1. **Problem Statement:** Calculate the covariance $\sigma_{xy}$ and correlation coefficient $\rho_{xy}$ for discrete random variables $X$ and $Y$ given their joint PMF. 2. **Given Data:** Joint PMF table: $$\begin{array}{c|ccc|c} X \backslash Y & 0 & 1 & 2 & f_1(x) \\ \hline 0 & 0.10 & 0.10 & 0.20 & 0.40 \\ 1 & 0 & 0.15 & 0.05 & 0.20 \\ 3 & 0.10 & 0.20 & 0.10 & 0.40 \\ f_2(y) & 0.20 & 0.45 & 0.35 & 1 \end{array}$$ 3. **Step 1: Calculate $E(X)$ and $E(Y)$** $$E(X) = \sum_x x f_1(x) = 0 \times 0.40 + 1 \times 0.20 + 3 \times 0.40 = 0 + 0.20 + 1.20 = 1.40$$ $$E(Y) = \sum_y y f_2(y) = 0 \times 0.20 + 1 \times 0.45 + 2 \times 0.35 = 0 + 0.45 + 0.70 = 1.15$$ 4. **Step 2: Calculate $E(XY)$** $$E(XY) = \sum_x \sum_y x y p(x,y)$$ Calculate each term: - For $x=0$: $0 \times (0 \times 0.10 + 1 \times 0.10 + 2 \times 0.20) = 0$ - For $x=1$: $1 \times (0 \times 0 + 1 \times 0.15 + 2 \times 0.05) = 1 \times (0 + 0.15 + 0.10) = 0.25$ - For $x=3$: $3 \times (0 \times 0.10 + 1 \times 0.20 + 2 \times 0.10) = 3 \times (0 + 0.20 + 0.20) = 3 \times 0.40 = 1.20$ Sum all: $E(XY) = 0 + 0.25 + 1.20 = 1.45$ 5. **Step 3: Calculate covariance $\sigma_{xy}$** Formula: $$\sigma_{xy} = E(XY) - E(X)E(Y)$$ Substitute values: $$\sigma_{xy} = 1.45 - (1.40)(1.15) = 1.45 - 1.61 = -0.16$$ 6. **Step 4: Calculate variances $\sigma_x^2$ and $\sigma_y^2$** Calculate $E(X^2)$: $$E(X^2) = \sum_x x^2 f_1(x) = 0^2 \times 0.40 + 1^2 \times 0.20 + 3^2 \times 0.40 = 0 + 0.20 + 3.6 = 3.80$$ Calculate $\sigma_x^2$: $$\sigma_x^2 = E(X^2) - [E(X)]^2 = 3.80 - (1.40)^2 = 3.80 - 1.96 = 1.84$$ Calculate $E(Y^2)$: $$E(Y^2) = \sum_y y^2 f_2(y) = 0^2 \times 0.20 + 1^2 \times 0.45 + 2^2 \times 0.35 = 0 + 0.45 + 1.40 = 1.85$$ Calculate $\sigma_y^2$: $$\sigma_y^2 = E(Y^2) - [E(Y)]^2 = 1.85 - (1.15)^2 = 1.85 - 1.3225 = 0.5275$$ 7. **Step 5: Calculate correlation coefficient $\rho_{xy}$** Formula: $$\rho_{xy} = \frac{\sigma_{xy}}{\sigma_x \sigma_y}$$ Calculate standard deviations: $$\sigma_x = \sqrt{1.84} \approx 1.356$$ $$\sigma_y = \sqrt{0.5275} \approx 0.726$$ Calculate $\rho_{xy}$: $$\rho_{xy} = \frac{-0.16}{1.356 \times 0.726} = \frac{-0.16}{0.984} \approx -0.1625$$ 8. **Interpretation:** The correlation coefficient $\rho_{xy} \approx -0.16$ indicates a weak negative linear relationship between $X$ and $Y$. This means as $X$ increases, $Y$ tends to decrease slightly, but the relationship is not strong. **Final answers:** - Covariance $\sigma_{xy} = -0.16$ - Correlation coefficient $\rho_{xy} \approx -0.16$