/*! 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 30 At time \(t=0,20\) identical com... [FREE SOLUTION] | 91Ó°ÊÓ

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At time \(t=0,20\) identical components are tested. The lifetime distribution of each is exponential with parameter \(\lambda\). The experimenter then leaves the test facility unmonitored. On his return 24 hours later, the experimenter immediately terminates the test after noticing that \(y=15\) of the 20 components are still in operation (so 5 have failed). Derive the mle of \(\lambda\). [Hint: Let \(Y=\) the number that survive 24 hours. Then \(Y \sim \operatorname{Bin}(n, p)\). What is the mle of \(p\) ? Now notice that \(p=P\left(X_{i} \geq 24\right)\), where \(X_{i}\) is exponentially distributed. This relates \(\lambda\) to \(p\), so the former can be estimated once the latter has been.]

Short Answer

Expert verified
The MLE of \(\lambda\) is approximately 0.012.

Step by step solution

01

Understanding the Problem

We are given a scenario where 20 components are tested, and after 24 hours, 15 are still functioning. We need to determine the maximum likelihood estimate (MLE) of the rate parameter \(\lambda\) of the exponential distribution, which models the component lifetimes.
02

Determine the Distribution of Survivors

Let \(Y\) be the number of components that survive 24 hours. \(Y\) follows a Binomial distribution: \(Y \sim \text{Binomial}(n=20, p)\), where \(p\) is the probability that a component survives 24 hours.
03

Relate Probability \(p\) to Exponential Distribution

A component survives at least 24 hours if its lifetime \(X_i\) is greater than or equal to 24. For an exponential distribution with parameter \(\lambda\), \(p = P(X_i \geq 24) = e^{-\lambda \cdot 24}\).
04

Find the MLE of \(p\)

The MLE of \(p\) for a Binomial distribution is \(\hat{p} = \frac{Y}{n}\). Substituting the given values: \(\hat{p} = \frac{15}{20} = 0.75\).
05

Solve for \(\lambda\) using \(\hat{p}\)

Set \(e^{-\lambda \cdot 24} = 0.75\). Taking the natural logarithm of both sides: \[-\lambda \cdot 24 = \ln(0.75)\].Solve for \(\lambda\): \[\lambda = -\frac{\ln(0.75)}{24}\].
06

Calculate \(\lambda\)

Substitute \(\ln(0.75) = -0.2877\) (approximately) to find \(\lambda\):\[\lambda = -\frac{-0.2877}{24} \approx 0.012\].

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Key Concepts

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

Exponential Distribution
The exponential distribution is a continuous probability distribution often used to model the time until an event occurs, such as the lifespan of an electronic component or the time until a radioactive particle decays. It is defined by its rate parameter, \( \lambda \), which determines the distribution's shape.

Key characteristics of the exponential distribution include:
  • Memoryless Property: The probability of surviving past a certain time is independent of how much time has already passed. This property is unique to the exponential distribution.
  • Probability Density Function (PDF): Given by \( f(x | \lambda) = \lambda e^{-\lambda x} \) for \( x \geq 0 \). Here, \( \lambda \) is the rate parameter, and \( e \) is the base of the natural logarithm.
  • Mean and Standard Deviation: Both the mean and standard deviation of an exponential distribution are \( \frac{1}{\lambda} \).
The distribution is frequently chosen in reliability engineering and queuing theory due to its simplicity and the memoryless property.
Maximum Likelihood Estimation
Maximum likelihood estimation (MLE) is a method used in statistics to estimate the parameters of a statistical model.
It works by finding the parameter values that maximize the likelihood function, thus making the observed data most probable.

For an exponential distribution with rate parameter \( \lambda \), the likelihood function \( L(\lambda) \) for a set of observations \( x_1, x_2, ..., x_n \) is given by:
\[ L(\lambda) = \prod_{i=1}^{n} (\lambda e^{-\lambda x_i}) \]

To obtain the MLE, take the natural logarithm of the likelihood function, known as the log-likelihood:
\[ \log L(\lambda) = n \log \lambda - \lambda \sum_{i=1}^{n} x_i \]

Solve for \( \lambda \) by differentiating with respect to \( \lambda \) and equating to zero.
This leads to an explicit solution for \( \lambda \) that is easy to compute, making MLE a powerful tool for parameter estimation.
MLE is appreciated for its consistency, meaning that as the sample size increases, the estimated parameters converge to the true parameters.
Binomial Distribution
The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent Bernoulli trials, each with the same probability of success, denoted by \( p \).
It is characterized by two parameters: \( n \), the number of trials, and \( p \), the probability of success on each trial.

The probability of getting exactly \( k \) successes in \( n \) trials is given by:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]

Key properties include:
  • Mean: \( n \cdot p \)
  • Variance: \( n \cdot p \cdot (1-p) \)
  • Supports: Any integer from 0 to \( n \)
In the context of the MLE problem, we consider the number of surviving components as a binomial random variable with probability \( p = e^{-\lambda \times 24} \), where each component's survival is a Bernoulli trial.
Rate Parameter Estimation
Rate parameter estimation in the context of an exponential distribution involves determining the value of \( \lambda \), which defines the distribution's behavior.
This is crucial in real-world applications such as estimating failure rates of machine components or predicting waiting times.

Using maximum likelihood estimation, \( \lambda \) is derived by relating it to a binomial probability, as was done in the given problem.
Here's a step-by-step approach:
  • Determine the probability \( p \) of survival (\( e^{-\lambda \times 24} \)) from the binomial distribution.
  • Calculate the MLE of \( p \) as \( \hat{p} = \frac{Y}{n} \) with given data.
  • Use the relationship \( \hat{p} = e^{-\lambda \times 24} \) to solve for \( \lambda \).
This provides a simple yet effective way of estimating the rate parameter by leveraging observed data and known probabilistic relationships.

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Most popular questions from this chapter

Of \(n_{1}\) randomly selected male smokers, \(X_{1}\) smoked filter cigarettes, whereas of \(n_{2}\) randomly selected female smokers, \(X_{2}\) smoked filter cigarettes. Let \(p_{1}\) and \(p_{2}\) denote the probabilities that a randomly selected male and female, respectively, smoke filter cigarettes. a. Show that \(\left(X_{1} / n_{1}\right)-\left(X_{2} / n_{2}\right)\) is an unbiased estimator for \(p_{1}-p_{2}\) - [Hint: \(E\left(X_{i}\right)=n_{i} p_{i}\) for \(i=1,2\).] b. What is the standard error of the estimator in part (a)? c. How would you use the observed values \(x_{1}\) and \(x_{2}\) to estimate the standard error of your estimator? d. If \(n_{1}=n_{2}=200, x_{1}=127\), and \(x_{2}=176\), use the estimator of part (a) to obtain an estimate of \(p_{1}-p_{2}\). e. Use the result of part (c) and the data of part (d) to estimate the standard error of the estimator.

Using a long rod that has length \(\mu\), you are going to lay out a square plot in which the length of each side is \(\mu\). Thus the area of the plot will be \(\mu^{2}\). However, you do not know the value of \(\mu\), so you decide to make \(n\) independent measurements \(X_{1}, X_{2}, \ldots, X_{n}\) of the length. Assume that each \(X_{i}\) has mean \(\mu\) (unbiased measurements) and variance \(\sigma^{2}\). a. Show that \(\bar{X}^{2}\) is not an unbiased estimator for \(\mu^{2}\). [Hint: For any rv \(Y, E\left(Y^{2}\right)=V(Y)+[E(Y)]^{2}\). Apply this with \(Y=\bar{X} .]\) b. For what value of \(k\) is the estimator \(\bar{X}^{2}-k S^{2}\) unbiased for \(\mu^{2}\) ? [Hint: Compute \(E\left(\bar{X}^{2}-k S^{2}\right)\).]

A sample of \(n\) captured Pandemonium jet fighters results in serial numbers \(x_{1}, x_{2}, x_{3}, \ldots, x_{\kappa}\). The CIA knows that the aircraft were numbered consecutively at the factory starting with \(\alpha\) and ending with \(\beta\), so that the total number of planes manufactured is \(\beta-\alpha+1\) (e.g., if \(\alpha=17\) and \(\beta=29\), then \(29-17+1=13\) planes having serial numbers \(17,18,19, \ldots, 28,29\) were manufactured). However, the CIA does not know the values of \(\alpha\) or \(\beta\). A CIA statistician suggests using the estimator \(\max \left(X_{i}\right)-\min \left(X_{i}\right)+1\) to estimate the total number of planes manufactured. a. If \(n=5, x_{1}=237, x_{2}=375, x_{3}=202, x_{4}=525\), and \(x_{5}=418\), what is the corresponding estimate? b. Under what conditions on the sample will the value of the estimate be exactly equal to the true total number of planes? Will the estimate ever be larger than the true total? Do you think the estimator is unbiased for estimating \(\beta-\alpha+1\) ? Explain in one or two sentences.

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Let \(X\) denote the proportion of allotted time that a randomly selected student spends working on a certain aptitude test. Suppose the pdf of \(X\) is $$ f(x ; \theta)=\left\\{\begin{array}{cl} (\theta+1) x^{\theta} & 0 \leq x \leq 1 \\ 0 & \text { otherwise } \end{array}\right. $$ where \(-1<\theta\). A random sample of ten students yields data \(x_{1}=.92, x_{2}=.79, x_{3}=.90, x_{4}=.65, x_{5}=.86\), \(x_{6}=.47, x_{7}=.73, x_{9}=.97, x_{9}=.94, x_{10}=.77\). a. Use the method of moments to obtain an estimator of \(\theta\), and then compute the estimate for this data. b. Obtain the maximum likelihood estimator of \(\theta\), and then compute the estimate for the given data.

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