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Chapter 12: Problem 5

Let \(X\) be uniformly distributed over \((-2,2) .\) Use Chebyshev's inequality to estimate \(P(|X| \geq 1)\), and compare your estimate with the exact answer.

Short Answer

Expert verified
Chebyshev's inequality estimates \(P(|X| \geq 1) \leq \frac{4}{3}\), whereas the exact probability is \(\frac{1}{2}\).

Step by step solution

01

Understanding Uniform Distribution

The variable \(X\) is uniformly distributed over the interval \((-2, 2)\). This means that \(X\) has a constant probability density function within this interval. The range of \(X\) is \(b-a = 2 - (-2) = 4\).
02

Compute Mean and Variance of Uniform Distribution

For a uniform distribution over \((a, b)\), the mean \(\mu\) is \(\frac{a+b}{2}\) and the variance \(\sigma^2\) is \(\frac{(b-a)^2}{12}\). Here, \(a = -2\) and \(b = 2\), so \(\mu = 0\) and \(\sigma^2 = \frac{4^2}{12} = \frac{16}{12} = \frac{4}{3}\).
03

Apply Chebyshev's Inequality

Chebyshev's inequality states that for any \(k > 0\), \(P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2}\). Here, we are finding \(P(|X| \geq 1)\), which is \(P(|X - 0| \geq 1)\). So, \(k = \frac{1}{\sigma} = \sqrt{\frac{3}{4}}\), and substituting we find the bound \(\frac{4}{3}\).
04

Calculate Exact Probability

The probability \(P(|X| \geq 1)\) can be found by integrating the PDF over the regions where \(X < -1\) and \(X > 1\). Each of these segments has length \(1\), so total length is \(2\), out of total \(4\). Therefore, \(P(|X| \geq 1) = \frac{2}{4} = \frac{1}{2}\).
05

Compare Chebyshev's Estimate to Exact Answer

Chebyshev's inequality gives a bound of \(\frac{4}{3} = 1.33\), which is greater than the literal probability \(\frac{1}{2} = 0.5\). Since Chebyshev's inequality provides an upper bound, it is expected to be equal to or greater than the exact probability.

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Key Concepts

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

Uniform Distribution
Uniform distribution is a type of probability distribution where every outcome in a continuous range is equally likely. In simple terms, if you imagine a block, the block has the same thickness at every point of its horizontal stretch. For this exercise, the variable \( X \) is uniformly distributed over the interval \((-2, 2)\). The length of this interval, or range, is the difference between the maximum and minimum values, which is \(4\).
  • All values within an interval are equally probable.
  • Probabilities are described with a constant probability density function across the range.
  • Interval length (or range) is calculated by subtracting the smallest value from the largest.
Understanding uniform distribution is crucial when calculating mean and variance, which are essential components for applying other statistical methods, like Chebyshev's inequality.
Probability
Probability measures how likely an event is to occur. In the context of a uniform distribution, probability can be visualized by the area under the probability density function (PDF) curve. For \(P(|X| \geq 1)\), the probability is found by calculating the areas where \(X\) is not within \(-1\) and \(1\).
  • Probability measures likelihood, represented numerically between \(0\) and \(1\).
  • For events within a specified interval, probabilities can be visualized as areas on a graph.
  • Uniform distribution makes calculations straightforward as each interval part holds equal likelihood.
By integrating the PDF over segments and understanding the uniform layout, students can calculate exact probabilities for given ranges, as performed in the solution.
Mean and Variance
Mean and variance form the backbone of statistical analysis, capturing central tendency and variability. For a uniform distribution over the interval \((a, b)\), the mean \(\mu\) is the midpoint, calculated as \((a + b)/2\). The variance \(\sigma^2\) measures how spread out the values are, given by \((b - a)^2/12\).
  • Mean (\(\mu\)): The central point of the distribution.
  • Variance (\(\sigma^2\)): Reflects spread or dispersion of distribution.
  • Calculation formulas vary based on the distribution type.
In this case, mean \(\mu = 0\) and variance \(\sigma^2 = 4/3\) are derived, which are essential for further statistical applications. These values are utilized within Chebyshev's inequality to estimate probabilities for given conditions.

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