Problem 18 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

Consider a vase containing balls numbered \(1,2, \ldots, N\). We draw \(n\) balls without replacement from the vase. Each ball is selected with equal probability, i.e., in the first draw each ball has probability \(1 / N\), in the second draw each of the \(N-1\) remaining balls has probability \(1 /(N-1)\), and so on. For \(i=1,2, \ldots, n\), let \(X_{i}\) denote the number on the ball in the \(i\) th draw. From Exercise \(9.18\) we know that the variance of \(X_{i}\) equals $$ \operatorname{Var}\left(X_{i}\right)=\frac{1}{12}(N-1)(N+1) $$ Show that $$ \operatorname{Cov}\left(X_{1}, X_{2}\right)=-\frac{1}{12}(N+1) $$ Before you do the exercise: why do you think the covariance is negative? Hint: use \(\operatorname{Var}\left(X_{1}+X_{2}+\cdots+X_{N}\right)=0\) (why?), and apply Exercise \(10.17\).

Step-by-Step Solution

Answer

The covariance is negative because the draws are dependent; knowing one ball's number affects the expectation of the others.

Step 1: Understanding Total Variance Zero

The key hint provided is that the variance of the sum of all possible draws, \(X_1 + X_2 + \cdots + X_N\), is zero. This means we'll distribute variance and covariance across these terms in such a way that they all cancel out, affirming the sum's variance is zero. Intuitively, since all outcomes are determined as the full set is drawn without replacement, knowing one ball's value gives information about the others, leading to their dependence.

Step 2: Calculation of Total Variance

Given the total variance is zero: \[\text{Var}(X_1 + X_2 + \cdots + X_N) = \sum_{i=1}^{N} \text{Var}(X_i) + 2 \sum_{1 \leq i < j \leq N} \text{Cov}(X_i, X_j) = 0\]This equation expresses that the sum of variances plus twice the sum of all covariances equals zero.

Step 3: Inserting Known Variance

From the problem, \(\text{Var}(X_i) = \frac{1}{12}(N-1)(N+1)\). Substituting this into the sum of variances we get:\[N \cdot \frac{1}{12}(N-1)(N+1)\]which simplifies to:\[\frac{N}{12}(N^2-1)\]

Step 4: Addressing Covariance Terms

We explore the covariance terms for the total variance equation:\[2 \sum_{1 \leq i < j \leq N} \text{Cov}(X_i, X_j) \equiv 2 \binom{N}{2} \cdot \text{Cov}(X_1, X_2)\]where all covariances are identical due to symmetry, leading to:\[2 \cdot \frac{N(N-1)}{2} \cdot \text{Cov}(X_1, X_2)\]This simplifies to:\[N(N-1) \cdot \text{Cov}(X_1, X_2)\]

Step 5: Solving for Covariance

From our setup, substituting into the total variance equation:\[\frac{N}{12}(N^2-1) + N(N-1) \cdot \text{Cov}(X_1, X_2) = 0\]isolate the covariance:\[N(N-1) \cdot \text{Cov}(X_1, X_2) = -\frac{N}{12}(N^2-1)\]simplifying further, we find:\[\text{Cov}(X_1, X_2) = -\frac{1}{12}(N+1)\]

Step 6: Final Conclusion

The covariance \( \text{Cov}(X_1, X_2) = -\frac{1}{12}(N+1) \) is derived from the total variance being zero, combining known variance with distributed covariance, confirming a dependent yet expected anti-correlation in draws due to without replacement scenario.

Key Concepts

Variance

Variance is a fundamental concept in probability and statistics that helps us measure how much a set of random variables differs from its expected value. In the context of the exercise, each ball drawn has a number, and the variance of the draw represents how much these numbers differ from the average.

When we calculate the variance of a draw like \( X_i \), we use the formula provided: \( \text{Var}(X_i) = \frac{1}{12}(N-1)(N+1) \). This gives us a numerical measure indicating the spread of the numbers on the drawn balls. A higher variance indicates a wider spread from the mean, while a lower variance means that the numbers are closer to the mean.

Understanding variance helps us anticipate fluctuations in our draws, and in broader contexts, can be applied to various fields beyond simple probability exercises, such as finance and quality control. The formula factors in the total number of balls \( N \) to assess how variance depends on the possible range of numbers, which is crucial for calculations like those in the original problem.

Random Variables

Random variables are essential in any statistical analysis as they define numerical outcomes of a random phenomenon. In this problem, \( X_1, X_2, \ldots, X_N \) are random variables representing the number on each of the \( N \) balls drawn from the vase.

A random variable essentially translates the outcome of a random event into a value that can be used in mathematical calculations, such as computing expectations, variances, and covariances. There are two main types of random variables: discrete and continuous, with this problem dealing with discrete random variables since we're drawing tangible balls each having a distinct number.

Each \( X_i \) adds a layer of analysis since they indicate specific outcomes of our draw. The idea is that the sequence of numbers you could draw makes probabilistic computations possible, including simulating potential outcomes and ranges. This random behavior is paramount not just in academic exercises but also in real-world scenarios where outcomes cannot be precisely predicted but are analyzed through statistical lenses.

Dependent Events

Dependent events occur when the outcome of one event affects the outcome of another. In our exercise, this concept is pivotal as we draw balls without replacement from a vase.

Since a ball is not replaced after it is drawn, the probability of drawing subsequent balls is influenced by the previous ones. This dependence is evidenced by the negative covariance derived between two draws, \( \text{Cov}(X_1, X_2) = -\frac{1}{12}(N+1) \). This covariance indicates that if one draw results in a higher number, it's likely that the subsequent draw results in a lower number, an anti-correlated relationship.

This effect is why understanding dependent events is crucial. While initial draws set the stage, each subsequent draw becomes predictable to some extent because it builds on the remaining pool of options. A beginner's understanding of dependent events can translate into better understanding systems where resource depletion affects future outcomes, such as card games or lottery draws, and helps illustrate the importance of accounting for event connections in probability.