Problem 45 - A First Course in Probability

Question

If \(X_{1}, X_{2}, X_{3},\) and \(X_{4}\) are (pairwise) uncorrelated random variables, each having mean 0 and variance \(1,\) compute the correlations of (a) \(X_{1}+X_{2}\) and \(X_{2}+X_{3}\) (b) \(X_{1}+X_{2}\) and \(X_{3}+X_{4}\)

Step-by-Step Solution

Answer

The correlation of \(X_1+X_2\) and \(X_2+X_3\) is \(\frac{1}{2}\), whereas the correlation of \(X_1+X_2\) and \(X_3+X_4\) is \(0\).

Step 1: (a) Covariance Calculation

First, we calculate the covariance for (a): \(\operatorname{Cov}(X_1 + X_2, X_2 + X_3) = \mathbb{E}[(X_1 + X_2) (X_2 + X_3)] - \mathbb{E}[X_1 + X_2] \mathbb{E}[X_2 + X_3]\) Since each of the random variables has mean 0, we get: \(\operatorname{Cov}(X_1 + X_2, X_2 + X_3) = \mathbb{E}[(X_1 + X_2) (X_2 + X_3)]\) Expanding the expectation, we have: \(\operatorname{Cov}(X_1 + X_2, X_2 + X_3) = \mathbb{E}[X_1 X_2 + X_1 X_3 + X_2^2 + X_2 X_3]\) Since all the random variables are pairwise uncorrelated, we have: \(\operatorname{Cov}(X_1 + X_2, X_2 + X_3) = \mathbb{E}[X_1 X_2] + \mathbb{E}[X_1 X_3] + \mathbb{E}[X_2^2] + \mathbb{E}[X_2 X_3]\) Applying the definition for pairwise uncorrelated variables, we obtain: \(\operatorname{Cov}(X_1 + X_2, X_2 + X_3) = 0 + 0 + \mathbb{E}[X_2^2] + 0 = \mathbb{E}[X_2^2]\) Since the variance of \(X_2\) is 1, we have \(\mathbb{E}[X_2^2] = 1\).

Step 2: (a) Correlation Calculation

To find the correlation for (a), we use the following formula: \(\rho(X_1 + X_2, X_2 + X_3) = \frac{\operatorname{Cov}(X_1 + X_2, X_2 + X_3)}{\sqrt{\operatorname{Var}(X_1 + X_2) \operatorname{Var}(X_2 + X_3)}}\) As \(X_1\), \(X_2\), and \(X_3\) are pairwise uncorrelated and their variance is 1, we obtain: \(\rho(X_1 + X_2, X_2 + X_3) = \frac{1}{\sqrt{(1 + 1)(1 + 1)}} = \frac{1}{2}\) Thus, the correlation between \(X_1 + X_2\) and \(X_2 + X_3\) is \(\frac{1}{2}\).

Step 3: (b) Covariance Calculation

Now, we calculate the covariance for (a): \(\operatorname{Cov}(X_1 + X_2, X_3 + X_4) = \mathbb{E}[(X_1 + X_2) (X_3 + X_4)] - \mathbb{E}[X_1 + X_2] \mathbb{E}[X_3 + X_4]\) Since each of the random variables has mean 0, we get: \(\operatorname{Cov}(X_1 + X_2, X_3 + X_4) = \mathbb{E}[(X_1 + X_2) (X_3 + X_4)]\) Expanding the expectation, we have: \(\operatorname{Cov}(X_1 + X_2, X_3 + X_4) = \mathbb{E}[X_1 X_3 + X_1 X_4 + X_2 X_3 + X_2 X_4]\) Since all the random variables are pairwise uncorrelated, we have: \(\operatorname{Cov}(X_1 + X_2, X_3 + X_4) = 0 + 0 + 0 + 0 = 0\)

Step 4: (b) Correlation Calculation

To find the correlation for (b), we use the following formula: \(\rho(X_1 + X_2, X_3 + X_4) = \frac{\operatorname{Cov}(X_1 + X_2, X_3 + X_4)}{\sqrt{\operatorname{Var}(X_1 + X_2) \operatorname{Var}(X_3 + X_4)}}\) As \(X_1\), \(X_2\), \(X_3\) and \(X_4\) are pairwise uncorrelated and their variance is 1, we obtain: \(\rho(X_1 + X_2, X_3 + X_4) = \frac{0}{\sqrt{(1 + 1)(1 + 1)}} = 0\) Thus, the correlation between \(X_1 + X_2\) and \(X_3 + X_4\) is \(0\).