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Find a sufficient statistic for the Rayleigh density, $$f(x | \theta)=\frac{x}{\theta^{2}} e^{-x^{2} /\left(2 \theta^{2}\right)}, \quad x \geq 0$$

Short Answer

Expert verified
The sufficient statistic for \( \theta \) is \( T(X) = \sum_{i=1}^{n} x_i^2 \).

Step by step solution

01

Understanding the Rayleigh Density Function

The given Rayleigh density function is \( f(x | \theta) = \frac{x}{\theta^{2}} e^{-x^{2} /(2 \theta^{2})} \), where \( x \geq 0 \) and \( \theta > 0 \). We need to find a sufficient statistic for the parameter \( \theta \). The sufficient statistic summarizes all the information about \( \theta \) in a dataset.
02

Applying the Factorization Theorem

The Factorization Theorem states that a statistic \( T(X) \) is sufficient for parameter \( \theta \) if the likelihood can be factored as \( L(\theta | x_1, x_2, ..., x_n) = g(T(x), \theta)h(x) \). Here, \( L(\theta | x) \) is the likelihood for a sample, and \( T(x) \) should capture all the information about \( \theta \).
03

Construct the Likelihood Function

For \( n \) independent observations \( X_1, X_2, ..., X_n \), the likelihood function is the product of individual densities: \[ L(\theta | x_1, x_2, ..., x_n) = \prod_{i=1}^{n} \frac{x_i}{\theta^{2}} e^{-x_i^{2}/(2\theta^{2})} = \left(\frac{1}{\theta^{2}}\right)^n \prod_{i=1}^{n} x_i \cdot e^{-\sum_{i=1}^{n} x_i^2/(2\theta^2)}. \]
04

Factorize the Likelihood

The likelihood can be rewritten as: \[ L(\theta | x_1, x_2, ..., x_n) = \left(\frac{1}{\theta^{2}}\right)^n \cdot e^{-\sum_{i=1}^{n} x_i^2 / (2\theta^2)} \prod_{i=1}^{n} x_i. \] Here, \( g(T(x), \theta) = \left(\frac{1}{\theta^{2}}\right)^n e^{-\sum_{i=1}^{n} x_i^2 / (2\theta^2)} \) and \( h(x) = \prod_{i=1}^{n} x_i \). This indicates \( T(X) = \sum_{i=1}^{n} x_i^2 \) is a sufficient statistic for \( \theta \).
05

Conclusion on Sufficient Statistic

By the factorization, \( T(X) = \sum_{i=1}^{n} x_i^2 \) captures all the needed information about \( \theta \). This makes it the sufficient statistic according to the Factorization Theorem.

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Key Concepts

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

Rayleigh Distribution
The Rayleigh distribution is commonly used in scenarios involving mechanical or signal processing systems. It is a continuous probability distribution that describes the magnitude of a two-dimensional vector whose components are independent and identically distributed. Typically, these components follow a normal distribution with the same standard deviation. One example is assessing the distribution of wind speeds.

The Rayleigh density function is given by:\[f(x | \theta) = \frac{x}{\theta^{2}} e^{-x^{2} /(2 \theta^{2})}, \]where \( x \geq 0 \) and \( \theta > 0 \). Here, \( \theta \) represents a scale parameter that is essential for defining the spread of the distribution. Since it deals with non-negative values, the Rayleigh distribution is perfect for modeling real-world magnitudes such as fading signals.
  • The shape of the distribution is always skewed to the right.
  • It is a special case of the Weibull distribution with the shape parameter equal to 2.
Likelihood Function
The likelihood function is a fundamental part of statistical inference. It represents the probability of observing the given data under specific parameter estimates, in our case, the parameter \( \theta \). The likelihood function for the Rayleigh distribution with \( n \) independent observations \( X_1, X_2, ..., X_n \) is:

\[L(\theta | x_1, x_2, ..., x_n) = \prod_{i=1}^{n} \frac{x_i}{\theta^{2}} e^{-x_i^{2}/(2\theta^{2})} = \left(\frac{1}{\theta^{2}}\right)^n \prod_{i=1}^{n} x_i \cdot e^{-\sum_{i=1}^{n} x_i^2/(2\theta^2)}.\]
  • The likelihood approaches zero rapidly for incorrect parameter values.
  • Maximizing this function helps in estimating the most probable parameter values—here, it is the parameter \( \theta \).
Factorization Theorem
The Factorization Theorem is a powerful tool in statistics. It provides a method for identifying sufficient statistics. A statistic is sufficient if it captures all necessary information about a parameter in the data. According to the theorem, a statistic \( T(X) \) is sufficient for \( \theta \) if the likelihood can be expressed as:
\[L(\theta | x_1, x_2, ..., x_n) = g(T(x), \theta)h(x)\]This implies a clear split between the parameter and non-parameter-containing factors.

In our exercise, rewriting the likelihood to align with:\[g(T(x), \theta) = \left(\frac{1}{\theta^{2}}\right)^n e^{-\sum_{i=1}^{n} x_i^2 / (2\theta^2)} \]indicates that \( T(X) = \sum_{i=1}^{n} x_i^2 \) is indeed a sufficient statistic.
  • This factorization isolates \( \theta \) dynamics, encapsulating the essential information within \( T(X) \).
  • It profoundly simplifies the problem of parameter estimation.
Parameter Estimation
Parameter estimation in statistics involves determining the values of parameters that maximize the likelihood function. In the context of the Rayleigh distribution, this often involves using methods like Maximum Likelihood Estimation (MLE).

The goal is to find the parameter \( \theta \, \) that maximizes the likelihood function, thus making our sample data most probable under the assumed model. This requires an understanding of both the mathematical function in play and the interpretation of the optimization results.
  • MLE often leads to efficient and unbiased parameter estimates asymptotically.
  • Software tools can efficiently handle the computational complexities involved.
Gaining insights through parameter estimation allows us to build reliable models and make informed decisions based on statistical evidence.

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Most popular questions from this chapter

Suppose that a random sample of size 20 is taken from a normal distribution with unknown mean and known variance equal to \(1,\) and the mean is found to be \(\bar{x}=10 .\) A normal distribution was used as the prior for the mean, and it was found that the posterior mean was 15 and the posterior standard deviation was 0.1. What were the mean and standard deviation of the prior?

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.

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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 \(\alpha .\)

Suppose that \(X_{1}, X_{2}, \ldots, X_{n}\) are i.i.d. \(N\left(\mu, \sigma^{2}\right)\). a. If \(\mu\) is known, what is the mle of \(\sigma ?\) b. If \(\sigma\) is known, what is the mle of \(\mu ?\) c. In the case above \((\sigma \text { known), does any other unbiased estimate of } \mu\) have smaller variance?

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