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Chapter 9: Problem 18

Let \(X\) be a r.v. with mean \(\mu\) and variance \(\sigma^{2}\), both finite. Show that $$ P\\{\mu-d \sigma1\).)

Short Answer

Expert verified
Use Chebyshev's inequality to show \(P(\mu - d\sigma < X < \mu + d\sigma) \geq 1 - \frac{1}{d^2}\).

Step by step solution

01

Understanding the Problem

We are asked to show that the probability of a random variable \(X\) being within \(d\) standard deviations of the mean \(\mu\) is at least \(1-\frac{1}{d^{2}}\). This is a result derived from Chebyshev's inequality.
02

Chebyshev's Inequality Statement

Recall Chebyshev's inequality which states that for any random variable \(X\) with mean \(\mu\) and variance \(\sigma^2\), the probability that \(X\) deviates from \(\mu\) by more than \(k\sigma\) is at most \(\frac{1}{k^2}\). That is: \[ P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2}. \]
03

Applying Chebyshev's Inequality

To find the probability that \(X\) is within \(d\) standard deviations of \(\mu\), we use the complement of Chebyshev's result: \[ P(|X - \mu| < d\sigma) = 1 - P(|X - \mu| \geq d\sigma) \geq 1 - \frac{1}{d^2}. \]
04

Rewriting the Probability

The probability \(P(|X - \mu| < d\sigma)\) can be expressed as \(P(\mu - d\sigma < X < \mu + d\sigma)\), which matches the given condition in the problem statement.
05

Conclusion

Thus, by applying Chebyshev鈥檚 inequality, we have shown that \[ P(\mu - d\sigma < X < \mu + d\sigma) \geq 1 - \frac{1}{d^2} \] for \(d > 1\), which concludes our solution. This confirms that \(X\) is within \(d\) standard deviations of the mean with at least the stated probability.

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Key Concepts

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

Random Variable
A random variable is a variable whose value depends on the outcome of a random event. It represents numerical outcomes as a result of probabilistic experiments. There are two main types of random variables: discrete and continuous.
  • Discrete random variables take on a countable number of values. For instance, rolling a die can result in one of six outcomes.
  • Continuous random variables can take on any value within a range. An example would be the exact time it takes for a computer to complete a task.
A random variable is typically denoted by a capital letter, like \(X\). It plays a central role in probability theory, offering a way to model uncertainty and variability in various contexts. Understanding random variables is crucial for learning about probability distributions, which describe how the values of a random variable are distributed.
Mean and Variance
Mean and variance are two key statistical measures that describe different properties of a random variable. The mean, often represented by \(\mu\), is the average or expected value of a random variable. It provides a measure of the central tendency of the variable's probability distribution.
  • The mean is calculated by taking the sum of all possible values of the random variable, each multiplied by its respective probability, for discrete cases.
  • For continuous random variables, it is the integral of the variable multiplied by its probability density function.
Variance, represented by \(\sigma^2\), measures the spread or dispersion of the random variable's values around the mean. It is the average of the squared differences from the mean.
  • Variance provides insights into the variability within the data.
  • A small variance indicates that the data points are close to the mean, while a large variance indicates more spread out data points.
Both mean and variance are foundational concepts for understanding more complex ideas like the probability distribution and standard deviation.
Probability Distribution
A probability distribution describes how the values of a random variable are distributed. It provides a complete description of the likelihood of different outcomes. There are several types of probability distributions, each suited to different types of data.
  • Discrete probability distributions, like the binomial distribution, are used for variables that take on a finite number of values.
  • Continuous probability distributions, such as the normal distribution, apply to variables that can take on any value within a range.
Each probability distribution has its own unique characteristics and parameters. For instance, the normal distribution is defined by its mean and standard deviation. This is useful because it allows us to calculate probabilities and understand the behavior of the random variable under specific conditions. Probability distributions are foundational to statistical analysis and serve as the underpinning for various statistical methods, including hypothesis testing and regression analysis.
Standard Deviation
Standard deviation is a statistical measure that quantifies the amount of variation or dispersion in a set of values. It is the square root of the variance, denoted as \(\sigma\). Standard deviation is essential because it expressively illustrates how much the values in a data set deviate from the mean.
  • A small standard deviation indicates that the data points are generally close to the mean.
  • A large standard deviation suggests that there is a greater spread of data around the mean.
This measure is crucial for understanding and applying Chebyshev's Inequality, which provides bounds on how probable it is that a random variable deviates from its mean. In practical terms, standard deviation allows for assessing the risk or volatility in fields ranging from finance to engineering. Understanding standard deviation can also facilitate better decision-making by anticipating potential variations within a probability distribution.

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Most popular questions from this chapter

Given \((\Omega, \mathcal{A}, P)\), suppose \(X\) is a r.v. with \(X \geq 0\) a.s. and \(E\\{X\\}=1\). Define \(Q: \mathcal{A} \rightarrow \mathbf{R}\) by \(Q(A)=E\left\\{X 1_{A}\right\\}\). Show that \(Q\) defines a probability measure on \((\Omega, \mathcal{A})\).

Let \(X:(\Omega, \mathcal{A}) \rightarrow(\mathbf{R}, \mathcal{B})\) be a r.v. Let $$ \mathcal{F}=\left\\{A: A=X^{-1}(B), \text { some } B \in \mathcal{B}\right\\}=X^{-1}(\mathcal{B}) $$ Show that \(X\) is measurable as a function from \((\Omega, \mathcal{F})\) to \((\mathbf{R}, \mathcal{B})\).

Let \(X\) be uniform over \((a, b)\). Show that \(E\\{X\\}=\frac{a+b}{2}\).

The gamma distribution is often simplified to a one parameter distribution. A r.v. \(X\) is said to have the standard gamma distribution with parameter \(\alpha\) if the density of its distribution measure is given by $$ f(x)= \begin{cases}\frac{x^{\alpha-1} e^{-x}}{\Gamma(\alpha)} & \text { if } x \geq 0 \\ 0 & \text { if } x<0\end{cases} $$ That is, \(\beta=1\). (Recall \(\Gamma(\alpha)=\int_{0}^{\infty} t^{\alpha-1} e^{-t} d t\).) Show that for \(X\) standard gamma with parameter \(\alpha\), then $$ E\left\\{X^{k}\right\\}=\frac{\Gamma(\alpha+k)}{\Gamma(\alpha)} \quad(k \geq 0) $$ Deduce that \(X\) has mean \(\alpha\) and also variance \(\alpha\).

Let \(X\) be normal (or Gaussian) with parameters \(\mu=0\) and \(\sigma^{2}=1\). Show that \(P(X>x) \leq \frac{1}{x \sqrt{2 \pi}} e^{-\frac{1}{2} x^{2}}\), for \(x>0\).
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