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91Ó°ÊÓ

Show that the cdf \(F(x)\) is a nondecreasing function; that is, \(x_{1}

Short Answer

Expert verified
A CDF is nondecreasing because \(F(x_1) \leq F(x_2)\) for \(x_1 < x_2\). Equality holds when there's no probability between \(x_1\) and \(x_2\).

Step by step solution

01

Understand the Definition of CDF

The cumulative distribution function (CDF) for a random variable \(X\) is defined by \(F(x) = P(X \leq x)\). It gives the probability that the random variable is less than or equal to \(x\).
02

State the Nondecreasing Property

To show that \(F(x)\) is nondecreasing, we need to show that for any two points \(x_1\) and \(x_2\) such that \(x_1 < x_2\), it holds that \(F(x_1) \leq F(x_2)\).
03

Use Probability Inequality

For any \(x_1 < x_2\), notice that the event \(X \leq x_1\) is a subset of the event \(X \leq x_2\). This implies \(P(X \leq x_1) \leq P(X \leq x_2)\) based on the monotonicity of probabilities. Hence, \(F(x_1) \leq F(x_2)\).
04

Determine the Condition for Equality

The condition for \(F(x_1) = F(x_2)\) occurs when \(P(x_1 < X \leq x_2) = 0\). This happens when there are no values of \(X\) in the interval \((x_1, x_2]\), indicating that the random variable has no mass or probability density between \(x_1\) and \(x_2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nondecreasing Function
In the realm of probability and statistics, a nondecreasing function is a critical concept, especially for the cumulative distribution function (CDF). A function, such as the CDF, is considered nondecreasing if it maintains its value or increases as the input increases, but never decreases. The idea here is simple: as we move to higher values on the x-axis, the value of the function should not drop.
\[F(x_1) \leq F(x_2) \text{ whenever } x_1 < x_2\]
This for the CDF means that as we observe larger values of the random variable, the probability of the variable being less than or equal to that value must cover a broader range, and never retract.
For the CDF, the nondecreasing property is intrinsic. It forms a building block in understanding how probabilities accumulate over possible values of a random variable.
Probability Inequality
Probability inequality is a powerful tool used to prove various properties of probability functions and distributions. It essentially deals with comparing probabilities of different events. When dealing with cumulative distribution functions or probabilities tied to order of events, understanding probability inequalities is key.
When we say the CDF is nondecreasing, what we imply is:
  • The event \(X \leq x_1\) is logically a subset of the event \(X \leq x_2\) whenever \(x_1 < x_2\).
  • This relation, \(P(X \leq x_1) \leq P(X \leq x_2)\), is the backbone of the probability inequality leveraged to show the nondecreasing property.
Simply put, if one event is encompassed within another, the probability of the broader event (larger bound) should naturally be greater or equal to the narrower one. This is also why the CDF is always nondecreasing.
Monotonicity of Probabilities
Monotonicity of probabilities relates to how probabilities evolve as conditions change by expansion of events. In the context of cumulative distribution functions, it means as our domain grows, the probability or mass it accumulates must never shrink. The monotonicity answers important questions about how sure we can be of events as our understanding window expands.
If two points, \(x_1\) and \(x_2\) lie on the real line such that \(x_1 < x_2\), any values the random variable \(X\) can take up to \(x_1\) will naturally be included in the values up to \(x_2\).
Accordingly,
  • the CDF at \(x_1\) must not exceed the CDF at \(x_2\),
  • and mathematically, this translates to \(F(x_1) \leq F(x_2)\).
The monotonicity is a direct result of the foundational probability principles and ensures the CDF reflects a consistent and logical accumulation of probability over its range.

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Most popular questions from this chapter

If the sample space \(S\) is an infinite set, does this necessarily imply that any rv \(X\) defined from \(\rho\) will have an infinite set of possible values? If yes, say why. If no, give an example.

The \(n\) candidates for a job have been ranked \(1,2,3, \ldots, n\). Let \(X=\) the rank of a randomly selected candidate, so that \(X\) has pmf $$ p(x)= \begin{cases}1 / n & x=1,2,3, \ldots, n \\ 0 & \text { otherwise }\end{cases} $$ (this is called the discrete uniform distribution). Compute \(E(X)\) and \(V(X)\) using the shortcut formula. [Hint: The sum of the first \(n\) positive integers is \(n(n+1) / 2\), whereas the sum of their squares is \(n(n+1)(2 n+1) / 6 .]\)

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The article "Expectation Analysis of the Probability of Failure for Water Supply Pipes" (J. of Pipeline Systems Engr. and Practice, May 2012: 36-46) proposed using the Poisson distribution to model the number of failures in pipelines of various types. Suppose that for cast-iron pipe of a particular length, the expected number of failures is 1 (very close to one of the cases considered in the article). Then \(X\), the number of failures, has a Poisson distribution with \(\mu=1\). a. Obtain \(P(X \leq 5)\) by using Appendix Table A.2. b. Determine \(P(X=2)\) first from the pmf formula and then from Appendix Table A.2. c. Determine \(P(2 \leq X \leq 4)\). d. What is the probability that \(X\) exceeds its mean value by more than one standard deviation?

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