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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.

Short Answer

Expert verified
The posterior means are 0.0392 for Case 1 and 0.0332 for Case 2. The choice of prior affects the mean estimate.

Step by step solution

01

Determine the Likelihood Function

When sampling items, the number of defective items follows a Binomial distribution. With 3 defective items out of 100, the likelihood function for the proportion of defective items \(\theta\), given by the binomial distribution, is:\[ \text{Likelihood} \propto \theta^3 (1-\theta)^{97} \].
02

Define the Prior Distributions for Each Case

Two beta distributions are given to represent different prior beliefs: 1. Case 1: Beta distribution \(\text{Beta}(1,1)\) which is uniform and represents no prior knowledge.2. Case 2: Beta distribution \(\text{Beta}(0.5,5)\) which suggests a stronger belief in fewer defective items.
03

Compute the Posterior Distribution for Case 1

For Case 1, using the prior \(\text{Beta}(1,1)\), the posterior distribution is given by:\[ \text{Posterior} \propto \theta^{3+1-1}(1-\theta)^{97+1-1} = \theta^3 (1-\theta)^{97} \]The resulting distribution is \(\text{Beta}(4,98)\).
04

Compute the Posterior Distribution for Case 2

For Case 2, using the prior \(\text{Beta}(0.5,5)\), the posterior distribution is:\[ \text{Posterior} \propto \theta^{3+0.5-1}(1-\theta)^{97+5-1} = \theta^{2.5} (1-\theta)^{101} \]This results in a \(\text{Beta}(3.5,102)\) distribution.
05

Determine the Posterior Means for Both Cases

To find the posterior means, use the formula for the mean of a beta distribution \(\frac{a}{a+b}\):- For Case 1 (\(\text{Beta}(4,98)\)): \[ E[\theta|X] = \frac{4}{4+98} = \frac{4}{102} \approx 0.0392 \].- For Case 2 (\(\text{Beta}(3.5,102)\)): \[ E[\theta|X] = \frac{3.5}{3.5+102} = \frac{3.5}{105.5} \approx 0.0332 \].
06

Plot the Posterior Distributions and Use Comparisons

Using the \(\text{Beta}(4,98)\) and \(\text{Beta}(3.5,102)\) distributions, plot them to visualize the differences. The first case suggests a higher expected proportion of defects due to the more neutral prior. The second case represents a more confident prior leading to a lower mean estimate.
07

Explain Differences in the Posterior Means

The difference in posterior means highlights how the prior belief influences the estimation. Case 1 has a higher mean suggesting less prior influence, whereas Case 2 shows a more 'conservative' estimation due to the prior belief in fewer defects.

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Key Concepts

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

Beta Distribution
The Beta distribution is a versatile and widely used probability distribution in the realm of Bayesian statistics. It is defined by two parameters, commonly denoted as \(a\) and \(b\), which shape the distribution, determining its skewness and variance. The Beta distribution is particularly beneficial for modeling variables that are constrained between 0 and 1, such as probabilities or proportions.
This makes it an excellent candidate for prior distributions in Bayesian inference when dealing with binomial proportions.

Let's break down its usefulness:
  • **Flexibility:** By adjusting the parameters \(a\) and \(b\), the Beta distribution can represent a wide variety of shapes. For example, \(\text{Beta}(1,1)\) yields a uniform distribution, suggesting no initial preference toward any value.
  • **Conjugate Prior:** The Beta distribution acts as a conjugate prior for the binomial distribution. This means that if you start with a Beta prior and update it with binomial data, the resulting posterior is also a Beta distribution. This mathematical harmony simplifies computation substantially.
In our exercise, two different Beta priors were used. The \(\text{Beta}(1,1)\) reflected a non-informative prior, indicating no preconceived notions.Meanwhile, \(\text{Beta}(0.5,5)\) suggested a belief favoring fewer defective items, affecting the posterior results significantly.
Posterior Distribution
In Bayesian statistics, the posterior distribution is the updated belief about an unknown parameter after considering new data. It combines prior information, represented by the prior distribution, with a likelihood function derived from observed data. The result is the posterior, which provides a comprehensive picture based on both prior beliefs and new evidence.

The general formula to compute a posterior distribution is given by Bayes' Theorem:\[P(\theta | X) \propto P(X | \theta) \cdot P(\theta)\]where:
  • \(P(\theta | X)\) is the posterior probability of \(\theta\) given data \(X\).
  • \(P(X | \theta)\) is the likelihood of observing data \(X\).
  • \(P(\theta)\) is the prior probability of \(\theta\).
In our exercise, once we applied Bayes' theorem, we attained two distinct posterior distributions for different priors.
The more neutral prior in Case 1 (\(\text{Beta}(1,1)\)) resulted in a higher mean estimate for defect proportions.In contrast, the prior with a stronger belief in fewer defects (\(\text{Beta}(0.5,5)\)) yielded a lower mean estimate. This highlights how different prior beliefs can lead to varied interpretations of the same dataset.
Binomial Distribution
The Binomial distribution is a foundational concept in probability theory, used to model situations where there are fixed numbers of independent trials, each with two possible outcomes: success or failure. It's defined by two parameters: the number of trials, \(n\), and the probability of success in a single trial, \(\theta\). The Binomial distribution serves as a perfect model for the likelihood function of our exercise, where we count the number of defective items.
  • **Key Characteristics:** The probability of observing \(k\) successes in \(n\) trials is given by the formula:
    \[ P(X = k) = \binom{n}{k} \theta^k (1-\theta)^{n-k} \]
  • **Real-world Application:** It applies to countless scenarios, such as quality control, where we might want to know the probability of finding a certain number of defective items in a batch.
In the exercise, the Binomial distribution describes the likelihood of finding defective items in the sample. Given 3 defective items out of 100, our likelihood was weighted by this distribution, informing how we update our beliefs within Bayesian context.This step was crucial because it integrated empirical evidence into the Bayesian framework, transforming prior distributions into informative posteriors.

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Most popular questions from this chapter

Let \(X_{1}, \ldots, X_{n}\) be an i.i.d. sample from a Rayleigh distribution with parameter \(\theta>0\) $$f(x | \theta)=\frac{x}{\theta^{2}} e^{-x^{2} /\left(2 \theta^{2}\right)}, \quad x \geq 0$$ (This is an alternative parametrization of that of Example A in Section 3.6.2.) a. Find the method of moments estimate of \(\theta\) b. Find the mle of \(\theta\) c. Find the asymptotic variance of the mle.

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?

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.]

The Poisson distribution has been used by traffic engineers as a model for light traffic, based on the rationale that if the rate is approximately constant and the traffic is light (so the individual cars move independently of each other), the distribution of counts of cars in a given time interval or space area should be nearly Poisson (Gerlough and Schuhl 1955 ). The following table shows the number of right turns during 300 3-min intervals at a specific intersection. Fit a Poisson distribution. Comment on the fit by comparing observed and expected counts. It is useful to know that the 300 intervals were distributed over various hours of the day and various days of the week. $$\begin{array}{cc} \hline n & \text { Frequency } \\ \hline 0 & 14 \\ 1 & 30 \\ 2 & 36 \\ 3 & 68 \\ 4 & 43 \\ 5 & 43 \\ 6 & 30 \\ 7 & 14 \\ 8 & 10 \\ 9 & 6 \\ 10 & 4 \\ 11 & 1 \\ 12 & 1 \\ 13+ & 0 \\ \hline \end{array}$$

Let the unknown probability that a basketball player makes a shot successfully be \(\theta .\) Suppose your prior on \(\theta\) is uniform on [0,1] and that she then makes two shots in a row. Assume that the outcomes of the two shots are independent. a. What is the posterior density of \(\theta ?\) b. What would you estimate the probability that she makes a third shot to be?

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