Explanation

The covariance between two random variables $X_1$ and $X_2$ is given by:

$\text{Cov}(X_1, X_2) = E(X_1 X_2) - E(X_1) \cdot E(X_2)$

To compute this, we need to find:

1. $E(X_1 X_2)$ - the expected value of the product
2. $E(X_1)$ - the expected value of $X_1`$3`. $E(X_2)$ - the expected value of $X_2$

Step 1: Verify if the function is a valid joint PDF

First, let's check if the given function integrates to 1 over the domain:

$\int_{0}^{1} \int_{0}^{4\sqrt{2}} \frac{1}{8} x_1 x_2 \, dx_2 \, dx_1$

$= \frac{1}{8} \int_{0}^{1} x_1 \left[ \frac{x_2^2}{2} \right]_{0}^{4\sqrt{2}} dx_1$

$= \frac{1}{8} \int_{0}^{1} x_1 \cdot \frac{(4\sqrt{2})^2}{2} dx_1$

$= \frac{1}{8} \int_{0}^{1} x_1 \cdot \frac{32}{2} dx_1$

$= \frac{1}{8} \int_{0}^{1} 16x_1 dx_1$

$= 2 \int_{0}^{1} x_1 dx_1 = 2 \cdot \frac{1}{2} = 1$

The function is indeed a valid joint PDF.

Step 2: Compute $E(X_1 X_2)$

$E(X_1 X_2) = \int_{0}^{1} \int_{0}^{4\sqrt{2}} x_1 x_2 \cdot \frac{1}{8} x_1 x_2 \, dx_2 \, dx_1$

$= \frac{1}{8} \int_{0}^{1} \int_{0}^{4\sqrt{2}} x_1^2 x_2^2 \, dx_2 \, dx_1$

$= \frac{1}{8} \int_{0}^{1} x_1^2 \left[ \frac{x_2^3}{3} \right]_{0}^{4\sqrt{2}} dx_1$

$= \frac{1}{8} \int_{0}^{1} x_1^2 \cdot \frac{(4\sqrt{2})^3}{3} dx_1$

$= \frac{1}{8} \int_{0}^{1} x_1^2 \cdot \frac{128\sqrt{2}}{3} dx_1$

$= \frac{16\sqrt{2}}{3} \int_{0}^{1} x_1^2 dx_1$

$= \frac{16\sqrt{2}}{3} \cdot \frac{1}{3} = \frac{16\sqrt{2}}{9}$

Step 3: Compute $E(X_1)$

$E(X_1) = \int_{0}^{1} \int_{0}^{4\sqrt{2}} x_1 \cdot \frac{1}{8} x_1 x_2 \, dx_2 \, dx_1$

$= \frac{1}{8} \int_{0}^{1} \int_{0}^{4\sqrt{2}} x_1^2 x_2 \, dx_2 \, dx_1$

$= \frac{1}{8} \int_{0}^{1} x_1^2 \left[ \frac{x_2^2}{2} \right]_{0}^{4\sqrt{2}} dx_1$

$= \frac{1}{8} \int_{0}^{1} x_1^2 \cdot \frac{(4\sqrt{2})^2}{2} dx_1$

$= \frac{1}{8} \int_{0}^{1} x_1^2 \cdot \frac{32}{2} dx_1$

$= \frac{1}{8} \int_{0}^{1} 16x_1^2 dx_1$

$= 2 \int_{0}^{1} x_1^2 dx_1 = 2 \cdot \frac{1}{3} = \frac{2}{3}$

Step 4: Compute $E(X_2)$

$E(X_2) = \int_{0}^{1} \int_{0}^{4\sqrt{2}} x_2 \cdot \frac{1}{8} x_1 x_2 \, dx_2 \, dx_1$

$= \frac{1}{8} \int_{0}^{1} \int_{0}^{4\sqrt{2}} x_1 x_2^2 \, dx_2 \, dx_1$

$= \frac{1}{8} \int_{0}^{1} x_1 \left[ \frac{x_2^3}{3} \right]_{0}^{4\sqrt{2}} dx_1$

$= \frac{1}{8} \int_{0}^{1} x_1 \cdot \frac{(4\sqrt{2})^3}{3} dx_1$

$= \frac{1}{8} \int_{0}^{1} x_1 \cdot \frac{128\sqrt{2}}{3} dx_1$

$= \frac{16\sqrt{2}}{3} \int_{0}^{1} x_1 dx_1$

$= \frac{16\sqrt{2}}{3} \cdot \frac{1}{2} = \frac{8\sqrt{2}}{3}$

Step 5: Compute Covariance

$\text{Cov}(X_1, X_2) = E(X_1 X_2) - E(X_1) \cdot E(X_2)$

$= \frac{16\sqrt{2}}{9} - \frac{2}{3} \cdot \frac{8\sqrt{2}}{3}$

$= \frac{16\sqrt{2}}{9} - \frac{16\sqrt{2}}{9} = 0$

Therefore, the covariance between the interest rate and the stock market return is 0.

Key Insight: The covariance is zero because the joint PDF can be factored as $f(x_1, x_2) = \frac{1}{8}x_1x_2 = \left(\frac{1}{2}x_1\right) \cdot \left(\frac{1}{4}x_2\right)$ over the rectangular domain, indicating that $X_1$ and $X_2$ are independent. When two random variables are independent, their covariance is always zero.