91Ó°ÊÓ

This exercise demonstrates that, in general, the results provided by Tchebysheff's theorem cannot be improved upon. Let \(Y\) be a random variable such that $$p(-1)=\frac{1}{18}, \quad p(0)=\frac{16}{18}, \quad p(1)=\frac{1}{18}$$ a. Show that \(E(Y)=0\) and \(V(Y)=1 / 9\) b. Use the probability distribution of \(Y\) to calculate \(P(|Y-\mu| \geq 3 \sigma) .\) Compare this exact probability with the upper bound provided by Tchebysheff's theorem to see that the bound provided by Tchebysheff's theorem is actually attained when \(k=3\) *.c. In part.(b) we guaranteed \(E(Y)=0\) by placing all probability mass on the values \(-1,0,\) and1, with \(p(-1)=p(1) .\) The variance was controlled by the probabilities assigned to \(p(-1)\) and \(p(1) .\) Using this same basic idea, construct a probability distribution for a random variable \(X\) that will yield \(P\left(\left|X-\mu_{X}\right| \geq 2 \sigma_{X}\right)=1 / 4\) * d. If any \(k>1\) is specified, how can a random variable \(W\) be constructed so that \(P\left(\left|W-\mu_{W}\right| \geq k \sigma_{W}\right)=1 / k^{2} ?\)

Step by step solution

01

Calculate Expected Value E(Y)

The expected value of a random variable is given by \( E(Y) = \sum y_i \cdot p(y_i) \). For \( Y \), \( E(Y) = (-1)\cdot \frac{1}{18} + 0\cdot \frac{16}{18} + 1\cdot \frac{1}{18} = 0 \). Therefore, \( E(Y) = 0 \).

02

Calculate Variance V(Y)

The variance is given by \( V(Y) = E(Y^2) - [E(Y)]^2 \). First, calculate \( E(Y^2) = (-1)^2 \cdot \frac{1}{18} + 0^2 \cdot \frac{16}{18} + 1^2 \cdot \frac{1}{18} = \frac{2}{18} = \frac{1}{9} \). Thus, \( V(Y) = \frac{1}{9} - 0 = \frac{1}{9} \).

03

Calculate P(|Y-μ| ≥ 3σ)

Here, \( \mu = 0 \) and \( \sigma = \sqrt{V(Y)} = \frac{1}{3} \). Therefore, \( 3\sigma = 1 \) and \( P(|Y - 0| \geq 1) = P(|Y| \geq 1) = P(Y = -1) + P(Y = 1) = \frac{1}{18} + \frac{1}{18} = \frac{2}{18} = \frac{1}{9} \).

04

Tchebysheff’s Theorem Upper Bound

Tchebysheff's theorem states that \( P(|Y - \mu| \geq k\sigma) \leq \frac{1}{k^2} \). For \( k=3 \), the bound is \( \frac{1}{3^2} = \frac{1}{9} \). Thus, the bound is actually attained, as \( P(|Y-\mu| \geq 3\sigma) = \frac{1}{9} \).

05

Construct X for P(|X-μ_X| ≥ 2σ_X) = 1/4

To construct \( X \), choose \( p(-1)=p(1)=\frac{1}{8}, \), \( p(0)=\frac{6}{8} \). Then, \( E(X)=0 \) and \( V(X) = \frac{1}{4} \). The standard deviation \( \sigma_{X} = \frac{1}{2} \), thus \( 2\sigma_{X} = 1 \) and \( P(|X - 0| \geq 1) = \frac{2}{8} = \frac{1}{4} \).

06

General Construction for W for Any k>1

Given \( k > 1 \), distribute probability such that \( p(-1)=p(1)=\frac{1}{2k^2} \), \( p(0)=1-\frac{1}{k^2} \). This ensures \( E(W)=0 \) and \( V(W)=\frac{1}{k^2} \), leading to \( P(|W-\mu| \geq k\sigma_{W}) = \frac{2}{k^2} = \frac{1}{k^2} \).

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Key Concepts

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

Probability Distribution

A probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes for a random variable. In this context, we have a random variable, \( Y \), with a very specific distribution: it only takes the values -1, 0, and 1. The corresponding probabilities are \( p(-1) = \frac{1}{18} \), \( p(0) = \frac{16}{18} \), and \( p(1) = \frac{1}{18} \).
This probability distribution allows us to compute important characteristics of the random variable, such as the expected value and variance. Probability distributions are like maps for random variables, guiding us on how likely each outcome is. They are fundamental in statistic and probability theory because they help us anticipate results and understand variations in data. The characteristics of the distribution, such as symmetry or skewness, significantly impact how calculations like expected value or variance are conducted.

Random Variable

A random variable is essentially a way to encapsulate outcomes in a numeric form. In probability and statistics, random variables allow us to quantify and work through uncertain situations. For example, in the exercise, the random variable \( Y \) can take values -1, 0, or 1.
This means every time \( Y \) is observed, it will be one of these values, and the associated probability distribution offers guidance on how likely each of these values is. It abstracts real-world randomness into a form that can be analyzed mathematically. When dealing with random variables, a fundamental task is to calculate the expected value and variance which reveal further insights into the behavior of the random variable.

Expected Value

The expected value, denoted \( E(Y) \), is a central concept in probability that gives us the average or "mean" value of the random variable if we were to sample it infinitely many times. It summarizes the center of a probability distribution.
For our specific example with \( Y \), the expected value is calculated as \( E(Y) = (-1) \cdot \frac{1}{18} + 0 \cdot \frac{16}{18} + 1 \cdot \frac{1}{18} = 0 \).
This indicates that on average, over a large number of trials, the value of \( Y \) would hover around 0. The expected value is crucial because it provides a single number that captures the central tendency of a distribution and is foundational for further concepts like variance and standard deviation.

Variance

Variance is a measure that explains how much the values of a random variable vary or spread out from the expected value. It's a way of quantifying uncertainty or risk. Mathematically, it represents the expected value of the squared deviations from the mean.
For \( Y \), the variance \( V(Y) \) is calculated as \( V(Y) = E(Y^2) - [E(Y)]^2 \). We found that \( E(Y^2) = \frac{1}{9} \) and since \( E(Y) = 0 \), \( V(Y) = \frac{1}{9} - 0 = \frac{1}{9} \).
This implies that the values of \( Y \) do not deviate much from its expected value on average. Variance is important because it gives context to the expected value; while the expected value tells us where the data center lies, variance tells us about the shape and spread around that center.