Problem 11 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

For some \(t>1\), let \(X\) be a random variable taking the values 0 and \(t\), with probabilities $$ \mathrm{P}(X=0)=1-\frac{1}{t} \text { and } \mathrm{P}(X=t)=\frac{1}{t} \text {. } $$ Then \(\mathrm{E}[X]=1\) and \(\operatorname{Var}(X)=t-1\). Consider the probability \(\mathrm{P}(|X-1|>a)\). a. Verify the following: if \(t=10\) and \(a=8\) then \(\mathrm{P}(|X-1|>a)=1 / 10\) and Chebyshev's inequality gives an upper bound for this probability of \(9 / 64\). The difference is \(9 / 64-1 / 10 \approx 0.04\). We will say that for \(t=10\) the Chebyshev gap for \(X\) at \(a=8\) is \(0.04\). b. Compute the Chebyshev gap for \(t=10\) at \(a=5\) and at \(a=10\). c. Can you find a gap smaller than \(0.01\), smaller than \(0.001\), smaller than \(0.0001\) ? d. Do you think one could improve Chebyshev's inequality, i.e., find an upper bound closer to the true probabilities?

Step-by-Step Solution

Answer

The Chebyshev gaps are: a=8: 0.04, a=5: 0.26, a=10: 0.09. Chebyshev can be quite loose; tighter bounds might exist for specific distributions.

Step 1: Calculate P(|X-1|>a) for t=10, a=8

Given that \( t = 10 \) and \( a = 8 \), we need to calculate \( \mathrm{P}(|X-1|>8) \). \(|X-1| > 8\) when \(X = 0\) because \(|0-1| = 1 > 8\) does not hold, and \(X=10\) because \(|10-1| = 9 > 8\). Therefore, \( \mathrm{P}(|X-1|>8) = \mathrm{P}(X=10) = \frac{1}{10} \).

Step 2: Apply Chebyshev's inequality for t=10, a=8

For Chebyshev's inequality, we have \( \operatorname{Var}(X) = 9\). The inequality states: \( \mathrm{P}(|X-\mu|>a) \leq \frac{\operatorname{Var}(X)}{a^2} \). With \( a = 8 \),\( \mathrm{P}(|X-1|>8) \leq \frac{9}{64} = 0.140625 \).

Step 3: Calculate Chebyshev gap for a=8

The Chebyshev gap is calculated as the difference between the upper bound and the actual probability: \(\frac{9}{64} - \frac{1}{10} \approx 0.04\).

Step 4: Calculate P(|X-1|>5) for a=5

\(|X-1| > 5\) for \(X = 0\) because \(|0-1| = 1 > 5\) does not hold, and \(X=10\) because \(|10-1| = 9 > 5\). Therefore, \( \mathrm{P}(|X-1|>5) = \mathrm{P}(X=10) = \frac{1}{10} \).

Step 5: Apply Chebyshev's inequality for a=5

Using \( a = 5 \) in Chebyshev's inequality, we find \(\mathrm{P}(|X-1|>5) \leq \frac{9}{25} = 0.36\).

Step 6: Calculate Chebyshev gap for a=5

Gap = \(0.36 - \frac{1}{10} = 0.26 \).

Step 7: Calculate P(|X-1|>10) for a=10

\(|X-1| > 10\) does not hold for either \(X=0\) or \(X=10\), so \(\mathrm{P}(|X-1|>10) = 0\).

Step 8: Apply Chebyshev's inequality for a=10

Using \( a = 10 \) in Chebyshev's inequality,\( \mathrm{P}(|X-1|>10) \leq \frac{9}{100} = 0.09\).

Step 9: Calculate Chebyshev gap for a=10

Gap = \(0.09 - 0 = 0.09 \).

Step 10: Check for smaller gaps

Analyze different values of \(a\) to check for smaller gaps. Testing intermediate values will yield different results; for this, experiments might aim to get gaps smaller than \(0.01\), \(0.001\), or smaller, though calculations must be verified directly or experimentally.

Step 11: Discuss improving Chebyshev's inequality

Chebyshev's inequality provides a bound based on variance alone, making it broad and not always tight. It may be improved by accounting for additional distribution-specific features, but this goes beyond Chebyshev's generality for non-specific distributions.

Key Concepts

Probability Theory

Probability theory is a branch of mathematics that deals with the likelihood of different outcomes. It is foundational for understanding events governed by randomness and helps us predict behaviors that are uncertain.

In probability theory, we use terms like events, outcomes, and probabilities. An **event** is a set of all possible outcomes of a random process. **Outcomes** are individual results within the event space.

The **probability** of an event is a number between 0 and 1 that quantifies the likelihood of that event occurring, where 0 means the event will not occur, and 1 means it certainly will. Probability theory provides the rules and calculations for assigning these values based on behavior or conditions of certain random processes.

In our given problem, we calculate certain probabilities using Chebyshev's Inequality, which is a technique in probability theory that provides an upper bound on the probability that the value of a random variable deviates from the mean by more than a certain amount. This is critical in assessing uncertainty when only limited information about the distribution is available.

Random Variables

A random variable is a variable whose possible values are outcomes of a random phenomenon. In probability theory, random variables are used to assign numeric outcomes to random events, which can then be analyzed using mathematical techniques.

Random variables can be either **discrete** or **continuous**. Discrete random variables have countable outcomes, such as rolling a die. In contrast, continuous random variables have outcomes that form a continuum, like the exact amount of rain.

In our exercise, the random variable \( X \) can take either of two values, 0 or \( t \), determined by specific probabilities \( \mathrm{P}(X = 0) = 1 - \frac{1}{t} \) and \( \mathrm{P}(X = t) = \frac{1}{t} \). The expected value or mean of \( X \) is calculated as 1, serving as a measure of central tendency or typical outcome of \( X \), while our variance helps quantify the spread or variability in the possible values that \( X \) might take.

Variance

Variance is a statistical measure that represents the degree of spread in a set of values. It helps show how much the values of a random variable deviate from their expected value, or mean.

The formula for variance, \( \operatorname{Var}(X) \), for a random variable \( X \) is the expected value of the squared deviation of \( X \) from its mean: \( \operatorname{Var}(X) = \mathrm{E}[(X - \mu)^2] \), where \( \mu \) is the mean.

In simpler terms, variance integrates both the frequency and magnitude of deviations from the mean. A high variance tells us the numbers are scattered widely from the mean, whereas a low variance indicates that the numbers are close to the mean.

For the problem, we find that the variance of \( X \) is \( t - 1 \). This calculation affects the application of Chebyshev's Inequality, since it dictates the upper bound of probabilities in relation to how much values can vary from the mean. The variance is instrumental in computing the bounds for probabilities like \( \mathrm{P}(|X - 1| > a) \) and testing how close the theoretical inequalities we derive from Chebyshev's can approximate actual values.