Problem 14
Question
Evaluate the integral $$ \int_{0}^{\infty} \frac{1}{1+x^{2}} d x $$ using the Student \(t\) distribution.
Step-by-Step Solution
Verified Answer
The value of the integral is \( \frac{1}{2} \).
1Step 1: Identify and Rewrite
First, identify the necessary parameters for the Student's t-distribution to match the integrand. The integrand is a special case of the pdf of the Student's t-distribution \( f(t) = \frac{\Gamma(\frac{v+1}{2})}{\sqrt{v\pi} \Gamma(\frac{v}{2})}(1+\frac{t^2}{v})^{-\frac{v+1}{2}} \) where \( v = 1 \). Rewrite the integral in terms of the pdf with \( v = 1 \) and \( t = x \) as \( \int_{0}^{\infty} f(t) dx \).
2Step 2: Use Properties of Distributions
The pdf integrates to 1 over its range, hence the result of the current integral equals the area starting from zero to positive infinity. As the pdf is symmetric around zero (\( f(-t) = f(t) \)), it's valid to suggest that the area from zero to positive infinity equals one-half.
3Step 3: Final Answer
Therefore, the integral evaluates to 0.5 or \( \frac{1}{2} \).
Key Concepts
Integral EvaluationProbability Density FunctionGamma Function
Integral Evaluation
Understanding how to evaluate integrals is essential in mathematics. Integrals can represent various concepts like area under a curve. In this problem, the integral \( \int_{0}^{\infty} \frac{1}{1+x^{2}} \, dx \) represents the area under the curve of \( \frac{1}{1+x^{2}} \) from 0 to infinity.
In the context of the Student's t-distribution, integral evaluation plays a critical role because it helps determine probabilities. The curve formed by the function resembles the probability density function (pdf) used in the distribution.
Evaluating this integral involves recognizing the structure of the Student's t-distribution with one degree of freedom. This is because the integrand \( \frac{1}{1+x^{2}} \) mirrors the form of the t-distribution when the degrees of freedom, \( v \), is 1. With this particular form, the integral equates to the area under half of the curve, due to symmetry. As a result, the integral evaluates to \( 0.5 \).
Working through integral evaluations often requires identifying such transformations or simplifications, especially when dealing with well-known distributions.
In the context of the Student's t-distribution, integral evaluation plays a critical role because it helps determine probabilities. The curve formed by the function resembles the probability density function (pdf) used in the distribution.
Evaluating this integral involves recognizing the structure of the Student's t-distribution with one degree of freedom. This is because the integrand \( \frac{1}{1+x^{2}} \) mirrors the form of the t-distribution when the degrees of freedom, \( v \), is 1. With this particular form, the integral equates to the area under half of the curve, due to symmetry. As a result, the integral evaluates to \( 0.5 \).
Working through integral evaluations often requires identifying such transformations or simplifications, especially when dealing with well-known distributions.
Probability Density Function
A Probability Density Function, or pdf, is a fundamental concept in probability and statistics. It describes how the probability of a continuous random variable is distributed over its possible values.
For the Student's t-distribution, the PDF can be represented as:\[ f(t) = \frac{\Gamma(\frac{v+1}{2})}{\sqrt{v\pi} \Gamma(\frac{v}{2})}\left(1+\frac{t^2}{v}\right)^{-\frac{v+1}{2}} \]
Here:
Understanding the pdf helps in calculating the probability that a random variable falls within a certain range. In the exercise, this equation aids in recognizing the integral as part of the Student's t-distribution. The symmetry of the pdf—from negative infinity to positive infinity—means that integrating from 0 to infinity will cover half the total area, resulting in a value of 0.5.
Learning about pdfs facilitates understanding their applications in statistics and how they describe data distributions.
For the Student's t-distribution, the PDF can be represented as:\[ f(t) = \frac{\Gamma(\frac{v+1}{2})}{\sqrt{v\pi} \Gamma(\frac{v}{2})}\left(1+\frac{t^2}{v}\right)^{-\frac{v+1}{2}} \]
Here:
- \( \Gamma \) denotes the gamma function, which generalizes factorials.
- \( v \) represents the degrees of freedom, controlling the spread of the distribution.
- This function becomes relevant to our exercise when \( v = 1 \).
Understanding the pdf helps in calculating the probability that a random variable falls within a certain range. In the exercise, this equation aids in recognizing the integral as part of the Student's t-distribution. The symmetry of the pdf—from negative infinity to positive infinity—means that integrating from 0 to infinity will cover half the total area, resulting in a value of 0.5.
Learning about pdfs facilitates understanding their applications in statistics and how they describe data distributions.
Gamma Function
The Gamma Function is an extension of the factorial function to complex numbers. It plays a crucial role in various areas like probability, statistics, and complex analysis.
Denoted as \( \Gamma(n) \), the gamma function for positive integers \( n \) is defined as\( \Gamma(n) = (n-1)! \), but it extends beyond integers. It is expressed through an integral:\[ \Gamma(z) = \int_{0}^{\infty} t^{z-1} e^{-t} \, dt \]
In the context of probability distributions, particularly the Student's t-distribution, the gamma function helps calculate the normalization constant in the pdf. It ensures the total area under the distribution's curve equals 1.
In the formula for the t-distribution's pdf, the gamma function appears as part of the expression that determines how the variable is scaled and shaped. This allows different degrees of freedom to adjust the distribution's spread.
The gamma function is thus a powerful tool, linking different areas of mathematics and providing valuable insight into complex problems.
Denoted as \( \Gamma(n) \), the gamma function for positive integers \( n \) is defined as\( \Gamma(n) = (n-1)! \), but it extends beyond integers. It is expressed through an integral:\[ \Gamma(z) = \int_{0}^{\infty} t^{z-1} e^{-t} \, dt \]
In the context of probability distributions, particularly the Student's t-distribution, the gamma function helps calculate the normalization constant in the pdf. It ensures the total area under the distribution's curve equals 1.
In the formula for the t-distribution's pdf, the gamma function appears as part of the expression that determines how the variable is scaled and shaped. This allows different degrees of freedom to adjust the distribution's spread.
The gamma function is thus a powerful tool, linking different areas of mathematics and providing valuable insight into complex problems.
Other exercises in this chapter
Problem 11
If the random variable \(F\) has an \(F\) distribution with \(m\) and \(n\) degrees of freedom, show that \(1 / F\) has an \(F\) distribution with \(n\) and \(m
View solution Problem 13
Show that as \(n \rightarrow \infty\), the pdf of a Student \(t\) random variable with \(n\) df converges to \(f_{Z}(z)\). (Hint: To show that the constant term
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For a Student \(t\) random variable \(T\) with \(n\) degrees of freedom and any positive integer \(k\), show that \(E\left(T^{2 k}\right)\) exists if \(2 k0, \b
View solution Problem 20
Suppose a random sample of size \(n=11\) is drawn from a normal distribution with \(\mu=15.0\). For what value of \(k\) is the following true? $$ P\left(\left|\
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