/*! 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 29 Consider a random sample \(X_{1}... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider a random sample \(X_{1}, X_{2}, \ldots, X_{n}\) from the shifted exponential pdf $$ f(x ; \lambda, \theta)=\left\\{\begin{array}{cc} \lambda e^{-\lambda(x-\theta)} & x \geq \theta \\ 0 & \text { otherwise } \end{array}\right. $$ 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
For the provided data, \(\hat{\theta} = 0.64\) and \(\hat{\lambda} \approx 0.2065\).

Step by step solution

01

Define the Likelihood Function

For the shifted exponential distribution, the likelihood function for a sample \( X_1, X_2, \ldots, X_n \) is given by the product of the individual densities:\[L(\lambda, \theta | X_1, X_2, \ldots, X_n) = \prod_{i=1}^{n} \lambda e^{-\lambda(X_i - \theta)} \quad \text{for } X_i \geq \theta\]This can be written as:\[L(\lambda, \theta) = \lambda^n e^{-\lambda \sum_{i=1}^{n} (X_i - \theta)}\]
02

Log-Likelihood Function

To simplify the process, we take the natural logarithm of the likelihood function. The log-likelihood function \( \ell(\lambda, \theta) \) is:\[\ell(\lambda, \theta) = n \ln \lambda - \lambda \sum_{i=1}^{n} (X_i - \theta)\]
03

Derive the Estimator for \(\lambda\)

Differentiate the log-likelihood function with respect to \(\lambda\):\[\frac{\partial \ell}{\partial \lambda} = \frac{n}{\lambda} - \sum_{i=1}^{n} (X_i - \theta)\]Setting \( \frac{\partial \ell}{\partial \lambda} = 0 \) yields:\[\lambda = \frac{n}{\sum_{i=1}^{n} (X_i - \theta)}\]
04

Derive the Estimator for \(\theta\)

Differentiate the log-likelihood function with respect to \(\theta\):\[\frac{\partial \ell}{\partial \theta} = \lambda \sum_{i=1}^{n} \left(\frac{\partial}{\partial \theta} -1\right) = \lambda n\]The constraint \(X_i \geq \theta\) suggests setting \(\theta\) to the smallest observed \(X_i\) since for larger \(\theta\), the likelihood becomes zero. Thus, the MLE for \(\theta\) is:\[\hat{\theta} = \min(X_1, X_2, \ldots, X_n)\]
05

Apply Estimators to Given Data

Given the data: \(3.11, 0.64, 2.55, 2.20, 5.44, 3.42, 10.39, 8.93, 17.82, 1.30\).1. Calculate the estimator for \(\theta\): \[ \hat{\theta} = \min(3.11, 0.64, 2.55, 2.20, 5.44, 3.42, 10.39, 8.93, 17.82, 1.30) = 0.64 \]2. Calculate the estimator for \(\lambda\): \[ \hat{\lambda} = \frac{10}{\sum_{i=1}^{10} (X_i - 0.64)} \] Find \(\sum_{i=1}^{10} (X_i - 0.64) = (3.11 + 0.64 + 2.55 + 2.20 + 5.44 + 3.42 + 10.39 + 8.93 + 17.82 + 1.30) - 6.4 = 54.82 - 6.4 = 48.42\). \[ \hat{\lambda} = \frac{10}{48.42} \approx 0.2065 \]

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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 modification of the regular exponential distribution. In mathematical terms, the shift parameter, \( \theta \), modifies the traditional exponential distribution to begin beyond zero. This functional adjustment allows for flexibility in modeling scenarios where the minimum of the distribution is not zero, but another value, \( \theta \).

For example, if you are measuring time between events in traffic flow as seen in Example 4.5, a shift might be needed since there is a minimal time gap (\( \theta \)) that is realistically possible.

Key characteristics of the shifted exponential distribution include:
  • The probability density function (pdf) is only non-zero for values greater than or equal to \( \theta \). This property ensures that the function \/p within permissible values.
  • The parameter \( \lambda \) remains the same as in the exponential distribution, controlling the rate of events occurring beyond \( \theta \).
Understanding the need for shifts helps tailor statistical models to real-world phenomena better.
Log-Likelihood Function
The log-likelihood function is a handy tool for simplifying the computation of likelihoods, particularly when dealing with products of probabilities.

In our case, given a shifted exponential distribution, the likelihood function for a sample involves a product, as each data point contributes a probability, which can quickly become cumbersome to compute.

Formulaically, the likelihood function \( L(\lambda, \theta) \) is:
  • \[ L(\lambda, \theta) = \lambda^n e^{-\lambda \sum_{i=1}^{n} (X_i - \theta)} \]
To simplify, we take its natural logarithm, turning products into sums and exponentials into linear terms. This leads us to the log-likelihood function \( \ell(\lambda, \theta) \):

  • \[ \ell(\lambda, \theta) = n \ln \lambda - \lambda \sum_{i=1}^{n} (X_i - \theta) \]
Utilizing the log-likelihood is particularly useful when deriving parameter estimates, as differentiating this expression often results in straightforward mathematical manipulations.
Estimating Parameters
Estimating parameters \( \lambda \) (rate) and \( \theta \) (shift) for a shifted exponential distribution revolves around determining values that maximize the likelihood of observing the given data.

### Estimating \( \lambda \)
The first step involves differentiating the log-likelihood function with respect to \( \lambda \) and setting the derivative to zero:
  • \[ \frac{\partial \ell}{\partial \lambda} = \frac{n}{\lambda} - \sum_{i=1}^{n} (X_i - \theta) \]
Solving this provides the maximum likelihood estimator (MLE) for \( \lambda \):
  • \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} (X_i - \theta)} \]
### Estimating \( \theta \)
The parameter \( \theta \) represents the shift and is estimated by considering the smallest observation in the data set, because the likelihood is zero for any observation smaller than \( \theta \):
  • \[ \hat{\theta} = \min(X_1, X_2, \ldots, X_n) \]
### Application
For our example with the given observations, by following these steps, we estimated \( \hat{\theta} = 0.64 \) and \( \hat{\lambda} \approx 0.2065 \). These estimators help tailor the shifted exponential model to the specific data, offering insights into the rate and starting point of the observed phenomenon.

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Most popular questions from this chapter

Suppose a certain type of fertilizer has an expected yield per acre of \(\mu_{1}\) with variance \(\sigma^{2}\), whereas the expected yield for a second type of fertilizer is \(\mu_{2}\) with the same variance \(\sigma^{2}\). Let \(S_{1}^{2}\) and \(S_{2}^{2}\) denote the sample variances of yields based on sample sizes \(n_{1}\) and \(n_{2}\), respectively, of the two fertilizers. Show that the pooled (combined) estimator $$ \hat{\sigma}^{2}=\frac{\left(n_{1}-1\right) S_{1}^{2}+\left(n_{2}-1\right) S_{2}^{2}}{n_{1}+n_{2}-2} $$ is an unbiased estimator of \(\sigma^{2}\).

An investigator wishes to estimate the proportion of students at a certain university who have violated the honor code. Having obtained a random sample of \(n\) students, she realizes that asking each, "Have you violated the honor code?" will probably result in some untruthful responses. Consider the following scheme, called a randomized response technique. The investigator makes up a deck of 100 cards, of which 50 are of type I and 50 are of type II. Type I: Have you violated the honor code (yes or no)? Type II: Is the last digit of your telephone number a 0 , 1 , or 2 (yes or no)? Each student in the random sample is asked to mix the deck, draw a card, and answer the resulting question truthfully. Because of the irrelevant question on type II cards, a yes response no longer stigmatizes the respondent, so we assume that responses are truthful. Let \(p\) denote the proportion of honor- code violators (i.e., the probability of a randomly selected student being a violator), and let \(\lambda=P\) (yes response). Then \(\lambda\) and \(p\) are related by \(\lambda=.5 p+(.5)(.3)\) a. Let \(Y\) denote the number of yes responses, so \(Y \sim\) Bin \((n, \lambda)\). Thus \(Y / n\) is an unbiased estimator of \(\lambda\). Derive an estimator for \(p\) based on \(Y\). If \(n=80\) and \(y=20\), what is your estimate? [Hinr: Solve \(\lambda=.5 p+.15\) for \(p\) and then substitute \(Y / n\) for \(\lambda .]\) b. Use the fact that \(E(Y / n)=\lambda\) to show that your estimator \(\hat{p}\) is unbiased. c. If there were 70 type I and 30 type II cards, what would be your estimator for \(p\) ?

Let \(X_{1}, \ldots, X_{n}\) be a random sample from a gamma distribution with parameters \(\alpha\) and \(\beta\). a. Derive the equations whose solutions yield the maximum likelihood estimators of \(\alpha\) and \(\beta\). Do you think they can be solved explicitly? b. Show that the mle of \(\mu=\alpha \beta\) is \(\hat{\mu}=\bar{X}\).

Urinary angiotensinogen (AGT) level is one quantitative indicator of kidney function. The article "Urinary Angiotensinogen as a Potential Biomarker of Chronic Kidney Diseases" ( \(J\). of the Amer. Society of Hypertension, 2008: 349-354) describes a study in which urinary AGT level \((\mu \mathrm{g})\) was determined for a sample of adults with chronic kidney disease. Here is representative data (consistent with summary quantities and descriptions in the cited article): \begin{tabular}{rrrrrrrrr} \multicolumn{7}{r}{ and descriptions in the cited article): } \\ \(2.6\) & \(6.2\) & \(7.4\) & \(9.6\) & \(11.5\) & \(13.5\) & \(14.5\) & \(17.0\) \\ \(20.0\) & \(28.8\) & \(29.5\) & \(29.5\) & \(41.7\) & \(45.7\) & \(56.2\) & \(56.2\) \\ \(66.1\) & \(66.1\) & \(67.6\) & \(74.1\) & \(97.7\) & \(141.3\) & \(147.9\) & \(177.8\) \\ \(186.2\) & \(186.2\) & \(190.6\) & \(208.9\) & \(229.1\) & \(229.1\) & \(288.4\) & \(288.4\) \\\ \(346.7\) & \(407.4\) & \(426.6\) & \(575.4\) & \(616.6\) & \(724.4\) & \(812.8\) & \(1122.0\) \end{tabular} An appropriate probability plot supports the use of the lognormal distribution (see Section \(4.5\) ) as a reasonable model for urinary AGT level (this is what the investigators did). 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 AGT level. [Hint: What is \(E(X)\) ?]

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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