifies calculations using tools like the memoryless property of the exponential distribution and the characteristics of sums of independent variables.
Expected Value
Expected value, often referred to as the mean, is a fundamental concept in statistics that represents the average or mean value of a random variable if you could repeat the experiment an infinite number of times. For a discrete random variable, the expected value \( E[X] \) is calculated as:\[ E[X] = \sum_{i=1}^{n} x_i \cdot P(x_i) \]For continuous random variables, the expected value is calculated using an integral:\[ E[X] = \int_{-\infty}^{\infty} x \cdot f(x) \, dx \]In the context of the problem, the expected value \(E[T]\) of the time until the stockpile is empty is crucial. It is determined by summing up the expected times until each of the \(n\) batteries fails. Knowing the expected value helps in planning and managing resources effectively since it provides an average measure of the time until an event or series of events occurs.

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Most popular questions from this chapter

For the conditional Poisson process, let \(m_{1}=E[L], m_{2}=E\left[L^{2}\right] .\) In terms of \(m_{1}\) and \(m_{2}\), find \(\operatorname{Cov}(N(s), N(t))\) for \(s \leqslant t .\)

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