/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 A population consists of \(k\) c... [FREE SOLUTION] | 91Ó°ÊÓ

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A population consists of \(k\) classes \(\theta_{1}, \ldots, \theta_{k}\) and it is required to classify an individual on the basis of an observation \(Y\) having density \(f_{i}\left(y \mid \theta_{i}\right)\) when the individual belongs to class \(i=1, \ldots, k\). The classes have prior probabilities \(\pi_{1}, \ldots, \pi_{k}\) and the loss in classifying an individual from class \(i\) into class \(j\) is \(l_{i j}\). (a) Find the posterior probability \(\pi_{i}(y)=\operatorname{Pr}\) (class \(\left.i \mid y\right)\) and the posterior risk of allocating the individual to class \(i\). (b) Now consider the case of \(0-1\) loss, that is, \(l_{i j}=0\) if \(i=j\) and \(l_{i j}=1\) otherwise. Show that the risk is the probability of misclassification. (b) Suppose that \(k=3\), that \(\pi_{1}=\pi_{2}=\pi_{3}=1 / 3\) and that \(Y\) is normally distributed with mean \(i\) and variance 1 in class \(i\). Find the Bayes rule for classifying an observation. Use it to classify the observation \(y=2.2\).

Short Answer

Expert verified
The observation \(y=2.2\) is classified into class 2.

Step by step solution

01

Find Posterior Probability

To find the posterior probability \(\pi_{i}(y) = \Pr(\text{class } i \mid y)\), we use Bayes' Theorem: \[\pi_{i}(y) = \frac{f_{i}(y \mid \theta_{i}) \pi_{i}}{\sum_{j=1}^{k} f_{j}(y \mid \theta_{j}) \pi_{j}}.\] This equation calculates the probability that an observation belongs to class \(i\) given \(Y = y\), based on the class-specific density function and prior probabilities.
02

Calculate Posterior Risk

The posterior risk of allocating an individual to class \(i\) is given by:\[R(i \mid y) = \sum_{j=1}^{k} l_{ij} \pi_{j}(y),\]where \(\pi_{j}(y)\) is the posterior probability. It measures the expected loss of incorrectly classifying an observation into class \(i\).
03

Consider 0-1 Loss

For a 0-1 loss, \(l_{ij} = 0\) if \(i = j\) and \(l_{ij} = 1\) otherwise. Applying this to the posterior risk, we find:\[R(i \mid y) = \sum_{j eq i} \pi_{j}(y) = 1 - \pi_{i}(y).\]Thus, the risk is equivalent to the probability of misclassification, \(1 - \pi_{i}(y)\).
04

Set Up the Problem for Bayes Rule (k=3)

Given: \(k=3\), \(\pi_1 = \pi_2 = \pi_3 = 1/3\), \(Y \sim N(i, 1)\) in class \(i\). The density function for class \(i\) is:\[f_{i}(y \mid \theta_{i}) = \frac{1}{\sqrt{2\pi}} e^{-(y-i)^2/2}.\]By Bayes' Rule, the posterior probability for class \(i\) becomes:\[\pi_{i}(y) = \frac{f_{i}(y \mid \theta_{i}) \cdot \pi_{i}}{\sum_{j=1}^{3} f_{j}(y \mid \theta_{j}) \cdot \pi_{j}}.\]
05

Apply Bayes Rule to Classify Observation

Using the density functions and given \(y = 2.2\), calculate \(f_{i}(2.2 \mid \theta_{i})\) for each class, assume mean \(\mu_i = i\):\[f_1(2.2) = \frac{1}{\sqrt{2\pi}} e^{-0.8^2/2}, \quad f_2(2.2) = \frac{1}{\sqrt{2\pi}} e^{-0.2^2/2}, \quad f_3(2.2) = \frac{1}{\sqrt{2\pi}} e^{-1.2^2/2}.\]Subsequently, compute posterior probabilities and classify \(y\) into the class with the highest posterior probability.
06

Calculate Values and Determine Class

Calculate the probabilities:\[f_1(2.2) = 0.3521, \quad f_2(2.2) = 0.3910, \quad f_3(2.2) = 0.1942.\]Calculate posterior probabilities:\[\pi_1(2.2) = 0.298, \quad \pi_2(2.2) = 0.331, \quad \pi_3(2.2) = 0.164.\]Thus, the individual is classified into class \(2\) as it has the highest posterior probability.

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Key Concepts

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

Posterior Probability
In statistics, the concept of posterior probability serves as a crucial element in Bayesian Classification. Simply put, the posterior probability represents the likelihood that a certain observation belongs to a specific class, given some observed data. It incorporates both the underlying patterns from the data via the probability density function, and the prior knowledge regarding the distribution of different classes, known as prior probabilities.

Bayes' Theorem plays a pivotal role in computing posterior probabilities as it helps in combining prior knowledge with observed evidence. The formula to determine the posterior probability is given by:
  • \[ \pi_{i}(y) = \frac{f_{i}(y \mid \theta_{i}) \pi_{i}}{\sum_{j=1}^{k} f_{j}(y \mid \theta_{j}) \pi_{j}}.\]
This formula articulates that the probability of an observation belonging to class \(i\) is proportional to the product of the probability density function for that class and the prior probability of the class, divided by the total probability of the observation across all classes. This hinges on both the characteristics of the class and the observations we make.
Bayes' Theorem
Bayes' Theorem is a foundational principle in probability theory and statistics. It allows us to update our beliefs in light of new evidence. In essence, it provides a way to revise existing predictions or hypotheses (prior probability) by incorporating new relevant information (likelihood).

This theorem is mathematically expressed as follows:
  • \[ \text{Posterior Probability} = \frac{\text{Likelihood} \times \text{Prior Probability}}{\text{Evidence}}.\]
In the context of classification, Bayes' Theorem helps in determining the class to which a particular observation is most likely to belong. The theorem essentially bridges prior probabilities of classes (our initial guess based on previous information) with the observed data (how well the data fits each class), producing a revised set of probabilities—posterior probabilities.
  • **Prior Probability:** Our initial understanding about the classes.
  • **Likelihood:** The probability of observing the data given the class.
  • **Posterior Probability:** The revised probability influenced by the evidence.
Misclassification Risk
When working with classification tasks, a potential challenge is misclassification. This occurs when an observation is assigned to the wrong class, and understanding this risk is fundamental in Bayesian Classification. The **misclassification risk** can be quantified using the concept of posterior risk.

The posterior risk defines the expected loss when a decision about a class allocation is made based on an observation. This is calculated by weighting the loss of misclassifying an observation by the posterior probabilities of that happening. Particularly in a 0-1 loss scenario, where the loss is 0 for correct classification and 1 for misclassification, the risk simplifies as follows:
  • \[ R(i \mid y) = 1 - \pi_{i}(y)\]
This formula tells us the risk is equivalent to the probability of misclassification. Practically, if the posterior probability of class \(i\) is high, the risk of misclassification is low, and vice versa. Understanding misclassification risk is pivotal for improving the robustness and reliability of classification models, such as deciding whether a new data-driven approach is necessary or adjusting existing thresholds.
Probability Density Function
The **Probability Density Function (PDF)** is a central concept when dealing with continuous data and forms the backbone of many statistical approaches, including Bayesian Classification. It describes the likelihood of a random variable falling within a particular range of values.

For normally distributed data, which is often a common assumption, the PDF is represented by:
  • \[ f_{i}(y \mid \theta_{i}) = \frac{1}{\sqrt{2\pi}} e^{-(y-i)^2/2}.\]
This equation models how data related to each class \(i\) is distributed, assuming it follows a normal distribution with its specific mean and variance. It indicates the likeliness of observing a particular value of \(y\) given the class parameters \(\theta_{i}\).

The PDF is crucial in determining posterior probabilities as it helps define the likelihood that is fundamental in calculating the posterior using Bayes' Theorem. When applying Bayes' Rule for classifying observations, like evaluating the example observation \(y = 2.2\), it is necessary to compute the PDF for each potential class to find which class the observation likely belongs to.

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Most popular questions from this chapter

Consider predicting the outcome of a future random variable \(Z\) on the basis of a random sample \(Y_{1}, \ldots, Y_{n}\) from density \(\lambda^{-1} e^{-u / \lambda}, u>0, \lambda>0\). Show that \(\pi(\lambda) \propto \lambda^{-1}\) gives posterior predictive density $$ f(z \mid y)=\frac{\int f(z, y \mid \lambda) \pi(\lambda) d \lambda}{\int f(y \mid \lambda) \pi(\lambda) d \lambda}=n s^{n} /(s+z)^{n+1}, \quad z>0 $$ where \(s=y_{1}+\cdots+y_{n}\) Show that when Laplace's method is applied to each integral in the predictive density the result is proportional to the exact answer, and assess how close the approximation is to a density when \(n=5\).

Show that the \((1-2 \alpha)\) HPD credible interval for a continuous unimodal posterior density \(\pi(\theta \mid y)\) is the shortest credible interval with level \((1-2 \alpha)\).

Suppose that \(Y_{1}, \ldots, Y_{n}\) are taken from an AR(1) process with innovation variance \(\sigma^{2}\) and correlation parameter \(\rho\) such that \(|\rho|<1\). Show that $$ \operatorname{var}(\bar{Y})=\frac{\sigma^{2}}{n^{2}\left(1-\rho^{2}\right)}\left\\{n+2 \sum_{j=1}^{n-1}(n-j) \rho^{j}\right\\} $$ and deduce that as \(n \rightarrow \infty\) for any fixed \(\rho, n \operatorname{var}(\bar{Y}) \rightarrow \sigma^{2} /(1-\rho)^{2}\). What happens when \(|\rho|=1 ?\) Discuss estimation of \(\operatorname{var}(\bar{Y})\) based on \((n-1)^{-1} \sum\left(Y_{j}-\bar{Y}\right)^{2}\) and an estimate \(\widehat{\rho}\).

According to the principle of insufficient reason probabilities should be ascribed uniformly to finite sets unless there is some definite reason to do otherwise. Thus the most natural way to express prior ignorance for a parameter \(\theta\) that inhabits a finite parameter space \(\theta_{1}, \ldots, \theta_{k}\) is to set \(\pi\left(\theta_{1}\right)=\cdots=\pi\left(\theta_{k}\right)=1 / k\). Let \(\pi_{i}=\pi\left(\theta_{i}\right)\). Consider a parameter space \(\left\\{\theta_{1}, \theta_{2}\right\\}\), where \(\theta_{1}\) denotes that there is life in orbit around the star Sirius and \(\theta_{2}\) that there is not. Can you see any reason not to take \(\pi_{1}=\pi_{2}=1 / 2 ?\) Now consider the parameter space \(\left[\omega_{1}, \omega_{2}, \omega_{3} \mid\right.\), where \(\omega_{1}, \omega_{2}\), and \(\omega_{3}\) denote the events that there is life around Sirius, that there are planets but no life, and that there are no planets. With this parameter space the principle of insufficient reason gives Pr(life around Sirius) \(=1 / 3\) Discuss this partitioning paradox. What solutions do you see? (Schafer. 1976. pp. 23-24)

An autoregressive process of order one with correlation parameter \(\rho\) is stationary only if \(|\rho|<1 .\) Discuss Bayesian inference for such a process. How might you (a) impose stationarity through the prior, (b) compute the probability that the process underlying data \(y\) is non-stationary, (c) compare the models of stationarity and non-stationarity?

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