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Magazine Chapter 7: Problem 18
Suppose a function \(F\) is given by $$ F(x)=\sum_{i=1}^{\infty} \frac{1}{2^{i}} 1_{\left[\frac{1}{i}, \infty\right)} . $$ Show that it is the distribution function of a probability on \(\mathbf{R}\). Let us define \(P\) by \(P((-\infty, x])=F(x)\). Find the probabilities of the following events: a) \(A=[1, \infty)\) b) \(B=\left[\frac{1}{10}, \infty\right)\) c) \(C=\\{0\\}\) d) \(D=\left[0, \frac{1}{2}\right)\) e) \(E=(-\infty, 0)\) f) \(G=(0, \infty)\)
Short Answer
Expert verified
P(A)=1, P(B)=0.998, P(C)=0, P(D)=0.5, P(E)=0, P(G)=1.
Step by step solution
01
Understand the Definition of Function F
The function \( F(x) \) is defined as an infinite series sum. It involves an indicator function \( 1_{[\frac{1}{i}, \infty)}(x) \), which is equal to 1 if \( x \) is in the interval \([\frac{1}{i}, \infty)\) and 0 otherwise. This creates a series where each term is either \( \frac{1}{2^i} \) or 0 depending on \( x \).
02
Determine F(x) as a Distribution Function
To show that \( F(x) \) is a distribution function, we need to verify that it is non-decreasing, right continuous, and \( \lim_{x \to +\infty} F(x) = 1 \) and \( \lim_{x \to -\infty} F(x) = 0 \).\[ F(x) \] is a series of non-negative terms, making it non-decreasing. For right continuity, observe continuity points occur at values \( x = \frac{1}{i} \) where indicators turn on and change the sum. As \( x \to +\infty \), the series approaches 0 summing, and as \( x \to -\infty \), all terms in the series vanish. Thus, all requirements are satisfied.
03
Define Probability P Based on F
The function \( P((a, b]) = F(b) - F(a) \) defines the probability for any interval \( (a, b] \). We'll use this setup to find probabilities for specified events (intervals).
04
Calculate Probability for Event A
For \( A = [1, \infty) \), we have each term \( 1_{[\frac{1}{i}, \infty)}(1) = 1 \) for \( i = 1 \) or even greater. Hence, the sum becomes \( \sum_{i=1}^{\infty} \frac{1}{2^i} = 1 \). Thus, \( P(A) = F(1) = 1 \).
05
Calculate Probability for Event B
For \( B = \left[ \frac{1}{10}, \infty \right) \), terms for \( i = 1 \) through \( i = 10 \) contribute, so \( F\left( \frac{1}{10} \right) = \sum_{i=1}^{10} \frac{1}{2^i}\). Compute and find the sum: \( \frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{1024} \).
06
Calculate Probability for Event C
Here \( C = \{ 0 \} \). Since \( 0 \) does not belong to any interval \([\frac{1}{i}, \infty)\), the sum contributes 0 for all terms. Hence, \( P(C) = F(0) = 0 \).
07
Calculate Probability for Event D
For \( D = \left[ 0, \frac{1}{2} \right)\), only terms \( i \geq 2 \) matter starting at \( i=2 \), so \( F\left( \frac{1}{2} \right) = \sum_{i=2}^{\infty} \frac{1}{2^i}\). Compute: \( \frac{1}{4} + \frac{1}{8} + \ldots \), which results to \( \frac{1}{2} \).
08
Calculate Probability for Event E
For the interval \( E = (-\infty, 0) \), no indicator functions from the series describe it true, hence \( F(x) = 0 \). Therefore, \( P(E) = 0\).
09
Calculate Probability for Event G
For \( G = (0, \infty) \), this is the entire set of the series representing valid indicators, equivalent to a cumulative complete sum, namely 1. Thus, \( P(G) = 1 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Probability Theory
Probability Theory is a mathematical framework used to quantify uncertainty. It helps us understand and predict outcomes in different scenarios. In our exercise, we deal with a probability distribution function, which represents the probabilities of all possible outcomes for a random variable.
- A probability distribution function, like the one in our problem, takes inputs (in this case, the variable \(x\)) and outputs corresponding probabilities.
- The function \(F(x)\) is a particular kind of probability distribution known as a cumulative distribution function (CDF). It shows the probability that a random variable is less than or equal to \(x\).
Indicator Function
An indicator function is a simple yet crucial concept in mathematics and probability theory. It helps determine whether certain conditions are met.
- In our problem, the indicator function \(1_{[\frac{1}{i}, \infty)}(x)\) is used within the infinite series. It evaluates to 1 if \(x\) is greater than or equal to \(\frac{1}{i}\), and 0 otherwise.
- This function is instrumental in building the probability distribution function, ensuring that the series only adds terms when \(x\) fits within the specified intervals.
- Thus, the indicator function acts like a filter, letting through only those values that contribute positively to the probability measure as per the condition \([\frac{1}{i}, \infty)\).
Infinite Series
An infinite series is a sum of infinitely many terms. It's a common concept in mathematics and plays a pivotal role in our exercise solution.
- The series \(\sum_{i=1}^{\infty} \frac{1}{2^{i}} 1_{[\frac{1}{i}, \infty)}\) is used to define our function \(F(x)\).
- Each term in this series is either \(\frac{1}{2^i}\) or 0, depending on whether the indicator function evaluates to 1 or 0.
- Infinite series can converge to finite values, which is crucial in our context, as the series converges to 1 as \(x\) approaches infinity, fulfilling one of the properties of a distribution function.
Distribution Function
The distribution function, especially the cumulative distribution function (CDF), is vital in probability theory.
- It provides a complete description of the probability distribution of a real-valued random variable.
- In our problem, \(F(x)\) is a CDF, representing the probability that a random number from the distribution is less than or equal to \(x\).
- To be a valid CDF, \(F(x)\) must satisfy several conditions: it must be non-decreasing (never decreases as \(x\) increases), right-continuous (does not jump abruptly for small changes in \(x\)), and must approach 0 as \(x\) approaches negative infinity, and 1 as \(x\) approaches positive infinity.
