1. **Problem Statement:** Given the joint pdf of continuous random variables $X$ and $Y$: $$f(x,y) = \frac{1}{8}(x+y), \quad 0 < x < 2, 0 < y < 2; \quad 0 \text{ otherwise}$$ Find: (i) Marginal densities $f_1(x)$ and $f_2(y)$. (ii) Whether $X$ and $Y$ are independent. (iii) Covariance $\sigma_{XY}$. (iv) Correlation coefficient $\rho_{XY}$. 2. **Marginal Densities:** The marginal density of $X$ is: $$f_1(x) = \int_0^2 f(x,y) dy = \int_0^2 \frac{1}{8}(x+y) dy$$ Calculate: $$f_1(x) = \frac{1}{8} \left( x \int_0^2 dy + \int_0^2 y dy \right) = \frac{1}{8} \left( x \cdot 2 + \frac{2^2}{2} \right) = \frac{1}{8} (2x + 2) = \frac{x+1}{4}$$ for $0 < x < 2$. Similarly, marginal density of $Y$: $$f_2(y) = \int_0^2 f(x,y) dx = \int_0^2 \frac{1}{8}(x+y) dx = \frac{1}{8} \left( y \int_0^2 dx + \int_0^2 x dx \right) = \frac{1}{8} \left( y \cdot 2 + \frac{2^2}{2} \right) = \frac{y+1}{4}$$ for $0 < y < 2$. 3. **Independence Check:** Two variables are independent if: $$f(x,y) = f_1(x) f_2(y)$$ Check: $$f_1(x) f_2(y) = \frac{x+1}{4} \cdot \frac{y+1}{4} = \frac{(x+1)(y+1)}{16}$$ Given joint pdf: $$f(x,y) = \frac{x+y}{8}$$ Since $\frac{x+y}{8} \neq \frac{(x+1)(y+1)}{16}$ for all $x,y$, $X$ and $Y$ are **not independent**. 4. **Covariance $\sigma_{XY}$:** Formula: $$\sigma_{XY} = E[XY] - E[X]E[Y]$$ Calculate $E[X]$: $$E[X] = \int_0^2 x f_1(x) dx = \int_0^2 x \frac{x+1}{4} dx = \frac{1}{4} \int_0^2 (x^2 + x) dx = \frac{1}{4} \left( \frac{2^3}{3} + \frac{2^2}{2} \right) = \frac{1}{4} \left( \frac{8}{3} + 2 \right) = \frac{1}{4} \cdot \frac{14}{3} = \frac{14}{12} = \frac{7}{6}$$ Calculate $E[Y]$ similarly: $$E[Y] = \int_0^2 y f_2(y) dy = \frac{7}{6}$$ Calculate $E[XY]$: $$E[XY] = \int_0^2 \int_0^2 xy f(x,y) dx dy = \int_0^2 \int_0^2 xy \frac{x+y}{8} dx dy = \frac{1}{8} \int_0^2 \int_0^2 (x^2 y + x y^2) dx dy$$ Calculate inner integrals: $$\int_0^2 x^2 y dx = y \int_0^2 x^2 dx = y \cdot \frac{8}{3}$$ $$\int_0^2 x y^2 dx = y^2 \int_0^2 x dx = y^2 \cdot 2$$ So, $$E[XY] = \frac{1}{8} \int_0^2 \left( y \frac{8}{3} + 2 y^2 \right) dy = \frac{1}{8} \left( \frac{8}{3} \int_0^2 y dy + 2 \int_0^2 y^2 dy \right) = \frac{1}{8} \left( \frac{8}{3} \cdot 2 + 2 \cdot \frac{8}{3} \right) = \frac{1}{8} \left( \frac{16}{3} + \frac{16}{3} \right) = \frac{1}{8} \cdot \frac{32}{3} = \frac{4}{3}$$ Finally, $$\sigma_{XY} = E[XY] - E[X]E[Y] = \frac{4}{3} - \frac{7}{6} \cdot \frac{7}{6} = \frac{4}{3} - \frac{49}{36} = \frac{48}{36} - \frac{49}{36} = -\frac{1}{36}$$ 5. **Correlation Coefficient $\rho_{XY}$:** Formula: $$\rho_{XY} = \frac{\sigma_{XY}}{\sigma_X \sigma_Y}$$ Calculate variances: $$\sigma_X^2 = E[X^2] - (E[X])^2$$ Calculate $E[X^2]$: $$E[X^2] = \int_0^2 x^2 f_1(x) dx = \int_0^2 x^2 \frac{x+1}{4} dx = \frac{1}{4} \int_0^2 (x^3 + x^2) dx = \frac{1}{4} \left( \frac{2^4}{4} + \frac{2^3}{3} \right) = \frac{1}{4} \left( 4 + \frac{8}{3} \right) = \frac{1}{4} \cdot \frac{20}{3} = \frac{5}{3}$$ So, $$\sigma_X^2 = \frac{5}{3} - \left( \frac{7}{6} \right)^2 = \frac{5}{3} - \frac{49}{36} = \frac{60}{36} - \frac{49}{36} = \frac{11}{36}$$ Similarly, $\sigma_Y^2 = \frac{11}{36}$. Therefore, $$\rho_{XY} = \frac{-\frac{1}{36}}{\sqrt{\frac{11}{36}} \sqrt{\frac{11}{36}}} = \frac{-\frac{1}{36}}{\frac{11}{36}} = -\frac{1}{11}$$ **Final answers:** - $f_1(x) = \frac{x+1}{4}$ for $0 < x < 2$ - $f_2(y) = \frac{y+1}{4}$ for $0 < y < 2$ - $X$ and $Y$ are **not independent** - $\sigma_{XY} = -\frac{1}{36}$ - $\rho_{XY} = -\frac{1}{11}$