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Informative prior distribution: as a modern-day Laplace, you have more definite beliefs about the ratio of male to female births than reflected by his uniform prior distribution. In particular, if \(\theta\) represents the proportion of female births in a given population, you are willing to place a Beta(100, 100) prior distribution on \(\theta\). (a) Show that this means you are more than \(95 \%\) sure that \(\theta\) is between \(0.4\) and \(0.6\), although you are ambivalent as to whether it is greater or less than \(0.5\). (b) Now you observe that out of a random sample of 1,000 births, 511 are boys. What is your posterior probability that \(\theta>0.5 ?\) Compute using the exact beta distribution or the normal approximation.

Short Answer

Expert verified
The probability that \(\theta > 0.5\) is slightly less than 0.5, indicating ambivalence.

Step by step solution

01

Understand the Prior Distribution

We are given that the prior distribution for the proportion of female births, \(\theta\), is Beta(100, 100). In a Beta distribution, the parameters \(\alpha = 100\) and \(\beta = 100\) imply that we have highly concentrated beliefs around the mean, which is \( \frac{\alpha}{\alpha + \beta} = \frac{100}{200} = 0.5\). This is symmetric around \(0.5\).
02

Calculate the 95% Credible Interval for the Prior

To find that we are more than 95% sure \(\theta\) is between \(0.4\) and \(0.6\), we calculate a 95% credible interval for the Beta(100, 100) distribution. Using statistical software or a Beta distribution table, we find the quantiles for 0.025 and 0.975. This will show that the interval \([0.4, 0.6]\) contains more than 95% of the probability mass.
03

Use Observations to Determine Likelihood

Given 511 boys out of 1000 births, the number of girls is 489. The likelihood given this data follows a binomial distribution: \( \text{Binomial}(1000, \theta)\).
04

Compute the Posterior Distribution

Using the beta-binomial conjugacy, the posterior distribution is of the form \( \text{Beta}(\alpha + \, \text{girls}, \beta + \, \text{boys}) = \text{Beta}(100 + 489, 100 + 511) = \text{Beta}(589, 611) \).
05

Calculate Posterior Probability \( \theta > 0.5 \)

With the posterior distribution Beta(589, 611), calculate the probability that \(\theta > 0.5\). This can be done using cumulative distribution functions of the Beta distribution in statistical software, which provides the probability that \(\theta\) is less than 0.5. Thus, \( P(\theta > 0.5) = 1 - \text{CDF}(0.5)\).
06

Interpret the Results

After calculation, \( P(\theta > 0.5)\) typically yields a value less than 0.5, showing a higher confidence in \(\theta\) being close to 0.5 than significantly greater than it. This makes intuitive sense given the prior distribution's concentration around 0.5 and the symmetric data results.

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Key Concepts

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

Informative Prior Distribution
In Bayesian statistics, an informative prior distribution is used when we have strong prior beliefs about a parameter, based on evidence or expertise, before understanding new data. In the context of births, the informative prior distribution expresses our beliefs about the proportion of female births.

A prior is considered informative if it provides a distinct concentration of probability, reflecting this pre-existing knowledge. In our exercise, the use of a
  • Beta(100, 100)
distribution as a prior implies that we expect the proportion of female births, denoted by \( \theta \), to be around 0.5. This signifies balanced knowledge, neither skewed towards male nor female births.

This balance appears through the symmetrical parameters of the beta distribution that both equal 100, which anchors our certainty in the region close to 50-50 odds. Before processing new data, this distribution suggests that we are "more than 95% sure" that \( \theta \) lies between 0.4 and 0.6, conveying a narrow band of expectation around the mean of 0.5.
Posterior Probability
Posterior probability is a crucial step in Bayesian analysis. It represents the probability of a parameter after considering the new data in light of the prior distribution. The exercise's query about the probability that \( \theta > 0.5 \) post-observation leads us to examine the posterior distribution.

Combining our prior belief (Beta(100, 100)) about the proportion of female births with the new data from a sample of 1,000 births, where there were 511 boys and therefore 489 girls, allows us to update our beliefs about \( \theta \). We do this using the concept of conjugate priors: when the prior and likelihood belong to the same family, here beta and binomial respectively, this simplifies the updating process.

In this example, after computing through the
  • binomial likelihood
  • and adjusting the beta parameters,
we derive a posterior distribution of Beta(589, 611). Subsequently, the probability \( P(\theta > 0.5) \) is calculated by finding 1 minus the cumulative distribution function (CDF) value at 0.5. This statistical measure quantifies our updated certainty, following the introduction of actual birth data.
Beta Distribution
The beta distribution is a versatile tool in Bayesian statistics, particularly well-known for modeling probabilities because it is bound within the interval [0,1]. By adjusting its two shape parameters, \( \alpha \) and \( \beta \), the distribution can take on a variety of shapes, reflecting different levels of prior certainty.

In the context of the exercise, the
  • prior distribution Beta(100, 100)
illustrates a concentrated level of belief, symmetric around 0.5, indicating no initial bias towards a higher probability of either male or female births. The compactness about the center stems from the high parameter values which produce a small variance.

After observing new data (489 females out of 1,000 births), the posterior becomes Beta(589, 611). This shift adjusts our confidence and introduces a refined perspective based on actual observations, while maintaining the beta's flexibility and convenience. The new shape reflects both initial beliefs and the influence of empirical data.
Binomial Distribution
The binomial distribution comes into play when dealing with binary outcomes, such as birth gender, where only two possibilities exist. This distribution is defined by two parameters: the number of trials and the probability of success in each trial. For our exercise, this is expressed in the random sample of 1,000 births, with the outcome of each birth being either a boy or a girl.

When analyzing the number of girls (489), the likelihood is modeled using the binomial distribution with \( \theta \) being the probability of a girl in the population. The choice of this likelihood reflects our understanding of the randomness and nature of births.

Crucially, the binomial distribution's alignment with the beta distribution as a conjugate prior allows for a straightforward update to our prior beliefs, transforming them into a posterior distribution. This seamless integration pacifies the mathematical complexity often faced in Bayesian updates, offering clear insights into how observed data sways our beliefs.

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Most popular questions from this chapter

Predictive distribution and tolerance intervals: a scientist using an apparatus of known standard deviation \(0.12\) takes nine independent measurements of some quantity. The measurements are assumed to be normally distributed, with the stated standard deviation and unknown mean \(\theta\), where the scientist is willing to place a vague prior distribution on \(\theta .\) If the sample mean obtained is \(17.653\), obtain limits between which a tenth measurement will lie with \(99 \%\) probability. This is called a \(99 \%\) tolerance interval.

Noninformative prior densities: (a) For the binomial likelihood, \(y \sim \operatorname{Bin}(n, \theta)\), show that \(p(\theta) \propto \theta^{-1}(1-\) \(\theta)^{-1}\) is the uniform prior distribution for the natural parameter of the exponential family. (b) Show that if \(y=0\) or \(n\), the resulting posterior distribution is improper.

Setting parameters for a beta prior distribution: suppose your prior distribution for \(\theta\), the proportion of Californians who support the death penalty, is beta with mean \(0.6\) and standard deviation \(0.3 .\) (a) Determine the parameters \(\alpha\) and \(\beta\) of your prior distribution. Sketch the prior density function. (b) A random sample of 1000 Californians is taken, and \(65 \%\) support the death penalty. What are your posterior mean and variance for \(\theta\) ? Draw the posterior density function.

Normal distribution with unknown mean: a random sample of \(n\) students is drawn from a large population, and their weights are measured. The average weight of the \(n\) sampled students is \(\bar{y}=150\) pounds. Assume the weights in the population are normally distributed with unknown mean \(\theta\) and known standard deviation 20 pounds. Suppose your prior distribution for \(\theta\) is normal with mean 180 and standard deviation 40 . (a) Give your posterior distribution for \(\theta .\) (Your answer will be a function of \(n .\) ) (b) A new student is sampled at random from the same population and has a weight of \(\tilde{y}\) pounds. Give a posterior predictive distribution for \(\tilde{y}\). (Your answer will still be a function of \(n .\) ) (c) For \(n=10\), give a \(95 \%\) posterior interval for \(\theta\) and a \(95 \%\) posterior predictive interval for \(\hat{y}\). (d) Do the same for \(n=100\).

Predictive distributions: let \(y\) be the number of 6 's in 1000 independent rolls of a particular real die, which may be unfair. Let \(\theta\) be the probability that the die lands on '6.' Suppose your prior distribution for \(\theta\) is as follows: $$ \begin{aligned} \operatorname{Pr}(\theta=1 / 12) &=0.25 \\ \operatorname{Pr}(\theta=1 / 6) &=0.5 \\ \operatorname{Pr}(\theta=1 / 4) &=0.25 \end{aligned} $$ (a) Using the normal approximation for the conditional distributions, \(p(y \mid \theta)\), sketch your approximate prior predictive distribution for \(y\). (b) Give approximate \(5 \%, 25 \%, 50 \%, 75 \%\), and \(95 \%\) points for the distribution of \(y .\) (Be careful here: \(y\) does not have a normal distribution, but you can still use the normal distribution as part of your analysis.)

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