1. **Problem Statement:** Given the joint density function $$f(x,y) = x + y$$ for $$0 < x \leq 1$$ and $$0 < y < 1$$, and zero elsewhere, we need to find: (i) $$E(X)$$ (ii) $$E(Y)$$ (iii) $$E(XY)$$ (iv) Determine if $$X$$ and $$Y$$ are independent. Also, find $$P(X + Y \leq 1)$$. --- 2. **Check if $$f(x,y)$$ is a valid joint density:** We verify $$\int_0^1 \int_0^1 (x + y) \, dy \, dx = 1$$. Calculate inner integral: $$\int_0^1 (x + y) \, dy = \int_0^1 x \, dy + \int_0^1 y \, dy = x(1 - 0) + \frac{1^2}{2} = x + \frac{1}{2}$$ Now outer integral: $$\int_0^1 \left(x + \frac{1}{2}\right) dx = \int_0^1 x \, dx + \int_0^1 \frac{1}{2} \, dx = \frac{1}{2} + \frac{1}{2} = 1$$ So, $$f(x,y)$$ is a valid joint density. --- 3. **Find $$E(X)$$:** Formula: $$E(X) = \int \int x f(x,y) \, dy \, dx$$ Calculate: $$E(X) = \int_0^1 \int_0^1 x (x + y) \, dy \, dx = \int_0^1 \int_0^1 (x^2 + xy) \, dy \, dx$$ Inner integral: $$\int_0^1 (x^2 + xy) \, dy = x^2 \int_0^1 dy + x \int_0^1 y \, dy = x^2 (1) + x \frac{1}{2} = x^2 + \frac{x}{2}$$ Outer integral: $$\int_0^1 \left(x^2 + \frac{x}{2}\right) dx = \int_0^1 x^2 dx + \frac{1}{2} \int_0^1 x dx = \frac{1}{3} + \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{3} + \frac{1}{4} = \frac{7}{12}$$ --- 4. **Find $$E(Y)$$:** Formula: $$E(Y) = \int \int y f(x,y) \, dy \, dx$$ Calculate: $$E(Y) = \int_0^1 \int_0^1 y (x + y) \, dy \, dx = \int_0^1 \int_0^1 (xy + y^2) \, dy \, dx$$ Inner integral: $$\int_0^1 (xy + y^2) \, dy = x \int_0^1 y \, dy + \int_0^1 y^2 \, dy = x \frac{1}{2} + \frac{1}{3} = \frac{x}{2} + \frac{1}{3}$$ Outer integral: $$\int_0^1 \left(\frac{x}{2} + \frac{1}{3}\right) dx = \frac{1}{2} \int_0^1 x dx + \frac{1}{3} \int_0^1 dx = \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{3} \cdot 1 = \frac{1}{4} + \frac{1}{3} = \frac{7}{12}$$ --- 5. **Find $$E(XY)$$:** Formula: $$E(XY) = \int \int xy f(x,y) \, dy \, dx$$ Calculate: $$E(XY) = \int_0^1 \int_0^1 xy (x + y) \, dy \, dx = \int_0^1 \int_0^1 (x^2 y + x y^2) \, dy \, dx$$ Inner integral: $$\int_0^1 (x^2 y + x y^2) \, dy = x^2 \int_0^1 y \, dy + x \int_0^1 y^2 \, dy = x^2 \frac{1}{2} + x \frac{1}{3} = \frac{x^2}{2} + \frac{x}{3}$$ Outer integral: $$\int_0^1 \left(\frac{x^2}{2} + \frac{x}{3}\right) dx = \frac{1}{2} \int_0^1 x^2 dx + \frac{1}{3} \int_0^1 x dx = \frac{1}{2} \cdot \frac{1}{3} + \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}$$ --- 6. **Check independence:** If $$X$$ and $$Y$$ are independent, then $$f(x,y) = f_X(x) f_Y(y)$$. Find marginal densities: $$f_X(x) = \int_0^1 (x + y) dy = x + \frac{1}{2}$$ $$f_Y(y) = \int_0^1 (x + y) dx = y + \frac{1}{2}$$ Check product: $$f_X(x) f_Y(y) = \left(x + \frac{1}{2}\right) \left(y + \frac{1}{2}\right) = xy + \frac{x}{2} + \frac{y}{2} + \frac{1}{4}$$ This is not equal to $$f(x,y) = x + y$$. Therefore, $$X$$ and $$Y$$ are **not independent**. --- 7. **Find $$P(X + Y \leq 1)$$:** Calculate: $$P(X + Y \leq 1) = \int_0^1 \int_0^{1 - x} (x + y) \, dy \, dx$$ Inner integral: $$\int_0^{1 - x} (x + y) dy = x(1 - x) + \frac{(1 - x)^2}{2} = x - x^2 + \frac{(1 - 2x + x^2)}{2} = x - x^2 + \frac{1}{2} - x + \frac{x^2}{2} = \frac{1}{2} - \frac{x^2}{2}$$ Outer integral: $$\int_0^1 \left(\frac{1}{2} - \frac{x^2}{2}\right) dx = \frac{1}{2} \int_0^1 (1 - x^2) dx = \frac{1}{2} \left(1 - \frac{1}{3}\right) = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}$$ --- **Final answers:** (i) $$E(X) = \frac{7}{12}$$ (ii) $$E(Y) = \frac{7}{12}$$ (iii) $$E(XY) = \frac{1}{3}$$ (iv) $$X$$ and $$Y$$ are **not independent**. $$P(X + Y \leq 1) = \frac{1}{3}$$