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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 \(\lambda\).

Short Answer

Expert verified
Use numerical methods to solve for \( \lambda \). The MLE equation doesn't have a simple analytical solution.

Step by step solution

01

Understand the Problem

We have 20 components being tested, each with an exponential lifetime distribution characterized by the rate parameter \( \lambda \). After 24 hours, 5 components have failed, and 15 are still operational. We need to find the maximum likelihood estimate (MLE) of \( \lambda \).
02

Define the Exponential Distribution

An exponential distribution is defined such that the probability of a single component not failing by time \( t \) is given by the survival function \( S(t) = e^{-\lambda t} \). The probability of a single component failing by time \( t \) is then \( 1 - S(t) = 1 - e^{-\lambda t} \).
03

Set Up the Likelihood Function

For 15 components to survive and 5 to fail, we use the survival function for 15 components and the probability of failure for 5 components. The likelihood is given by: \[L(\lambda) = \left(e^{-\lambda \times 24}\right)^{15} \times \left(1 - e^{-\lambda \times 24}\right)^{5}.\]
04

Take the Log of the Likelihood Function

To simplify calculations, we take the natural log of the likelihood function:\[\ln L(\lambda) = 15 \times (-\lambda \times 24) + 5 \times \ln(1 - e^{-\lambda \times 24}).\]
05

Differentiate and Solve for \( \lambda \)

Differentiate \( \ln L(\lambda) \) with respect to \( \lambda \) and set the derivative equal to zero to find the MLE. Solve: \[\frac{d}{d\lambda}\left[15 \times (-24\lambda) + 5 \times \ln(1 - e^{-\lambda \times 24})\right] = 0.\]On differentiating and simplifying, the MLE equation can be numerically solved. This will typically be done using computational tools as the equation does not have a closed-form analytical solution.
06

Estimate \( \lambda \) Using Numerical Methods

Given that the differentiated equation is complex, use software or a numerical method such as the Newton-Raphson method to solve for \( \lambda \). For rough estimates, the failure rate can be initially assumed over 24 hours and adjusted as per the solution from methods used.

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Key Concepts

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

Maximum Likelihood Estimation
To begin with, Maximum Likelihood Estimation (MLE) is an important method used in statistics to estimate the parameter of a statistical model. In simple terms, it tries to find the parameter values that make the observed data most probable.
In the context of our problem, we are dealing with components whose lifetimes are described by an exponential distribution characterized by the parameter \( \lambda \). By observing the system for a period and recording the number of failures, we aim to determine the best estimate for \( \lambda \).
Using the exponential distribution's properties, we formulate a likelihood function that incorporates both the probability of failure and survival.
The likelihood maximization involves constructing a function dependent on \( \lambda \) and finding where this function peaks. This peak represents our maximum likelihood estimate. After setting up the function, the logarithm of the likelihood is taken to simplify calculations before differentiating to find \( \lambda \).
Ultimately, MLE provides a systematic way to use your observations to make precise and useful inferences about the underlying parameters of your model.
Survival Function
The survival function is a key concept when dealing with lifetime distributions and describes the probability that a particular component or entity will survive beyond a specific time. In statistical terms, it is denoted as \( S(t) \) and calculated as:
\[ S(t) = e^{-\lambda t} \]
For exponential distributions, it directly relates to the parameter \( \lambda \), representing how rapidly failures occur. In the example at hand, since 15 components survived the 24-hour test period, we utilize the survival function in our likelihood expression.
This function gives us the probability that an individual component avoids failure up to that time point, and by raising it to the number of surviving components, we account for all that survived. The survival and failure probabilities combined enable us to set an equation reflecting the most likely parameter for our observations.
Hence, the survival function is pivotal in understanding and computing probabilities within reliability and life data analyses.
Probability of Failure
While the survival function gives insight into how likely an item is to survive a certain duration, it’s equally valuable to consider the probability of failure. For exponential distributions, the failure probability at time \( t \) can be formulated as:
\[ 1 - S(t) = 1 - e^{-\lambda t} \]
This expression calculates the chance that an event like a component failure has occurred by some specified time. Our exercise document considers five failed components through this perspective.
In crafting our likelihood function, we therefore multiply the probability of five components failing, by the probability of the rest surviving. This balance gives a holistic view of what happened over the monitoring period, allowing us to productively solve for \( \lambda \).
The probability of failure is an integral component in arriving at an MLE and enables you to measure and predict the vulnerabilities in any system or set of processes.
Numerical Methods
Numerical methods are essential when analytical solutions are either infeasible or impossible to derive. Often, equations resulting from complex differentiation, such as those involving MLE, do not resolve neatly. For these situations, numerical techniques can solve for parameters like \( \lambda \).
One common and powerful numerical method is the Newton-Raphson method. This iterative method helps find successively better approximations to the roots (or zeroes) of a real-valued function. Applying such a method lets us home in on the solution for our problem where the analytical path doesn’t exist.
When using numerical methods, initial estimates are typically required, and the algorithm refines these guesses. This makes numerical methods highly adaptable across various problems in statistical analysis.
They are integral to real-world applications when dealing with complex models and datasets, ensuring that solutions are both practical and grounded in the observed data.

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Most popular questions from this chapter

Consider the accompanying observations on stream flow (1000s of acre-feet) recorded at a station in Colorado for the period April 1-August 31 over a 31-year span (from an article in the 1974 volume of Water 91Ó°ÊÓ Research). $$ \begin{array}{rrrrr} 127.96 & 210.07 & 203.24 & 108.91 & 178.21 \\ 285.37 & 100.85 & 89.59 & 185.36 & 126.94 \\ 200.19 & 66.24 & 247.11 & 299.87 & 109.64 \\ 125.86 & 114.79 & 109.11 & 330.33 & 85.54 \\ 117.64 & 302.74 & 280.55 & 145.11 & 95.36 \\ 204.91 & 311.13 & 150.58 & 262.09 & 477.08 \\ 94.33 & & & & \end{array} $$ An appropriate probability plot supports the use of the lognormal distribution (see Section 4.5) as a reasonable model for stream flow. a. Estimate the parameters of the distribution. [Hint: Remember that \(X\) has a lognormal distribution with parameters \(\mu\) and \(\sigma^{2}\) if \(\ln (X)\) is normally distributed with mean \(\mu\) and variance \(\sigma^{2}\).] b. Use the estimates of part (a) to calculate an estimate of the expected value of stream flow.

A sample of \(n\) captured Pandemonium jet fighters results in serial numbers \(x_{1}, x_{2}, x_{3}, \ldots, x_{n}\). 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.

At time \(t=0\), there is one individual alive in a certain population. A pure birth process then unfolds as follows. The time until the first birth is exponentially distributed with parameter \(\lambda\). After the first birth, there are two individuals alive. The time until the first gives birth again is exponential with parameter \(\lambda\), and similarly for the second individual. Therefore, the time until the next birth is the minimum of two exponential \((\lambda)\) variables, which is exponential with parameter \(2 \lambda\). Similarly, once the second birth has occurred, there are three individuals alive, so the time until the next birth is an exponential rv with parameter \(3 \lambda\), and so on (the memoryless property of the exponential distribution is being used here). Suppose the process is observed until the sixth birth has occurred and the successive birth times are \(25.2\), \(41.7,51.2,55.5,59.5,61.8\) (from which you should calculate the times between successive births). Derive the mle of \(\lambda\).

Each of \(n\) specimens is to be weighed twice on the same scale. Let \(X_{i}\) and \(Y_{i}\) denote the two observed weights for the \(i\) th specimen. Suppose \(X_{i}\) and \(Y_{i}\) are independent of one another, each normally distributed with mean value \(\mu_{i}\) (the true weight of specimen \(i\) ) and variance \(\sigma^{2}\). a. Show that the maximum likelihood estimator of \(\sigma^{2}\) is \(\hat{\sigma}^{2}=\sum\left(X_{i}-Y_{i}\right)^{2} /(4 n)\). [Hint: If \(\bar{z}=\left(z_{1}+z_{2}\right) / 2\), then \(\left.\Sigma\left(z_{i}-\bar{z}\right)^{2}=\left(z_{1}-z_{2}\right)^{2} / 2 .\right]\) b. Is the mle \(\hat{\sigma}^{2}\) an unbiased estimator of \(\sigma^{2}\) ? Find an unbiased estimator of \(\sigma^{2}\).

a. A random sample of 10 houses in a particular area, each of which is heated with natural gas, is selected and the amount of gas (therms) used during the month of January is determined for each house. The resulting observations are \(103,156,118,89,125,147,122,109,138,99\). Let \(\mu\) denote the average gas usage during January by all houses in this area. Compute a point estimate of \(\mu\). b. Suppose there are 10,000 houses in this area that use natural gas for heating. Let \(\tau\) denote the total amount of gas used by all of these houses during January. Estimate \(\tau\) using the data of part (a). What estimator did you use in computing your estimate? c. Use the data in part (a) to estimate \(p\), the proportion of all houses that used at least 100 therms. d. Give a point estimate of the population median usage (the middle value in the population of all houses) based on the sample of part (a). What estimator did you use?

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