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Consider a random sample \(X_{1}, X_{2}, \ldots, X_{n}\) from the shifted exponential pdf $$ f(x ; \lambda, \theta)=\left\\{\begin{array}{cl} \lambda e^{-\lambda(x-\theta)} & x \geq \theta \\ 0 & \text { otherwise } \end{array}\right. $$ Taking \(\theta=0\) gives the pdf of the exponential distribution considered previously (with positive density to the right of zero). An example of the shifted exponential distribution appeared in Example \(4.5\), in which the variable of interest was time headway in traffic flow and \(\theta=.5\) was the minimum possible time headway. a. Obtain the maximum likelihood estimators of \(\theta\) and \(\lambda\). b. If \(n=10\) time headway observations are made, resulting in the values \(3.11, .64,2.55,2.20,5.44,3.42,10.39,8.93\), \(17.82\), and \(1.30\), calculate the estimates of \(\theta\) and \(\lambda\).

Short Answer

Expert verified
MLE of \( \theta \) is 0.64, and MLE of \( \lambda \) is approximately 0.234.

Step by step solution

01

Understanding the likelihood function

The pdf of the shifted exponential distribution is given by \( f(x; \lambda, \theta) = \lambda e^{-\lambda(x-\theta)} \) for \( x \geq \theta \). For a sample \( X_1, X_2, \ldots, X_n \), the likelihood function \( L(\lambda, \theta) \) is \( \prod_{i=1}^{n} \lambda e^{-\lambda(X_i-\theta)}\). We can simplify it to \( \lambda^n \exp\left(-\lambda\sum_{i=1}^{n}(X_i-\theta)\right) \).
02

Finding the MLE for \( \theta \)

The likelihood involves \( \theta \) within the exponential function and as a starting point of \( x \). To maximize it concerning \( \theta \), consider the condition \( x \geq \theta \). Thus, to ensure the highest likelihood, since \( \sum_{i=1}^{n}(X_i-\theta) \) is minimized when \( \theta \) is maximized, we use \( \theta = \min(X_1, X_2, \ldots, X_n) \).
03

Calculating the MLE for \( \theta \)

Given the observations \( 3.11, 0.64, 2.55, 2.20, 5.44, 3.42, 10.39, 8.93, 17.82, 1.30 \), we identify \( \theta = 0.64 \) as it is the smallest value among the observations.
04

Finding the MLE for \( \lambda \)

The derivative of the log-likelihood with respect to \( \lambda \) is\\( \frac{\partial}{\partial \lambda}\left( n \ln \lambda - \lambda \sum_{i=1}^{n}(X_i - \theta)\right) = \frac{n}{\lambda} - \sum_{i=1}^{n}(X_i - \theta) \)\Setting the derivative to zero gives \( \hat{\lambda} = \frac{n}{\sum_{i=1}^{n}(X_i - \theta)} \).
05

Calculating the MLE for \( \lambda \)

Using \( \theta = 0.64 \) and the sample data, calculate \( \sum_{i=1}^{n}(X_i - 0.64) = 3.11 + 0 + 2.55+ 2.20 + 5.44 + 3.42 + 10.39 + 8.93 + 17.82 + 1.30 - 10 \times 0.64 \). This simplifies to \( 42.72 \). Then \( \hat{\lambda} = \frac{10}{42.72} \approx 0.234 \).

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Key Concepts

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

Shifted Exponential Distribution
The shifted exponential distribution is a fascinating variant of the exponential distribution. In this distribution, there's an additional parameter called the shift parameter \( \theta \). This shift moves the starting point of the distribution along the x-axis. The probability density function (pdf) for this distribution is:
  • \( f(x ; \lambda, \theta) = \lambda e^{-\lambda(x-\theta)} \) for \( x \geq \theta \)
  • \( 0 \) otherwise
The parameter \( \theta \) represents a minimum threshold below which no values can occur. This can be particularly useful in real-world situations where negative values aren't possible, such as measuring time intervals. By setting \( \theta = 0 \), the shifted exponential distribution becomes a standard exponential distribution, which is used to model time until a specific event occurs, assuming non-negative times.
Probability Density Function
The probability density function (pdf) is a crucial concept in statistics. It describes the likelihood of a random variable to take on a particular value. For the shifted exponential distribution, the pdf is given by:
  • \( f(x ; \lambda, \theta) = \lambda e^{-\lambda(x-\theta)} \)
This formula shows us that the probability of \( x \) is controlled by the parameters \( \lambda \) and \( \theta \). Here, \( \lambda \) is the rate parameter, which inversely affects the spread or scale of the distribution. The higher the \( \lambda \), the steeper the curve of the distribution, representing a more rapid decay of density as \( x \) increases. The parameter \( \theta \) determines where the distribution starts on the x-axis, which implies all values \( x \geq \theta \) are possible under this distribution.
Statistical Parameter Estimation
Statistical parameter estimation is a technique used to find the parameters of a probability distribution that best fit a given set of sample data. In the context of the shifted exponential distribution, two main parameters need estimation: \( \theta \) and \( \lambda \).For \( \theta \), which is the shift parameter, the estimation involves maximizing the likelihood. Since it dictates the starting point of the distribution, it's chosen as the smallest observed value in the sample data. For \( \lambda \), the rate parameter, it involves calculus techniques, specifically setting the derivative of the log-likelihood with respect to \( \lambda \) to zero. The result is:
  • \( \hat{\lambda} = \frac{n}{\sum_{i=1}^{n}(X_i - \theta)} \)
This gives a fair estimate of \( \lambda \) based on the spread of values adjusted by the shift \( \theta \).
Sample Data Analysis
Sample data analysis is a process where you work with a set of observed data to uncover insights about the behavior of the related population. When performed on time headway observations in a traffic flow scenario using a shifted exponential distribution, it helps calculate the MLEs for both \( \theta \) and \( \lambda \).To analyze the provided data: \( X_1 = 3.11, X_2 = 0.64, ..., X_{10} = 1.30 \), start by identifying the smallest value, \( \theta = 0.64 \). This minimal value forms the basis for the shift parameter as no time headway can be lower.Then, using these observations, calculate the sum \( \sum_{i=1}^{n}(X_i - \theta) \) to find \( \hat{\lambda} \). It's a formulaic approach, but the insights gleaned from this application can be powerful for predictive modeling and real-time decisions in similar fields.

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Most popular questions from this chapter

Each of 150 newly manufactured items is examined and the number of scratches per item is recorded (the items are supposed to be free of scratches), yielding the following data: Number of scratches \begin{tabular}{lllllllll} per item & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline Observed frequency & 18 & 37 & 42 & 30 & 13 & 7 & 2 & 1 \end{tabular} Let \(X=\) the number of scratches on a randomly chosen item, and assume that \(X\) has a Poisson distribution with parameter \(\lambda\). a. Find an unbiased estimator of \(\lambda\) and compute the estimate for the data. [Hint: \(E(X)=\lambda\) for \(X\) Poisson, so \(E(\bar{X})=\) ?] b. What is the standard deviation (standard error) of your estimator? Compute the estimated standard error. [Hint: \(\sigma_{X}^{2}=\lambda\) for \(X\) Poisson.]

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?

A sample of 20 students who had recently taken elementary statistics yielded the following information on brand of calculator owned \((\mathrm{T}=\) Texas Instruments, \(\mathrm{H}=\) Hewlett Packard, \(\mathrm{C}=\) Casio, \(\mathrm{S}=\) Sharp): \(\begin{array}{llllllllll}\text { T } & \text { T } & H & \text { T } & \text { C } & \text { T } & \text { T } & \text { S } & \text { C } & \text { H }\end{array}\) \(\begin{array}{lllllllllll}S & S & T & H & C & T & T & T & H & T\end{array}\) a. Estimate the true proportion of all such students who own a Texas Instruments calculator. b. Of the 10 students who owned a TI calculator, 4 had graphing calculators. Estimate the proportion of students who do not own a TI graphing calculator.

Let \(X_{1}, X_{2}, \ldots, X_{n}\) represent a random sample from a Rayleigh distribution with pdf $$ f(x ; \theta)=\frac{x}{\theta} e^{-x^{2} /(2 \theta)} \quad x>0 $$ a. It can be shown that \(E\left(X^{2}\right)=2 \theta\). Use this fact to construct an unbiased estimator of \(\theta\) based on \(\sum X_{i}^{2}\) (and use rules of expected value to show that it is unbiased). b. Estimate \(\theta\) from the following \(n=10\) observations on vibratory stress of a turbine blade under specified conditions: \(\begin{array}{lllll}16.88 & 10.23 & 4.59 & 6.66 & 13.68 \\ 14.23 & 19.87 & 9.40 & 6.51 & 10.95\end{array}\)

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 \(\max \left(X_{j}\right)-\min \left(X_{j}\right)+1\) 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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