Chapter 7: Problem 17
Suppose that \(X\) has distribution function $$ F(x)=1-\exp \left(-\int_{0}^{x} g(u) d u\right) $$ for some function \(g(.)\). Show that this is possible if and only if \(g(u) \geq 0\) and \(\int_{0}^{\infty} g(u) d u=\infty\).
Short Answer
Expert verified
The conditions are necessary for the CDF properties; so, \(g(u) \geq 0\) and \(\int_{0}^{\infty} g(u) du = \infty\) ensure a valid CDF.
Step by step solution
01
Understand the distribution function
The given distribution function is \(F(x)=1-\exp \left(-\int_{0}^{x} g(u) \, du\right)\). This represents the cumulative distribution function (CDF) for a random variable \(X\). A valid CDF should satisfy certain properties like being non-decreasing, right-continuous, \(F(-\infty) = 0\), and \(F(\infty) = 1\). We need to analyze these properties under the conditions for \(g(u)\).
02
Verify condition \(g(u) \geq 0\)
For the CDF \(F(x)\) to be non-decreasing, the function under the exponent, \(\int_{0}^{x} g(u) \, du\), must be non-decreasing. This requires \(g(u) \geq 0\) for almost all \(u\). Otherwise, if \(g(u) < 0\) for any \(u\), the integral will decrease, leading \(F(x)\) to not be non-decreasing.
03
Verify condition \(\int_{0}^{\infty} g(u) \, du = \infty\)
The CDF \(F(x)\) should satisfy \(F(\infty) = 1\). For this to hold, \(1 - \exp \left(-\int_{0}^{\infty} g(u) \, du\right) = 1\). This requires \(\exp \left(-\int_{0}^{\infty} g(u) \, du\right) = 0\), leading to the condition \(\int_{0}^{\infty} g(u) \, du = \infty\). If this integral were finite, \(F(\infty)\) would be less than 1, invalidating \(F(x)\) as a CDF.
04
Conclude the proof
By ensuring \(g(u) \geq 0\), the CDF \(F(x)\) remains non-decreasing. Additionally, requiring \(\int_{0}^{\infty} g(u) \, du = \infty\) ensures that \(F(x)\) approaches 1 as \(x\) approaches infinity, satisfying the fundamental properties of a CDF. Therefore, these conditions are necessary and sufficient for the distribution function \(F(x)\) to be valid.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cumulative Distribution Function (CDF)
A Cumulative Distribution Function (CDF) is a fundamental concept in probability theory. It's a function, often denoted as \( F(x) \), that gives the probability that a random variable \( X \) takes a value less than or equal to \( x \). In other words, it tells us how the probabilities accumulate as the values of \( X \) increase. This means for any given \( x \), \( F(x) \ = P(X \leq x) \).CDFs have a few key characteristics:
- They are non-decreasing, meaning as we move to the right on the \( x \)-axis, the value of \( F(x) \) does not decrease.
- They are right-continuous, meaning they do not exhibit sudden jumps when approached from the right.
- At the extreme ends, they have boundary behaviors: \( F(-\infty) = 0 \) and \( F(\\infty) = 1 \).
Non-Decreasing Function
A non-decreasing function is a function that maintains or increases its value as its input increases. This property is crucial for cumulative distribution functions (CDFs). Since a CDF represents the accumulated probability of a random variable, if it were to decrease at any point, it would imply a negative probability, which is inconsistent with probability laws.In the context of the exercise, we have the integral of \( g(u) \) within the exponent of the CDF formula:\[F(x) = 1 - \exp \left(-\int_{0}^{x} g(u) du\right)\]To ensure this function is non-decreasing:
- The integrand \( g(u) \) must be non-negative (\( g(u) \geq 0 \)). If \( g(u) \) were negative in any part, the integral \( \int_{0}^{x} g(u) \, du \) could decrease for increasing \( x \), leading to a decrease in the CDF value \( F(x) \).
Integral Divergence
Integral divergence, in this context, refers to the behavior of an improper integral which diverges instead of reaching a finite limit. For the Cumulative Distribution Function (CDF) to satisfy \( F(\infiy) = 1 \), it requires that the exponent becomes so large that \( \exp \left(-\int_{0}^{\finity} g(u) \, du\right) \) approaches zero.This is achieved when:
- The integral \( \int_{0}^{\finity} g(u) du \) diverges to infinity. This divergence ensures that the exponential term approaches zero, making \( 1 - \exp \left(-\int_{0}^{\finity} g(u) du\right) = 1 \).