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Let \(X_{1}, \ldots, X_{n}\) be a random sample of component lifetimes from an exponential distribution with parameter \(\lambda\). Use the factorization theorem to show that \(\sum X_{i}\) is a sufficient statistic for \(\lambda\).

Short Answer

Expert verified
The statistic \(\sum X_i\) is sufficient for \(\lambda\) by the factorization theorem.

Step by step solution

01

Define the Probability Density Function (PDF)

The probability density function of an exponential distribution with parameter \(\lambda\) is given by:\[ f(x; \lambda) = \lambda e^{-\lambda x}, \quad x \geq 0 \]
02

Determine the Joint PDF

For a random sample \(X_{1}, X_{2}, \ldots, X_{n}\) from an exponential distribution, the joint PDF is given by the product of the individual PDFs:\[ f(x_1, x_2, \ldots, x_n ; \lambda) = \prod_{i=1}^{n} (\lambda e^{-\lambda x_i}) = \lambda^n e^{-\lambda \sum_{i=1}^{n} x_i} \]
03

Apply the Factorization Theorem

The factorization theorem states that a statistic \(T(X)\) is sufficient for \(\lambda\) if the joint PDF can be expressed as a product of two functions: one that depends on the sample data only through \(T(X)\), and another that does not depend on \(\lambda\).Rewriting the joint PDF, we get:\[ f(x_1, x_2, \ldots, x_n ; \lambda) = \left(e^{-\lambda \sum_{i=1}^{n} x_i}\right) \left(\lambda^n\right) \]Here, \( e^{-\lambda \sum_{i=1}^{n} x_i} \) depends on \(\lambda\) and the sample only through \(\sum_{i=1}^{n} x_i\), and \(\lambda^n\) is independent of the data.
04

Identify the Sufficient Statistic

From the factorization, it is evident that \(\sum_{i=1}^{n} x_i\) is a function of the sample data and alone captures all necessary information about \(\lambda\). Thus, \(\sum X_i\) is a sufficient statistic for \(\lambda\).

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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 commonly used to model the time between independent events that happen at a constant average rate. It is characterized by its parameter \( \lambda \), known as the rate parameter. This type of distribution is often used to describe things like the lifespan of electronic components or the time until a radioactive atom decays.

The probability density function (PDF) for an exponential distribution is given by:
\[ f(x; \lambda) = \lambda e^{-\lambda x}, \quad x \geq 0 \]
This formula shows that the probability of the event occurring decreases exponentially as time \( x \) increases. The parameter \( \lambda \) helps define how fast this probability decays with time.
  • High \( \lambda \) value: Shorter expected wait times.
  • Low \( \lambda \) value: Longer expected wait times.
Its simplicity and memoryless property, where the distribution of the remaining time is the same regardless of how much time has already passed, make it very useful in various applications.
Factorization Theorem
The factorization theorem is a powerful tool in statistics used to identify sufficient statistics, which compress sample data into a summary that captures all necessary information about a parameter. According to this theorem, a statistic is sufficient for a parameter if the joint PDF can be expressed as a product of two functions:

  • One function depends on the data only through a statistic \( T(X) \).
  • The other function is independent of the parameter.
This theorem helps you determine the minimal summary statistic \( T(X) \) that can be used to make inference about an unknown parameter of interest. It simplifies problems by focusing only on the necessary data, avoiding irrelevant details.

In the context of an exponential distribution, applying the factorization theorem to the joint PDF, you can rewrite it as two parts such that:
\[ f(x_1, x_2, \ldots, x_n ; \lambda) = g(T(x); \lambda) \cdot h(x_1, x_2, \ldots, x_n) \]
where \( g(T(x); \lambda) = e^{-\lambda \sum x_i} \) and \( h(x_1, x_2, \ldots, x_n) = \lambda^n \), showing that the statistic \( \sum x_i \) is sufficient for \( \lambda \).
Joint Probability Density Function
When dealing with a group of random variables, the joint probability density function (joint PDF) describes the likelihood of different outcomes occurring simultaneously across all these variables. For a collection of independent, identically distributed (i.i.d) random samples from an exponential distribution, each sample has its own PDF, and the joint PDF is simply the product of these individual PDFs.

Consider \( X_1, X_2, \ldots, X_n \) as a random sample from an exponential distribution. The joint PDF is given by:
\[ f(x_1, x_2, \ldots, x_n ; \lambda) = \prod_{i=1}^{n} (\lambda e^{-\lambda x_i}) = \lambda^n e^{-\lambda \sum x_i} \]
This equation illustrates how the joint PDF combines the information from all the sample data in a coordinated manner. Each \( X_i \) value contributes to the overall product, but they are all influenced by the parameter \( \lambda \).
  • Reflects the aggregate effect of all variables.
  • Helps in determining the likelihood for the entire sample group.
The joint PDF provides a unified perspective on how the random variables interact under the given distribution.
Parameter Estimation
Parameter estimation involves calculating parameters like \( \lambda \) of a statistical model, using sample data. These estimates are crucial for predicting trends and behaviors in data-driven contexts. In the case of an exponential distribution, the parameter \( \lambda \) is often estimated to understand the timing of events, such as system failures or arrival times.

The sufficient statistic, \( \sum X_i \), simplifies this process since it condenses the sample information into a single value that contains all the necessary information about \( \lambda \). This value can then be used for making robust estimates of the parameter:

  • Common approach: Maximum Likelihood Estimation (MLE).
  • MLE for \( \lambda \): \( \hat{\lambda} = \frac{n}{\sum X_i} \).
This method maximizes the likelihood function to find the parameter value that makes the observed data most probable. With the MLE approach, the computational task becomes finding the estimate that fits the sufficient statistic precisely, leading to efficient and unbiased parameter estimation.

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Most popular questions from this chapter

Consider a random sample \(X_{1}, \ldots, X_{n}\) from the pdf $$ f(x ; \theta)=.5(1+\theta x) \quad-1 \leq x \leq 1 $$ where \(-1 \leq \theta \leq 1\) (this distribution arises in particle physics). Show that \(\hat{\theta}=3 \bar{X}\) is an unbiased estimator of \(\theta\). [Hint: First determine \(\mu=E(X)=E(\bar{X}) .\).]

Let \(X_{1}, \ldots, X_{n}\) be a random sample from a gamma distribution with parameters \(\alpha\) and \(\beta\). a. Derive the equations whose solution yields 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 \(\hat{\mu}=\bar{X}\).

The long run proportion of vehicles that pass a certain emissions test is \(p\). Suppose that three vehicles are independently selected for testing. Let \(X_{i}=1\) if the \(i\) th vehicle passes the test and \(X_{i}=0\) otherwise \((i=1,2,3)\), and let \(X=X_{1}+\) \(X_{2}+X_{3}\). Use the definition of sufficiency to show that \(X\) is sufficient for \(p\) by obtaining the conditional distribution of the \(X_{i}\) 's given that \(X=x\) for each possible value \(x\). Then generalize by giving an analogous argument for the case of \(n\) vehicles.

The accompanying data on IQ for first-graders at a university lab school was introduced in Example 1.2. \(\begin{array}{rrrrrrrrrrr}82 & 96 & 99 & 102 & 103 & 103 & 106 & 107 & 108 & 108 & 108 \\ 108 & 109 & 110 & 110 & 111 & 113 & 113 & 113 & 113 & 115 & 115 \\\ 118 & 118 & 119 & 121 & 122 & 122 & 127 & 132 & 136 & 140 & 146\end{array}\) a. Calculate a point estimate of the mean value of IQ for the conceptual population of all first graders in this school, and state which estimator you used. [Hint: \(\mathbf{\Sigma} x_{i}=3753\) ] b. Calculate a point estimate of the IQ value that separates the lowest \(50 \%\) of all such students from the highest \(50 \%\), and state which estimator you used. c. Calculate and interpret a point estimate of the population standard deviation \(\sigma\). Which estimator did you use? [Hint: \(\Sigma x_{i}^{2}=432,015\) ] d. Calculate a point estimate of the proportion of all such students whose IQ exceeds 100 . [Hint: Think of an observation as a "success" if it exceeds 100.] e. Calculate a point estimate of the population coefficient of variation \(\sigma / \mu\), and state which estimator you used.

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 \(Y \sim \operatorname{Bin}(n, \lambda)\). 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? [Hint: 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 \(p\) ?

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