Problem 39 - A First Course in Probability

Question

Let \(X_{1}, \ldots\) be independent with common mean \(\mu\) and common variance \(\sigma^{2},\) and set \(Y_{n}=X_{n}+\) \(X_{n+1}+X_{n+2} .\) For \(j \geq 0,\) find \(\operatorname{Cov}\left(Y_{n}, Y_{n+j}\right)\)

Step-by-Step Solution

Answer

The covariance between \(Y_n\) and \(Y_{n+j}\) is given by: \(\operatorname{Cov}(Y_n, Y_{n+j})=\begin{cases} 3\sigma^2 & \text{ if } j = 0\\ 2\sigma^2 & \text{ if } j = 1\\ 0 & \text{ if } j \geq 2 \end{cases}\)

Step 1: Define Variables for \(Y_n\) and \(Y_{n+j}\)

First, let's define the variables for \(Y_n\) and \(Y_{n + j}\) more clearly: \(Y_n = X_n + X_{n + 1} + X_{n + 2}\) \(Y_{n + j} = X_{n + j} + X_{n + j + 1} + X_{n + j + 2}\)

Step 2: Compute the Covariance Using the Main Formula

To find the covariance between \(Y_n\) and \(Y_{n+j}\), we use the main formula: \(\operatorname{Cov}(Y_n, Y_{n + j}) = E(Y_nY_{n + j}) - E(Y_n)E(Y_{n + j})\)

Step 3: Expand the Expected Values Using the Definitions of Yn and (Yn+j)

We can now replace \(Y_n\) and \(Y_{n+j}\) in the above formula using the definitions obtained in Step 1: \(E(\left(X_n + X_{n + 1} + X_{n + 2}\right)\left(X_{n + j} + X_{n + j + 1} + X_{n + j + 2}\right)) - E(X_n + X_{n + 1} + X_{n + 2})E(X_{n + j} + X_{n + j + 1} + X_{n + j + 2})\)

Step 4: Simplify Using Properties of Expected Values and Covariances

Now, we must simplify the expression obtained using the properties of expected values and covariances for independent variables. For example, we'll use the facts that \(E(X_i X_j) = E(X_i)E(X_j)\) and \(\operatorname{Cov}(X_i, X_j) = 0\) for \(i \neq j\). The result will depend on the value of \(j\), so let's consider three cases:

Step 5: Case 1: \(j = 0\)

\(\operatorname{Cov}(Y_n, Y_n) = \operatorname{Var}(Y_n)\) \(= \operatorname{Var}(X_n + X_{n + 1} + X_{n + 2})\) \(= \operatorname{Var}(X_n) + \operatorname{Var}(X_{n + 1}) + \operatorname{Var}(X_{n + 2})\) \(= \sigma^2 + \sigma^2 + \sigma^2 = 3\sigma^2\)

Step 6: Case 2: \(j = 1\)

\(\operatorname{Cov}(Y_n, Y_{n + 1}) = \operatorname{Cov}(X_{n} + X_{n + 1} + X_{n + 2},X_{n + 1} + X_{n + 2} + X_{n + 3})\) \(= \operatorname{Cov}(X_{n}, X_{n + 1}) + \operatorname{Cov}(X_{n}, X_{n + 2}) + \cdots + \operatorname{Cov}(X_{n + 2}, X_{n + 3})\) \(= 0 + \sigma^2 + \sigma^2 + 0 = 2\sigma^2\)

Step 7: Case 3: \(j \geq 2\)

\(\operatorname{Cov}(Y_n, Y_{n + j}) = \operatorname{Cov}(X_{n} + X_{n + 1} + X_{n + 2},X_{n + j} + X_{n + j + 1} + X_{n + j + 2})\) There is no overlap of indices greater than or equal to 2, hence: \(\operatorname{Cov}(Y_n, Y_{n + j}) = 0\)

Step 8: Final Answer

So, the covariance between \(Y_n\) and \(Y_{n+j}\) is as follows: \(\operatorname{Cov}(Y_n, Y_{n+j})=\begin{cases} 3\sigma^2 & \text{ if } j = 0\\ 2\sigma^2 & \text{ if } j = 1\\ 0 & \text{ if } j \geq 2 \end{cases}\)

Key Concepts

Covariance calculation

Covariance measures the degree to which two random variables change together. If the variables tend to show similar behavior, the covariance is positive, while if one variable tends to increase when the other decreases, the covariance is negative. A zero covariance indicates no linear relationship between the variables.

To calculate covariance, you use the formula \[\begin{equation} \operatorname{Cov}(X, Y) = E[(X - E[X])(Y - E[Y])]. \end{equation}\]

This formula can be expanded as \[\begin{equation} \operatorname{Cov}(X, Y) = E[XY] - E[X]E[Y], \end{equation}\]

where E[X] and E[Y] are the expected values of X and Y, respectively. In the provided exercise, the covariance between particular sums of random variables is sought, highlighting how this concept is used to measure the relationship between complex combinations of variables.

Independent random variables

Independent random variables are a pair of variables where the occurrence of one does not affect the probability of the occurrence of the other. In other words, the knowledge of one variable’s outcome provides no information about the other’s outcome.

One of the key properties of independent variables is that the covariance between them is equal to zero: \[\begin{equation} \operatorname{Cov}(X_i, X_j) = 0 \text{ for } i eq j, \end{equation}\]

if the variables are independent. This property simplifies the computation of covariance for sums of independent random variables, as seen in the case with variables \( Y_n \) and \( Y_{n+j} \). When the variables are independent, the variance of their sum is simply the sum of their variances, used in the solution to compute the final answer.

Expected value properties

The expected value, or mean, of a random variable is a measure of its central tendency. The expected value has several properties that make it a fundamental tool in probability and statistics:

• \( E[aX + bY] = aE[X] + bE[Y] \) - The expected value is linear, allowing us to combine and scale variables easily.
• If X and Y are independent, then \( E[XY] = E[X]E[Y] \).
• The expected value of a constant is the constant itself: \( E[c] = c \).

These properties are utilized in the step-by-step solution when evaluating the expected values of the sums of the individual terms of \(Y_n\) and \(Y_{n+j}\).

Variance properties

Variance is a measure of how much a set of values are spread out from their average value. Variance has several properties that are particularly useful when dealing with multiple random variables:

• The variance of a sum of independent random variables is the sum of their variances: \( \operatorname{Var}(X+Y) = \operatorname{Var}(X) + \operatorname{Var}(Y) \) if X and Y are independent.
• \( \operatorname{Var}(aX) = a^2 \operatorname{Var}(X) \), which indicates variance is scaled by the square of the constant term when the variable is multiplied by a constant.
• The variance of a constant is zero: \( \operatorname{Var}(c) = 0 \).

These properties are illustrated in the exercise's solution. For instance, the computation of the variance of \(Y_n\) as \(3\sigma^2\) uses the fact that the variance of the sum of independent variables is the sum of their variances.