Problem 48
Question
The time between forest fires is often modeled using a Weibull distribution. According to this distribution the likelihood that a time \(t\) elapses between the end of one forest fire and the start of the next one is proportional to \(p(t)=\) \(t^{k-1} \exp \left(-t^{k}\right)\) where \(k\) is a positive constant. It can be shown using the laws of probability (see Chapter 12) that the probability of a second forest fire starting at any time following the first is proportional to \(\int_{0}^{\infty} p(t) d t\). (a) Assuming \(k=1\), show that \(\int_{0}^{\infty} p(t) d t\) is convergent and calculate the value of the integral. (b) Now assume \(k=2\). Show that \(\int_{0}^{\infty} p(t) d t\) is convergent and calculate the value of the integral (Hint: you will need to use integration by substitution). (c) Now adapt your argument from (b) to show that the integral is convergent for any value of \(k\) obeying \(k \geq 1\).
Step-by-Step Solution
Verified Answer
The integral \( \int_{0}^{\infty} p(t) \, dt \) converges to 1 for \( k=1 \), \( \frac{1}{2} \) for \( k=2 \), and \( \frac{1}{k} \) for \( k \geq 1 \).
1Step 1: Understanding the Problem
We are asked to evaluate the convergence of the integral \( \int_{0}^{\infty} p(t) \, dt \) where \( p(t) = t^{k-1} \exp(-t^k) \). In parts (a) and (b), specific values for \( k \) (i.e., 1 and 2) are considered. In part (c), we explore the general case \( k \geq 1 \).
2Step 2: Evaluate the Integral for k=1
Assume \( k = 1 \). The integral becomes \( \int_{0}^{\infty} \exp(-t) \, dt \). Recognize this as the standard exponential integral. Evaluate it by calculating \( -[\exp(-t)]_{0}^{\infty} = -(0 - 1) = 1 \). Thus, the integral is convergent and equals 1.
3Step 3: Utilize Integration by Substitution for k=2
Assume \( k = 2 \). Consider the integral: \( \int_{0}^{\infty} t \exp(-t^2) \, dt \). Let \( u = t^2 \), so \( du = 2t \, dt \) or \( t \, dt = \frac{1}{2} du \). Substitute to get \( \int_{0}^{\infty} \frac{1}{2} \exp(-u) \, du = \frac{1}{2} [\exp(-u)]_{0}^{\infty} = \frac{1}{2}(0 - (-1)) = \frac{1}{2} \). The integral converges to \( \frac{1}{2} \).
4Step 4: General Case for k ≥ 1
For any \( k \geq 1 \), analyze \( p(t) = t^{k-1} \exp(-t^k) \) through substitution. Let \( u = t^k \), then \( du = k t^{k-1} \, dt \), leading to \( \int_{0}^{\infty} \frac{1}{k} \exp(-u) \, du \). This simplifies to \( \frac{1}{k} \int_{0}^{\infty} \exp(-u) \, du = \frac{1}{k} \), ensuring convergence for any \( k \geq 1 \).
Key Concepts
Convergence of IntegralsProbabilityIntegration by Substitution
Convergence of Integrals
When dealing with integrals extending to infinity, it's crucial to understand when they converge. Convergence means that as the upper limit of an integral approaches infinity, the value of the integral settles to a specific number. This concept is pivotal in probability and statistical applications where the total probability must be finite.
For the given problem, the function, which is part of the Weibull distribution, is expressed as \( p(t) = t^{k-1} \exp(-t^k) \). The challenge is to determine the convergence of the integral \( \int_{0}^{\infty} p(t) \, dt \).
In part (a) of the original exercise, for \( k=1 \), the integrand simplifies to an exponential integral \( \exp(-t) \), which converges to 1. This tells us that this probability distribution is well-defined over the set of all positive real numbers for \( k=1 \).
When \( k=2 \), the integral is \( \int_{0}^{\infty} t \exp(-t^2) \, dt \). Here, integration by substitution is used to show convergence, demonstrating an expected result as for \( k=1 \). The substitution ultimately simplifies the integral and guarantees convergence.
For the general case \( k \geq 1 \), understanding that the integral \( \int_{0}^{\infty} t^{k-1} \exp(-t^k) \, dt \) remains convergent is essential. Using a substitution method, it shows that the integral converges to \( \frac{1}{k} \), confirming that regardless of the choice of \( k \), if \( k \geq 1 \), the total probability becomes finite.
For the given problem, the function, which is part of the Weibull distribution, is expressed as \( p(t) = t^{k-1} \exp(-t^k) \). The challenge is to determine the convergence of the integral \( \int_{0}^{\infty} p(t) \, dt \).
In part (a) of the original exercise, for \( k=1 \), the integrand simplifies to an exponential integral \( \exp(-t) \), which converges to 1. This tells us that this probability distribution is well-defined over the set of all positive real numbers for \( k=1 \).
When \( k=2 \), the integral is \( \int_{0}^{\infty} t \exp(-t^2) \, dt \). Here, integration by substitution is used to show convergence, demonstrating an expected result as for \( k=1 \). The substitution ultimately simplifies the integral and guarantees convergence.
For the general case \( k \geq 1 \), understanding that the integral \( \int_{0}^{\infty} t^{k-1} \exp(-t^k) \, dt \) remains convergent is essential. Using a substitution method, it shows that the integral converges to \( \frac{1}{k} \), confirming that regardless of the choice of \( k \), if \( k \geq 1 \), the total probability becomes finite.
Probability
Probability in the context of integrals and distributions showcases the likelihood of events occurring within specific intervals. For continuous variables, the probability of an event occurring within an interval is represented by an integral of a probability density function.
The Weibull distribution, a versatile tool in modeling times between events like forest fires, is defined by the density function \( p(t) = t^{k-1} \exp(-t^k) \). Its application in probability allows us to compute the likelihood of a second event, or fire in this case, occurring at any particular time frame since the last event.
The integral \( \int_{0}^{\infty} p(t) \, dt \) over the entire domain gives us the total probability, which must be equal to 1 for it to be a valid probability distribution. This ensures that we account for all possible outcomes or times since the last event. When \( k=1 \), it intuitively confirms that the probability of the next event, happening after a randomly chosen wait time, sums to a complete probability distribution.
When analyzing such distributions, the convergence of the integral confirms that for the model to be useful, the total probability on the chosen interval must amount to exactly 1.
The Weibull distribution, a versatile tool in modeling times between events like forest fires, is defined by the density function \( p(t) = t^{k-1} \exp(-t^k) \). Its application in probability allows us to compute the likelihood of a second event, or fire in this case, occurring at any particular time frame since the last event.
The integral \( \int_{0}^{\infty} p(t) \, dt \) over the entire domain gives us the total probability, which must be equal to 1 for it to be a valid probability distribution. This ensures that we account for all possible outcomes or times since the last event. When \( k=1 \), it intuitively confirms that the probability of the next event, happening after a randomly chosen wait time, sums to a complete probability distribution.
When analyzing such distributions, the convergence of the integral confirms that for the model to be useful, the total probability on the chosen interval must amount to exactly 1.
Integration by Substitution
Integration by substitution is a mathematical technique used to simplify the process of finding antiderivatives, much like factoring simplifies algebraic expressions. It is especially helpful when dealing with integrals that are difficult to manage directly.
In the case of the Weibull distribution for \( k=2 \), the integral \( \int_{0}^{\infty} t \exp(-t^2) \, dt \) is evaluated using this method. By letting \( u = t^2 \), the variable substitution significantly modifies the bounds and the integrand's expression to something more manageable.
This substitution confirms that calculations retain their rigorous results, ensuring the interminable sum of probabilities maintains its total at unity for each challenge posed by varying \( k \). This technique not only simplifies solving the integral but also verifies it adheres to the expectations of probability distribution computations.
In the case of the Weibull distribution for \( k=2 \), the integral \( \int_{0}^{\infty} t \exp(-t^2) \, dt \) is evaluated using this method. By letting \( u = t^2 \), the variable substitution significantly modifies the bounds and the integrand's expression to something more manageable.
- The differential, \( du = 2t \, dt \), suggests \( t \, dt = \frac{1}{2} du \).
- After substitution, the integral becomes \( \frac{1}{2} \int_{0}^{\infty} \exp(-u) \, du \), which is easily solved.
This substitution confirms that calculations retain their rigorous results, ensuring the interminable sum of probabilities maintains its total at unity for each challenge posed by varying \( k \). This technique not only simplifies solving the integral but also verifies it adheres to the expectations of probability distribution computations.
Other exercises in this chapter
Problem 47
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