Problem 8
Question
Let \(X\) have an \(\operatorname{Exp}(1)\) distribution, and let \(\alpha\) and \(\lambda\) be positive numbers. Determine the distribution function of the random variable $$ W=\frac{X^{1 / \alpha}}{\lambda} $$ The distribution of the random variable \(W\) is called the Weibull distribution with parameters \(\alpha\) and \(\lambda\).
Step-by-Step Solution
Verified Answer
The distribution function of \(W\) is \(F_W(w) = 1 - e^{-(\lambda w)^\alpha}\), which is the CDF of the Weibull distribution.
1Step 1: Define the Distribution of X
The random variable \(X\) with an \(\operatorname{Exp}(1)\) distribution has the probability density function (pdf) given by \(f_X(x) = e^{-x}\) for \(x \geq 0\). The cumulative distribution function (CDF) is \(F_X(x) = 1 - e^{-x}\).
2Step 2: Define the Transformation
We are given the transformation \(W = \frac{X^{1/\alpha}}{\lambda}\). We need to find the CDF of \(W\), denoted as \(F_W(w)\).
3Step 3: Express CDF of W in Terms of X
The CDF of \(W\) is given by \(F_W(w) = P(W \leq w) = P\left( \frac{X^{1/\alpha}}{\lambda} \leq w \right)\). Simplifying, we have \(X^{1/\alpha} \leq \lambda w\).
4Step 4: Solve the Inequality
The inequality \(X^{1/\alpha} \leq \lambda w\) implies \(X \leq (\lambda w)^\alpha\). Thus, \(F_W(w) = P(X \leq (\lambda w)^\alpha)\).
5Step 5: Apply the Distribution of X
Since \(X\) has an Exponential(1) distribution, use its CDF to find \(F_W(w) = 1 - e^{-(\lambda w)^\alpha}\). This gives us the CDF of the Weibull distribution with parameters \(\alpha\) and \(\lambda\).
6Step 6: Conclusion
The distribution function of the random variable \(W\) is given by \(F_W(w) = 1 - e^{-(\lambda w)^\alpha}\). This matches the CDF of the Weibull distribution with parameters \(\alpha\) and \(\lambda\).
Key Concepts
Exponential DistributionWeibull DistributionCumulative Distribution Function
Exponential Distribution
The exponential distribution is one of the simplest and most widely used probability distributions. It's frequently used to model time until an event happens, like how long you'll wait for your coffee. The time until failure of mechanical systems, like light bulbs, is another classic example.
Here are some key points about the Exponential Distribution:
Here are some key points about the Exponential Distribution:
- The probability density function (pdf) is denoted by \( f_X(x) = e^{-x} \) for \( x \geq 0 \). This shows the rapidly decreasing likelihood as x increases, which is characteristic of exponential decay.
- The cumulative distribution function (CDF), which provides the probability of the random variable being less than a specific value, is given by \( F_X(x) = 1 - e^{-x} \). This means as time (x) increases, the probability of the event occurring approaches 1, simplifying the analysis of time to failure.
- The exponential distribution is determined by a single parameter, often presented as the rate \( \lambda \), when \( X \sim \operatorname{Exp}(\lambda) \). In our problem, \( \lambda = 1 \), simplifying calculations.
Weibull Distribution
The Weibull distribution generalizes the exponential distribution by introducing a shape parameter \( \alpha \), offering greater versatility. Because of its flexibility, it's employed in a wide variety of fields. Let's delve into how it works and why it's so useful.
Here's what makes the Weibull Distribution special:
Here's what makes the Weibull Distribution special:
- It can model various types of data depending on the shape parameter \( \alpha \). For example, when \( \alpha = 1 \), the Weibull distribution resembles the exponential distribution.
- The inclusion of the parameter \( \lambda \), called the scale parameter, adjusts the spread of the distribution. By transforming \( X \) as \( W = \frac{X^{1/\alpha}}{\lambda} \), we effectively scale the data and modify its shape.
- The CDF of the Weibull distribution with parameters \( \alpha \) and \( \lambda \) is given as \( F_W(w) = 1 - e^{-(\lambda w)^\alpha} \), which controls both the scaling and the shape of data.
Cumulative Distribution Function
The cumulative distribution function (CDF) is a fundamental concept in probability and statistics. It provides a comprehensive way to understand the probability distribution of a random variable.
Here’s why the CDF is important:
Here’s why the CDF is important:
- The CDF of a random variable \( X \), such as \( F_X(x) \), shows the probability that \( X \) will take a value less than or equal to \( x \). It aggregates probabilities from the initial value up to \( x \).
- For continuous distributions, the CDF is an integral of the pdf, essentially summing up area under the curve of the probability density function until \( x \).
- In the case of a Weibull distribution, the CDF \( F_W(w) = 1 - e^{-(\lambda w)^\alpha} \) highlights how likely it is for the modeled event to occur by a certain time \( w \). This function offers insights into failure times, among others.
Other exercises in this chapter
Problem 5
Let \(X\) be a continuous random variable with probability density function $$ f_{X}(x)= \begin{cases}\frac{3}{4} x(2-x) & \text { for } 0 \leq x \leq 2 \\\ 0 &
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Let \(X\) be a continuous random variable with probability density \(f_{X}\) that takes only positive values and let \(Y=1 / X\). a. Determine \(F_{Y}(y)\) and
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Let \(X\) be a continuous random variable. Express the distribution function and probability density of the random variable \(Y=-X\) in terms of those of \(X\).
View solution Problem 11
Let \(X\) be a random variable, and let \(g\) be a twice differentiable function with \(g^{\prime \prime}(x) \leq 0\) for all \(x\). Such a function is called a
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