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Suppose waiting time for delivery of an item is uniform on the interval from \(\theta_{1}\) to \(\theta_{2}\) (so \(f\left(x ; \theta_{1}, \theta_{2}\right.\) ) \(=1 /\left(\theta_{2}-\theta_{1}\right)\) for \(\theta_{1}

Short Answer

Expert verified
\(\min(X_i)\) and \(\max(X_i)\) are jointly sufficient statistics for \(\theta_1\) and \(\theta_2\).

Step by step solution

01

Identify the Probability Density Function

Since the waiting time for delivery of an item is uniform on the interval from \(\theta_1\) to \(\theta_2\), the probability density function (PDF) is given by: \[ f(x; \theta_1, \theta_2) = \frac{1}{\theta_2 - \theta_1} \text{ for } \theta_1 < x < \theta_2 \] and 0 otherwise.
02

Formulate the Likelihood Function

The likelihood function for a random sample \(X_1, X_2, \ldots, X_n\) is the product of the PDFs, since the random variables are independent: \[ L(\theta_1, \theta_2; X_1, \ldots, X_n) = \prod_{i=1}^{n} \frac{1}{\theta_2 - \theta_1} \cdot I(\theta_1 < X_i < \theta_2) \] where \(I(\cdot)\) is the indicator function that equals 1 if the statement is true and 0 otherwise.
03

Simplify the Indicator Function

The joint indicator function becomes a single constraint because \(\min(X_i)\) and \(\max(X_i)\) determine the range: \ \(I(\theta_1 < X_i < \theta_2) = I(\theta_1 < \min(X_i)) \cdot I(\max(X_i) < \theta_2)\). \
04

Reformulate the Likelihood Function with Sufficient Statistics

Replace the individual indicators by: \[ L(\theta_1, \theta_2; X_1, \ldots, X_n) = \left(\frac{1}{\theta_2 - \theta_1}\right)^n \, I(\theta_1 < \min(X_i)) \, I(\max(X_i) < \theta_2) \] This shows the likelihood depends only on \(\min(X_i)\) and \(\max(X_i)\), indicating they are sufficient.
05

Apply the Factorization Theorem

According to the factorization theorem, a statistic is sufficient if the likelihood can be expressed as a product of terms: one depending only on the data through the statistic, and the other depending only on \(\theta_1\) and \(\theta_2\). Here: \[ g(X_1, \ldots, X_n; \theta_1, \theta_2) = I(\theta_1 < \min(X_i)) \, I(\max(X_i) < \theta_2) \]shows \(\min(X_i)\) and \(\max(X_i)\) are sufficient statistics.

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Key Concepts

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

Uniform Distribution
The uniform distribution is a simple and intuitive probability distribution. It models situations where all outcomes in a given interval are equally likely. For example, if you think about waiting times for delivery that can occur anytime between two times
  • say, \( \theta_1 \) (the earliest time)
  • and \( \theta_2 \) (the latest time)
then each moment between \( \theta_1 \)and \( \theta_2 \) has equal probability of being the waiting time. It’s like when you have a perfectly fair dice, where each number has the same chance of coming up. Within a uniform distribution, the probability density function (PDF) is simple: it’s constant over the defined interval and zero everywhere else. This means the PDF is \( \frac{1}{\theta_2 - \theta_1} \) over the interval \( \theta_1 < x < \theta_2 \).

So if we’ve got a uniform distribution in an exercise or problem, expect it to have this constant nature within a certain range. It’s straightforward because all outcomes are balanced.
Factorization Theorem
The factorization theorem is a crucial tool in statistics. It helps us determine what are known as sufficient statistics. Sufficient statistics summarize all the information needed from the sample about the parameter. Say we have a parameterized probability density function (PDF) and related likelihood function.

The theorem states: A statistic is sufficient for a parameter if the likelihood can be factored into two parts:
  • One part depends only on the observations through the statistic.
  • The other part depends only on the parameter.
Let’s take our uniform distribution problem: We aim to find sufficient statistics for parameters \( \theta_1 \) and \( \theta_2 \). By applying the factorization theorem, a statistic or a pair of statistics are sufficient if we can separate the likelihood function based on them. In this exercise, \( \min(X_i) \) and \( \max(X_i) \) turned out to be those sufficient statistics. Thus, by focusing on just the minimum and maximum in our random sample, we gather all necessary information about the interval of the distribution. It's efficient and elegant!
Likelihood Function
The likelihood function is a core concept in statistics that stems from the probability density function (PDF) of a statistical model. It measures how likely a particular set of parameters is, given the observed data. If you think of the probability density function as describing the possible range of "data given parameters," then the likelihood function flips it to "parameters given data."

To construct the likelihood function, you take the product of individual probabilities for all data points. Particularly for independent random variables, the likelihood is simply the product of their respective PDFs.
In our problem, with each data point having the PDF: \( \frac{1}{\theta_2 - \theta_1} \), we calculated the likelihood function across all "n" samples by: \( L(\theta_1, \theta_2; X_1, \ldots, X_n) = \prod_{i=1}^{n} \frac{1}{\theta_2 - \theta_1} \cdot I(\theta_1 < X_i < \theta_2) \).
Thus, the likelihood function gives us insights on how well different values of parameters \( \theta \) fit the given data.
Probability Density Function
The probability density function (PDF) is a fundamental concept in continuous probability distributions. It describes the relative likelihood for a continuous random variable to take on a given value. With uniform distribution, the PDF is remarkably simple and constant across the specified interval, reinforcing the key idea that each value within this range is equally probable.

In terms of mathematics:
  • For a uniform distribution from \( \theta_1 \) to \( \theta_2 \), the PDF is \( f(x; \theta_1, \theta_2) = \frac{1}{\theta_2 - \theta_1} \) if \( \theta_1 < x < \theta_2 \)
  • and 0 otherwise.
This means that you can find the probability between any two values within the interval by calculating the area under the PDF over that range. Unlike discrete probabilities, this concept helps in understanding continuous random variables, making it an essential tool in statistics and probability theory. Its simplicity in uniform distributions makes it an excellent starting point for understanding more complex PDFs.

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Most popular questions from this chapter

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 \(\lambda\).

As an example of a situation in which several different statistics could reasonably be used to calculate a point estimate, consider a population of \(N\) invoices. Associated with each invoice is its "book value," the recorded amount of that invoice. Let \(T\) denote the total book value, a known amount. Some of these book values are erroneous. An audit will be carried out by randomly selecting \(n\) invoices and determining the audited (correct) value for each one. Suppose that the sample gives the following results (in dollars). \begin{tabular}{lrcrrr} \hline & \multicolumn{5}{c}{ Invoice } \\ \cline { 2 - 6 } & \(\mathbf{1}\) & \(\mathbf{2}\) & \(\mathbf{3}\) & \(\mathbf{4}\) & \(\mathbf{5}\) \\ \hline Book value & 300 & 720 & 526 & 200 & 127 \\ Audited value & 300 & 520 & 526 & 200 & 157 \\ Error & 0 & 200 & 0 & 0 & \(-30\) \\ \hline \end{tabular} Let \(\bar{X}=\) the sample mean audited value, \(\bar{Y}=\) the sample mean book value, and \(\bar{D}=\) the sample mean error. Propose three different statistics for estimating the total audited (i.e. correct) value \(\theta\) - one involving just \(N\) and \(\bar{X}\), another involving \(N, T\), and \(\bar{D}\), and the last involving \(T\) and \(\bar{X} / \bar{Y}\). Then calculate the resulting estimates when \(N=5,000\) and \(T=1,761,300\) (The article "Statistical Models and Analysis in Auditing,", Statistical Science, 1989: 2 - 33 discusses properties of these estimators).

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 \(\mu^{2}\) ?

In Chapter 3 , we defined a negative binomial rv as the number of failures that occur before the \(r\) th success in a sequence of independent and identical success/failure trials. The probability mass function (pmf) of \(X\) is a. Suppose that \(r \geq 2\). Show that $$ \hat{p}=(r-1) /(X+r-1) $$ is an unbiased cstimator for \(p\). [Hint: Writc out \(E(\hat{p})\) and cancel \(x+r-1\) inside the sum.] b. A reporter wishing to interview five individuals who support a certain candidate begins asking people whether \((S)\) or not \((F)\) they support the candidate. If the sequence of responses is SFFSFFFSSS, estimate \(p=\) the true proportion who support the candidate.

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 each other, 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 \(\sum\left(z_{i}-\bar{z}\right)^{2}=\) \(\left.\left(z_{1}-z_{2}\right)^{2} / 2 .\right]\) b. Is the mle \(\hat{\sigma}^{2}\) an unbiased estimator of \(\sigma^{2}\) ? Find an unbiased estimator of \(\sigma^{2}\). [Hint: For any rv \(Z, E\left(Z^{2}\right)=V(Z)+[E(Z)]^{2}\). Apply this to \(Z=X_{i}-Y_{i}\).]

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