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Magazine Chapter 4: Problem 4
Let \(X\) have a Weibull distribution with the pdf from Expression (4.11). Verify that \(\mu=\beta \Gamma(1+1 / \alpha)\). [Hint: In the integral for \(E(X)\), make the change of variable \(y=(x / \beta)^{\alpha}\), so that \(x=\beta y^{1 / \alpha}\).]
Short Answer
Expert verified
The expected value \( \mu = \beta \Gamma(1 + 1/\alpha) \) is verified through the variable substitution and integration steps.
Step by step solution
01
Understand the Problem
The problem involves verifying that the expected value (mean) of a Weibull distribution is given by \( \mu = \beta \Gamma(1+1/\alpha) \). We will use integration and transformation techniques to show this.
02
Identify the Weibull PDF
A Weibull distribution with scale parameter \( \beta \) and shape parameter \( \alpha \) has the probability density function (pdf): \( f(x) = \frac{\alpha}{\beta} \left(\frac{x}{\beta}\right)^{\alpha - 1} e^{-(x/\beta)^\alpha} \) for \( x \geq 0 \).
03
Set Up the Expectation Integral
The expected value \( E(X) \) of a continuous random variable is given by the integral \( E(X) = \int_0^{\infty} x f(x) \, dx \). Substituting the Weibull pdf, we get \( E(X) = \int_0^{\infty} x \frac{\alpha}{\beta} \left(\frac{x}{\beta}\right)^{\alpha - 1} e^{-(x/\beta)^\alpha} \, dx \).
04
Make the Substitution
Use the substitution \( y = (x/\beta)^\alpha \), which implies \( x = \beta y^{1/\alpha} \) and \( dx = \frac{\beta}{\alpha} y^{1/\alpha - 1} \, dy \). Substitute these into the integral: \( E(X) = \int_0^{\infty} \beta y^{1/\alpha} \frac{\alpha}{\beta} \beta y^{1/\alpha - 1} e^{-y} \frac{\beta}{\alpha} \, dy \).
05
Simplify the Integral
After substitution, the integral becomes \( E(X) = \int_0^{\infty} \beta y^{1/\alpha} e^{-y} \, dy \). Simplify this to \( E(X) = \beta \int_0^{\infty} y^{1/\alpha} e^{-y} \, dy \). This is the definition of the Gamma function, \( \Gamma(n) = \int_0^{\infty} y^{n-1} e^{-y} \, dy \).
06
Identify the Gamma Function
In the integral \( \int_0^{\infty} y^{1/\alpha} e^{-y} \, dy \), the expression matches the Gamma function where \( n = 1 + 1/\alpha \). Therefore, \( \int_0^{\infty} y^{1/\alpha} e^{-y} \, dy = \Gamma(1 + 1/\alpha) \).
07
Write the Final Expression for \( E(X) \)
Thus, substituting back, we have \( E(X) = \beta \Gamma(1 + 1/\alpha) \), which matches the expected value formula we wanted to verify: \( \mu = \beta \Gamma(1 + 1/\alpha) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Expected Value
The concept of expected value is fundamental to understanding probability and statistics.
It represents the average or mean value that one would anticipate receiving if an experiment or situation was repeated many times. In mathematical terms, the expected value of a continuous random variable, such as the Weibull distribution, is computed through integration.
For the Weibull distribution, the expected value is expressed using the formula: \[ E(X) = \int_0^{\infty} x f(x) \, dx \] Where \( f(x) \) defines the probability density function (pdf).
By calculating this integral, we effectively find the average outcome of the distribution.
In the problem at hand, the expected value of the Weibull distribution is given as \( \mu = \beta \Gamma(1 + 1 / \alpha) \). The parameter \( \beta \) is known as the scale parameter, while \( \alpha \) is the shape parameter. With the use of substitution and the definition of the Gamma Function, the expected value can be verified to match this formula.
It represents the average or mean value that one would anticipate receiving if an experiment or situation was repeated many times. In mathematical terms, the expected value of a continuous random variable, such as the Weibull distribution, is computed through integration.
For the Weibull distribution, the expected value is expressed using the formula: \[ E(X) = \int_0^{\infty} x f(x) \, dx \] Where \( f(x) \) defines the probability density function (pdf).
By calculating this integral, we effectively find the average outcome of the distribution.
In the problem at hand, the expected value of the Weibull distribution is given as \( \mu = \beta \Gamma(1 + 1 / \alpha) \). The parameter \( \beta \) is known as the scale parameter, while \( \alpha \) is the shape parameter. With the use of substitution and the definition of the Gamma Function, the expected value can be verified to match this formula.
Gamma Function
The Gamma function is a crucial concept in higher mathematics, generalizing the idea of factorials to non-integer values.
It is denoted by \( \Gamma(n) \) and defined as: \[ \Gamma(n) = \int_0^{\infty} y^{n-1} e^{-y} \, dy \] This formula represents an integral that computes values for factorial-like calculations within continuous variable distributions.
In the context of the Weibull distribution and the expected value formula, the Gamma function appears when simplifying the integral for \( E(X) \).
A notable property of the Gamma function is when \( n = 1 + 1/\alpha \), enabling us to relate it directly to the given expression in the problem \( \mu = \beta \Gamma(1 + 1/\alpha) \). This relationship allows the conversion from the integral form of \( E(X) \) to the simplified and calculable formula that incorporates the Gamma function.
It is denoted by \( \Gamma(n) \) and defined as: \[ \Gamma(n) = \int_0^{\infty} y^{n-1} e^{-y} \, dy \] This formula represents an integral that computes values for factorial-like calculations within continuous variable distributions.
In the context of the Weibull distribution and the expected value formula, the Gamma function appears when simplifying the integral for \( E(X) \).
A notable property of the Gamma function is when \( n = 1 + 1/\alpha \), enabling us to relate it directly to the given expression in the problem \( \mu = \beta \Gamma(1 + 1/\alpha) \). This relationship allows the conversion from the integral form of \( E(X) \) to the simplified and calculable formula that incorporates the Gamma function.
Probability Density Function
The probability density function (pdf) is an essential tool in statistics, helping to define the likelihood of a random variable within certain bounds.
For continuous variables like the one in our Weibull distribution problem, the pdf provides the necessary framework to compute probabilities over a continuous range of outcomes.
In this function, \( \alpha \) is the shape parameter controlling the distribution's form, and \( \beta \) is the scale parameter altering the width.
The behavior of this pdf is crucial for deriving other statistical properties, like the expected value.
By integrating \( x f(x) \) over its range, you can determine the mean, verify its formulation, and acquire insights into the behavior of the corresponding random variable.
The pdf not only assists in finding the expected value but also in determining other characteristics, such as variance, quantiles, and modal intervals.
For continuous variables like the one in our Weibull distribution problem, the pdf provides the necessary framework to compute probabilities over a continuous range of outcomes.
- The pdf for a Weibull distribution with parameters \( \beta \) and \( \alpha \) is given by: \[ f(x) = \frac{\alpha}{\beta} \left(\frac{x}{\beta}\right)^{\alpha - 1} e^{-(x/\beta)^\alpha} \]
In this function, \( \alpha \) is the shape parameter controlling the distribution's form, and \( \beta \) is the scale parameter altering the width.
The behavior of this pdf is crucial for deriving other statistical properties, like the expected value.
By integrating \( x f(x) \) over its range, you can determine the mean, verify its formulation, and acquire insights into the behavior of the corresponding random variable.
The pdf not only assists in finding the expected value but also in determining other characteristics, such as variance, quantiles, and modal intervals.
