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Consider a random sample \(X_{1}, X_{2}, \ldots, X_{n}\) from the normal distribution with mean 0 and precision \(\tau\) (use \(\tau\) as a parameter instead of \(\sigma^{2}=1 / \tau\) ). Assume a gamma-distributed prior for \(\tau\) and show that the posterior distribution of \(\tau\) is also gamma. What are its parameters?

Short Answer

Expert verified
The posterior distribution of \( \tau \) is Gamma with parameters \(a + n/2, b + \frac{1}{2}\sum_{i=1}^n X_i^2\).

Step by step solution

01

Define the Prior and Likelihood

The prior distribution for precision \( \tau \) is given as a Gamma distribution: \[\tau \sim \text{Gamma}(a, b) \] where \( a \) and \( b \) are the shape and rate parameters, respectively. The likelihood of the sample \( X = (X_1, X_2, \ldots, X_n) \) given \(\tau\) is: \[L(X|\tau) = \prod_{i=1}^{n} \left( \frac{\sqrt{\tau}}{\sqrt{2\pi}} \right) \exp{-\frac{\tau X_i^2}{2}} \] This simplifies to: \[L(X|\tau) \propto \tau^{n/2} \exp\left(-\frac{\tau}{2} \sum_{i=1}^{n}X_i^2\right) \] This is because each \( X_i \) is normally distributed with mean 0 and precision \( \tau \).
02

Find the Posterior Distribution

Using Bayes' theorem, the posterior is proportional to the product of the prior and likelihood: \[P(\tau | X) \propto L(X|\tau) \cdot P(\tau) \]Substitute in the equations for the prior and likelihood:\[P(\tau | X) \propto \tau^{n/2} \exp\left(-\frac{\tau}{2} \sum_{i=1}^{n}X_i^2\right) \cdot \tau^{a-1}e^{-b\tau} \]Combine the powers of \( \tau \) and the exponents:\[P(\tau|X) \propto \tau^{a-1+n/2} \exp\left(-\tau\left(b + \frac{1}{2}\sum_{i=1}^n X_i^2\right)\right) \]
03

Recognize the Posterior as a Gamma Distribution

The resulting posterior can be recognized as a gamma distribution:\[\tau | X \sim \text{Gamma}(a + n/2, b + \frac{1}{2}\sum_{i=1}^n X_i^2) \] This is because the gamma distribution has the form:\[\text{Gamma}(c, d) = \tau^{c-1} e^{-d\tau} \] Thus, the parameters of the posterior distribution are \((a + n/2, b + \frac{1}{2}\sum_{i=1}^n X_i^2)\).

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Key Concepts

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

Gamma Distribution
The Gamma Distribution is a continuous probability distribution often used to model waiting times or lifetimes of processes. It is characterized by two parameters: the shape parameter \(a\) and the rate parameter \(b\). These parameters help define the behavior of the distribution.
  • Shape Parameter (\(a\)): Determines the shape of the distribution. Larger values make it more symmetric and spread out.
  • Rate Parameter (\(b\)): Influences the scale or spread of the distribution. Smaller values lead to wider spread values.
The probability density function (pdf) for a Gamma distribution is expressed as:
\[ f(x|a, b) = \frac{b^a x^{a-1} e^{-bx}}{\Gamma(a)},\]where \( x > 0 \) and \(\Gamma(a)\) is the gamma function. In Bayesian statistics, the Gamma distribution is often used as a prior distribution for unknown precision or rate parameters, such as those found in normal distributions.
Normal Distribution
The Normal Distribution, also known as the Gaussian distribution, is a fundamental statistical distribution that is symmetric and follows a bell-shaped curve. It is commonly used in statistics due to the Central Limit Theorem, which states that sums of random variables tend to follow a normal distribution.
  • Mean (\(\mu\)): The average or central value of the distribution.
  • Precision (\(\tau\)): Instead of variance, precision is considered here \(\tau = 1/\sigma^2\). Higher precision implies that the data is closely concentrated around the mean.
In this exercise, we consider a normal distribution with a mean of 0. The probability density function is expressed as:
\[ f(x) = \sqrt{\frac{\tau}{2\pi}} e^{-\tau x^2/2},\]where \(\tau\) is the precision. Understanding how precision affects the distribution helps in comprehending how the data clusters around the mean.
Posterior Distribution
The Posterior Distribution in Bayesian analysis is the updated probability of a parameter after considering new evidence or data. It is derived using Bayes' theorem, which updates prior beliefs with the likelihood of observed data to obtain a new distribution.
  • Prior Distribution: Initial beliefs about a parameter before observing data, here a Gamma distribution for \(\tau\).
  • Likelihood: The probability of the observed data given the parameter, derived from the normal distribution of the data.
Through Bayes' theorem, the posterior distribution \(P(\tau | X)\) is obtained by scaling the product of the prior and likelihood functions. In this exercise, it has been shown that the posterior distribution of \(\tau\) remains a Gamma distribution:
\[\tau | X \sim \text{Gamma}(a + n/2, b + \frac{1}{2}\sum_{i=1}^n X_i^2).\] This shows that even after observing the data, the form of the prior distribution is maintained in the posterior, with updated parameters based on the data.
Random Sample
A Random Sample refers to a subset of individuals or observations taken from a larger population, where each member of the population has an equal chance of being selected. It is essential for ensuring unbiased estimates of population parameters.
  • Independence: Each observation in the sample is independent of the others.
  • Uniform Selection: Every member of the population has an equal likelihood of being part of the sample.
In the context of this exercise, the random sample, \(X_1, X_2, \ldots, X_n\), is taken from a normal distribution with zero mean. The sample helps in computing the likelihood function, which in turn plays a crucial role in updating the posterior distribution. This concept highlights the importance of randomness in sampling to achieve representativeness and fairness in statistical analysis.

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Most popular questions from this chapter

The single-factor ANOVA model considered in Chapter 11 assumed the observations in the \(i\) th sample were selected from a normal distribution with mean \(\mu_{i}\) and variance \(\sigma^{2}\), that is, \(X_{i j}=\mu_{i}+\varepsilon_{i j}\) where the \(\varepsilon\) 's are normal with mean 0 and variance \(\sigma^{2}\). The normality assumption implies that the \(F\) test is not distribution-free. We now assume that the \(\varepsilon\) 's all come from the same continuous, but not necessarily normal, distribution, and develop a distribution-free test of the null hypothesis that all \(I \mu_{i}\) 's are identical. Let \(N=\sum J_{i}\), the total number of observations in the data set (there are \(J_{i}\) observations in the \(i\) th sample). Rank these \(N\) observations from 1 (the smallest) to \(N\), and let \(\bar{R}_{i}\) be the average of the ranks for the observations in the ith sample. When \(H_{0}\) is true, we expect the rank of any particular observation and therefore also \(\bar{R}_{i}\) to be \((N+1) / 2\). The data argues against \(H_{0}\) when some of the \(\bar{R}_{i}\) 's differ considerably from \((N+1) / 2\). The Kruskal-Wallis test statistic is $$ K=\frac{12}{N(N+1)} \sum J_{i}\left(\bar{R}_{i}-\frac{N+1}{2}\right)^{2} $$ When \(H_{0}\) is true and either (1) \(I=3\), all \(J_{i} \geq 6\) or (2) \(I>3\), all \(J_{i} \geq 5\), the test statistic has approximately a chi-squared distribution with \(I-1\) df. The accompanying observations on axial stiffness index resulted from a study of metal-plate connected trusses in which five different plate lengths-4 in., 6 in., 8 in., 10 in., and 12 in. were used ("Modeling Joints Made with LightGauge Metal Connector Plates," Forest Products \(J ., 1979: 39-44)\). \(\begin{array}{lllll}i=1(4 \text { in. }): & 309.2 & 309.7 & 311.0 & 316.8 \\\ & 326.5 & 349.8 & 409.5 & \\ i=2(6 \text { in. }): & 331.0 & 347.2 & 348.9 & 361.0 \\ & 381.7 & 402.1 & 404.5 & \\ i=3(8 \text { in. }): & 351.0 & 357.1 & 366.2 & 367.3 \\ & 382.0 & 392.4 & 409.9 & \\ i=4(10 \text { in. }): & 346.7 & 362.6 & 384.2 & 410.6 \\ & 433.1 & 452.9 & 461.4 & \\ i=5(12 \text { in. }): & 407.4 & 410.7 & 419.9 & 441.2 \\ & 441.8 & 465.8 & 473.4 & \end{array}\) Use the \(K-W\) test to decide at significance level \(.01\) whether the true average axial stiffness index depends somehow on plate length.

The model for the data from a randomized block experiment for comparing \(I\) treatments was \(X_{i j}=\mu+\alpha_{i}+\beta_{j}+\varepsilon_{i j}\), where the \(\alpha\) 's are treatment effects, the \(\beta\) 's are block effects, and the \(\varepsilon\) 's were assumed normal with mean 0 and variance \(\sigma^{2}\). We now replace normality by the assumption that the \(\varepsilon\) 's have the same continuous distribution. A distribution-free test of the null hypothesis of no treatment effects, called Friedman's test, involves first ranking the observations in each block separately from 1 to \(I\). The rank average \(\bar{R}_{i}\) is then calculated for each of the \(I\) treatments. If \(H_{0}\) is true, the expected value of each rank average is \((I+1) / 2\). The test statistic is $$ F_{r}=\frac{12 J}{I(I+1)} \sum\left(\bar{R}_{i}-\frac{I+1}{2}\right)^{2} $$ For even moderate values of \(J\), the test statistic has approximately a chi- squared distribution with \(I-1 \mathrm{df}\) when \(H_{0}\) is true. The article "Physiological Effects During Hypnotically Requested Emotions" (Psychosomatic Med., 1963: 334-343) reports the following data \(\left(x_{i j}\right)\) on skin potential in millivolts when the emotions of fear, happiness, depression, and calmness were requested from each of eight subjects. $$ \begin{array}{lllrl} \text { Fear } & 23.1 & 57.6 & 10.5 & 23.6 \\ \text { Happiness } & 22.7 & 53.2 & 9.7 & 19.6 \\ \text { Depression } & 22.5 & 53.7 & 10.8 & 21.1 \\ \text { Calmness } & 22.6 & 53.1 & 8.3 & 21.6 \\ & \mathbf{5} & \mathbf{6} & \mathbf{7} & \mathbf{8} \\ \cline { 2 - 5 } \text { Fear } & 11.9 & 54.6 & 21.0 & 20.3 \\ \text { Happiness } & 13.8 & 47.1 & 13.6 & 23.6 \\ \text { Depression } & 13.7 & 39.2 & 13.7 & 16.3 \\ \text { Calmness } & 13.3 & 37.0 & 14.8 & 14.8 \\ \hline \end{array} $$ Use Friedman's test to decide whether emotion has an effect on skin potential.

The accompanying data is a subset of the data reported in the article "Synovial Fluid \(\mathrm{pH}\), Lactate, Oxygen and Carbon Dioxide Partial Pressure in Various Joint Diseases" (Arthritis Rheum., 1971: 476-477). The observations are \(\mathrm{pH}\) values of synovial fluid (which lubricates joints and tendons) taken from the knees of individuals suffering from arthritis. Assuming that true average \(\mathrm{pH}\) for non-arthritic individuals is \(7.39\), test at level \(.05\) to see whether the data indicates a difference between average \(\mathrm{pH}\) values for arthritic and nonarthritic individuals. $$ \begin{array}{lllllll} 7.02 & 7.35 & 7.34 & 7.17 & 7.28 & 7.77 & 7.09 \\ 7.22 & 7.45 & 6.95 & 7.40 & 7.10 & 7.32 & 7.14 \end{array} $$

The sign test is a very simple procedure for testing hypotheses about a population median assuming only that the underlying distribution is continuous. To illustrate, consider the following sample of 20 observations on component lifetime (hr): \(\begin{array}{rrrrrrr}1.7 & 3.3 & 5.1 & 6.9 & 12.6 & 14.4 & 16.4 \\ 24.6 & 26.0 & 26.5 & 32.1 & 37.4 & 40.1 & 40.5 \\ 41.5 & 72.4 & 80.1 & 86.4 & 87.5 & 100.2 & \end{array}\) We wish to test the hypotheses \(H_{0}: \widetilde{\mu}=25.0\) versus \(H_{\mathrm{a}}: \widetilde{\mu}>25.0\) The test statistic is \(Y=\) the number of observations that exceed \(25 .\) a. Consider rejecting \(H_{0}\) if \(Y \geq 15\). What is the value of \(\alpha\) (the probability of a type I error) for this test? [Hint: Think of a "success" as a lifetime that exceeds \(25.0\). Then \(Y\) is the number of successes in the sample. What kind of a distribution does \(Y\) have when \(\widetilde{\mu}=25.0\) ?] b. What rejection region of the form \(Y \geq c\) specifies a test with a significance level as close to \(.05\) as possible? Use this region to carry out the test for the given data. [Note: The test statistic is the number of differences \(X_{i}-25.0\) that have positive signs, hence the name sign test.]

Here are the IQ scores for the 15 first-grade girls from the study mentioned in Example 14.8. \(\begin{array}{lllllll}102 & 96 & 106 & 118 & 108 & 122 & 115 \\ 109 & 113 & 82 & 110 & 121 & 110 & 99\end{array}\) Assume the same prior distribution used in Example 14.8, and assume that the data is a random sample from a normal distribution with mean \(\mu\) and \(\sigma=15\). a. Find the posterior distribution of \(\mu\). b. Find a \(95 \%\) credibility interval for \(\mu\). c. Add four observations with average 110 to the data and find a \(95 \%\) confidence interval for \(\mu\) using the 19 observations. Compare with the result of (b). d. Change the prior so the prior precision is very small but positive, and then recompute (a) and (b). e. Find a \(95 \%\) confidence interval for \(\mu\) using the 15 observations and compare with the credibility interval of (d).

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