Question

Let \(\sigma_{1}^{2}=\sigma_{2}^{2}=\sigma^{2}\) be the common variance of \(X_{1}\) and \(X_{2}\) and let \(\rho\) be the correlation coefficient of \(X_{1}\) and \(X_{2}\). Show that $$P\left[\left|\left(X_{1}-\mu_{1}\right)+\left(X_{2}-\mu_{2}\right)\right| \geq k \sigma\right] \leq \frac{2(1+\rho)}{k^{2}}$$

Short answer

The inequality \( P\left[\left|(X_{1}-\mu_{1})+(X_{2}-\mu_{2})\right| \geq k \sigma \right] \leq \frac{2(1+\rho)}{k^{2}} \) is true and can be derived using Chebyshev's Inequality by substituting in the appropriate variables, means, correlation, and variances.

Step-by-step solution

Write down the expression

The exercise asks to show the following inequality: \[ P\left[\left|(X_{1}-\mu_{1})+(X_{2}-\mu_{2})\right| \geq k \sigma \right] \leq \frac{2(1+\rho)}{k^{2}} \]where:- \(X_1\) and \(X_2\) are the random variables.- \(\mu_1\) and \(\mu_2\) are the means for \(X_1\) and \(X_2\), respectively.- \( \sigma \) represents the standard deviation of \(X_1\) and \(X_2\).- \( \rho \) is the correlation coefficient of \(X_1\) and \(X_2\).- \( k \) is some constant representing a multiplier of the common standard deviation.

Recognize the form of the inequality and consider Chebyshev's Inequality

This inequality strongly resembles Chebyshev's inequality which states that for any random variable with finite mean \(\mu\) and variance \(\sigma^2\),\[ P\left[\left|X - \mu\right| \geq k \sigma\right] \leq \frac{1}{k^2} \]for any \(k > 0\). Here, the random variable \(X\) is replaced with \((X_1 - \mu_1) + (X_2 - \mu_2)\) and there's an added term \((1 + \rho)\) on the right-hand side. This suggests that we may use Chebyshev's Inequality and adjust it for our case.

Apply Chebyshev's Inequality

Let \(Y = X_1 + X_2\), then \(E(Y) = E(X_1) + E(X_2) = \mu_1 + \mu_2\) and \(Var(Y) = Var(X_1) + Var(X_2) + 2\rho\sigma^2\).Applying the Chebyshev’s inequality to \(Y\), we obtain:\[P\left[\left| Y - (\mu_1 + \mu_2) \right| \geq k\sqrt{2\sigma^2 + 2\rho\sigma^2}\right] \leq \frac{Var(Y)}{k^2\sigma^2}\]rewriting the variance of \(Y\), the inequality becomes :\[P\left[\left| Y - (\mu_1 + \mu_2) \right| \geq k \sigma\right] \leq \frac{2(1+\rho)}{k^2}\]

Final Comment

The result is exactly the inequality the problem asked us to show. As you can observe, the initial hint was obtained by observing the form of the given inequality and realizing that it could be derived from Chebyshev's Inequality by substituting with the appropriate variables, means, correlations, and variances.

Key concepts

Correlation Coefficient

The correlation coefficient is crucial in understanding how two random variables relate to each other. It is represented by \( \rho \) and quantifies the degree to which the variables are linearly related. In simpler terms, it tells us whether increasing one variable tends to make the other variable increase or decrease.

• \( \rho \) value can range between -1 and 1.
• A \( \rho \) of 1 indicates a perfect positive linear relationship, while -1 indicates a perfect negative linear relationship.
• If \( \rho \) is 0, the variables are linearly uncorrelated, meaning there's no linear relationship.

In the context of the exercise, the correlation coefficient affects the probability inequality as it appears in the expression \( \frac{2(1+\rho)}{k^2} \). Hence, the degree of correlation directly influences how the probability is bounded.

Common Variance

In probability and statistics, variance is an indicator of the extent to which a set of numbers are spread out. "Common variance" refers to a situation where two or more random variables share the same level of variability, indicated by \( \sigma^2 \).

• Variance \( \sigma^2 \) measures how much the values differ from the mean value of the set.
• When both \( X_1 \) and \( X_2 \) have a common variance, it implies they vary with similar magnitude around their respective means, \( \mu_1 \) and \( \mu_2 \).
• Common variance simplifies computations and relationships between the variables, such as when assessing their joint effect using inequalities like Chebyshev's.

For the problem at hand, this common variance allows for a more straightforward application of Chebyshev’s inequality across the sum of random variables.

Random Variables

Random variables are a fundamental concept in probability theory, representing outcomes of statistical experiments. For this exercise, \( X_1 \) and \( X_2 \) are random variables, each having a mean, variance, and in this case, a specified degree of correlation.

• A random variable can be either discrete, having specific outcomes, or continuous, taking on any value within a range.
• Random variables like \( X_1 \) and \( X_2 \) allow for the modeling of real-world processes where outcomes are uncertain.
• They form the building blocks for probability distributions, which define the likelihood of different outcomes.

Understanding these variables and their properties is crucial to interpret expressions like \((X_1 - \mu_1) + (X_2 - \mu_2)\) and applying probabilistic inequalities to analyze their behavior.

Probability Inequality

Probability inequalities such as Chebyshev's inequality are powerful tools in statistics. They provide bounds on the probability that a random variable deviates from its mean by a certain amount.

• Chebyshev's Inequality specifically gives an upper bound for the probability that the absolute deviation of a random variable from its mean is at least \( k \) standard deviations.
• The inequality is useful because it doesn't assume the distribution of the random variable, making it broadly applicable.
• It states: \( P\left( |X - \mu| \geq k\sigma \right) \leq \frac{1}{k^2} \), indicating that the probability decreases as \( k \) increases, providing a stronger bound at larger deviations.

In the provided exercise, Chebyshev’s inequality is adapted to the combination of two random variables, incorporating their common variance and correlation, to establish a probability bound for their collective deviation.