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The double exponential distribution is $$f(x | \theta)=\frac{1}{2} e^{-|x-\theta|}, \quad-\infty< x<\infty$$ For an i.i.d. sample of size \(n=2 m+1,\) show that the mle of \(\theta\) is the median of the sample. (The observation such that half of the rest of the observations are smaller and half are larger.) [Hint: The function \(g(x)=|x|\) is not differentiable. Draw a picture for a small value of \(n\) to try to understand what is going on.]

Short Answer

Expert verified
The MLE of \( \theta \) is the median of the sample.

Step by step solution

01

Understand the Likelihood Function

The given probability density function (pdf) for the double exponential distribution is \( f(x | \theta) = \frac{1}{2} e^{-|x-\theta|} \). For \( n \) i.i.d. samples \( x_1, x_2, \ldots, x_n \), the likelihood function \( L(\theta) \) is the product of the individual density functions: \( L(\theta) = \prod_{i=1}^{n} \frac{1}{2} e^{-|x_i - \theta|} = \left(\frac{1}{2}\right)^n e^{- extstyle\sum_{i=1}^{n} |x_i - \theta|} \).
02

Simplify the Likelihood Function

The \( \frac{1}{2} \) term contributes only to the multiplicative constant, so we focus on the sum in the exponent: \( L(\theta) \) can be re-expressed as \( e^{- extstyle\sum_{i=1}^{n} |x_i - \theta|} \). This means the log-likelihood function is: \( \log L(\theta) = -\sum_{i=1}^{n} |x_i - \theta| + \text{constant} \). We seek the value of \( \theta \) that maximizes this log-likelihood function.
03

Interpret the Objective Function

To maximize the log-likelihood function, we equivalently want to minimize the sum \( \sum_{i=1}^{n} |x_i - \theta| \). This is known as the \( L_1 \) norm.
04

Recognize the Properties of the Median

The problem becomes a classic median problem. The median minimizes the sum of absolute deviations \( \sum_{i=1}^{n} |x_i - \theta| \). By definition, the median of a sample divides the data set such that half of the observations are below it, and half are above it.
05

Check with Small Sample Visualization

Consider a small sample. For instance, for \( n = 3 \), rank the sample points: \( x_{(1)} \leq x_{(2)} \leq x_{(3)} \). It is evident geometrically and through symmetry that \( x_{(2)} \), the median, minimizes the \( L_1 \) distance to the other points.
06

Conclusion from the Steps

Since minimizing \( \sum_{i=1}^{n} |x_i - \theta| \) achieves the maximum likelihood estimation of \( \theta \), and the median achieves this minimization, the MLE of \( \theta \) is indeed the median of the sample.

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Key Concepts

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

Double Exponential Distribution
The double exponential distribution, also known as the Laplace distribution, is a statistical distribution often used in different fields such as economics, finance, and engineering. This distribution is called "double exponential" because its probability density function (pdf) decreases exponentially on either side of the median, resulting in two tails. The pdf is given by: \[ f(x | \theta) = \frac{1}{2} e^{-|x-\theta|} \] In this formula, \( \theta \) is a parameter that represents the median of the distribution. The absolute value function in the exponent introduces a kink at \( x = \theta \), giving it a unique shape.
  • Key characteristic: The distribution is symmetric around \( \theta \).
  • Heavy tails: It is more robust to outliers than the normal distribution because of its heavier tails.
  • Applications: It is commonly used to model data with sudden changes or as a robust alternative when outliers are present in the data.
Median Estimation
When dealing with data from a double exponential distribution, the median plays a central role, especially in estimation problems. The median is a measure of central tendency that divides a data set into two equal parts. In the context of the double exponential distribution, the median, \( \theta \), minimizes the sum of absolute deviations. This is a different objective compared to the mean, which minimizes squared deviations.
  • The median is robust: It is not heavily influenced by extreme values or outliers, making it a good estimator in situations with abnormal data points.
  • In the problem, for an i.i.d. sample, the median of the sample serves as the maximum likelihood estimator (MLE) of \( \theta \). This means that among all possible estimators, the sample median is the one that is most likely to have produced the observed data.
Likelihood Function
The likelihood function is a fundamental concept in statistical inference, particularly in maximum likelihood estimation (MLE). It measures how likely a set of parameters would make the observed data occur. For the double exponential distribution and an i.i.d. sample, the likelihood function is: \[ L(\theta) = \prod_{i=1}^{n} \frac{1}{2} e^{-|x_i - \theta|} \] After simplification, focusing on maximizing this function is equivalent to minimizing the expression: \[ \sum_{i=1}^{n} |x_i - \theta| \] This simplification leads us directly to the heart of the double exponential problem: finding \( \theta \) that minimizes this sum. This problem is continuous and does not have a smooth derivative everywhere due to the absolute value.
  • Importance in MLE: Maximizing the likelihood helps us find the parameter values that are most supported by the data.
  • Connection to the median: In this specific distribution, the median minimizes the objective function, making it the MLE.
i.i.d. Samples
The acronym i.i.d. stands for "independent and identically distributed." It describes a common assumption in statistical models where each data sample is independently drawn and follows the same probability distribution. For the double exponential distribution, assuming i.i.d. samples means:
  • Each sample point, \( x_i \), is an independent observation, meaning the data do not influence each other.
  • All observations are drawn from the same double exponential distribution parameterized by the same \( \theta \).
This assumption simplifies the analysis and is essential for making generalizations about the data. In our exercise, the likelihood function is derived under the i.i.d. assumption. This allows us to treat each sample individually and multiply their probabilities to find the overall likelihood. Understanding i.i.d. helps in grasping why certain statistical properties, such as estimation techniques, can be applied.
  • Foundation for statistical inference: It gives reliability to the repeatability of experiments and the consistency of estimators.
  • Core assumption: Most foundational theorems in statistics, like the Central Limit Theorem, rely on the i.i.d. assumption to hold.

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Most popular questions from this chapter

Suppose that \(X\) is a discrete random variable with $$P(X=0)=\frac{2}{3} \theta$$ $$\begin{aligned} &P(X=1)=\frac{1}{3} \theta\\\ &\begin{array}{l} P(X=2)=\frac{2}{3}(1-\theta) \\ P(X=3)=\frac{1}{3}(1-\theta) \end{array} \end{aligned}$$ where \(0 \leq \theta \leq 1\) is a parameter. The following 10 independent observations were taken from such a distribution: (3,0,2,1,3,2,1,0,2,1) a. Find the method of moments estimate of \(\theta\) b. Find an approximate standard error for your estimate. c. What is the maximum likelihood estimate of \(\theta ?\) d. What is an approximate standard error of the maximum likelihood estimate? e. If the prior distribution of \(\Theta\) is uniform on \([0,1],\) what is the posterior density? Plot it. What is the mode of the posterior?

Suppose that 100 items are sampled from a manufacturing process and 3 are found to be defective. A beta prior is used for the unknown proportion \(\theta\) of defective items. Consider two cases: \((1) a=b=1,\) and \((2) a=0.5\) and \(b=5 .\) Plot the two posterior distributions and compare them. Find the two posterior means and compare them. Explain the differences.

Let \(X_{1}, \ldots, X_{n}\) be i.i.d. uniform on \([0, \theta].\) a. Find the method of moments estimate of \(\theta\) and its mean and variance. b. Find the mle of \(\theta\) c. Find the probability density of the mle, and calculate its mean and variance. Compare the variance, the bias, and the mean squared error to those of the method of moments estimate. d. Find a modification of the mle that renders it unbiased.

Suppose that \(X_{1}, X_{2}, \ldots, X_{n}\) are i.i.d. random variables on the interval [0,1] with the density function $$f(x | \alpha)=\frac{\Gamma(3 \alpha)}{\Gamma(\alpha) \Gamma(2 \alpha)} x^{\alpha-1}(1-x)^{2 \alpha-1}$$ where \(\alpha>0\) is a parameter to be estimated from the sample. It can be shown that $$\begin{aligned} E(X) &=\frac{1}{3} \\ \operatorname{Var}(X) &=\frac{2}{9(3 \alpha+1)} \end{aligned}$$ a. How could the method of moments be used to estimate \(\alpha ?\) b. What equation does the mle of \(\alpha\) satisfy? c. What is the asymptotic variance of the mle? d. Find a sufficient statistic for \(\alpha .\)

In Example A of Section \(8.4,\) we used knowledge of the exact form of the sampling distribution of \(\hat{\lambda}\) to estimate its standard error by $$s_{\hat{\lambda}}=\sqrt{\frac{\hat{\lambda}}{n}}$$ This was arrived at by realizing that \(\sum X_{i}\) follows a Poisson distribution with parameter \(n \lambda_{0} .\) Now suppose we hadn't realized this but had used the bootstrap, letting the computer do our work for us by generating \(B\) samples of size \(n=23\) of Poisson random variables with parameter \(\lambda=24.9,\) forming the mle of \(\lambda\) from each sample, and then finally computing the standard deviation of the resulting collection of estimates and taking this as an estimate of the standard error of \(\hat{\lambda}\) Argue that as \(B \rightarrow \infty,\) the standard error estimated in this way will tend to \(s_{\hat{\lambda}}\).

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