1. **Problem Statement:** Given the joint probability density function (pdf) of random variables $X$ and $Y$: $$f(x,y) = \begin{cases} 1 & 0 \leq x \leq 1, 0 \leq y \leq 1 \\ 0 & \text{elsewhere} \end{cases}$$ Compute: (i) $E(X - Y)$ (ii) $E(XY)$ (iii) $E(X^2 + Y^2)$ (iv) $\mathrm{Var}(XY)$ 2. **Important Notes:** - Since $f(x,y) = 1$ over the unit square, $X$ and $Y$ are independent and uniformly distributed on $[0,1]$. - Expectation for continuous variables: $$E[g(X,Y)] = \int \int g(x,y) f(x,y) \, dx \, dy$$ - Variance formula: $$\mathrm{Var}(Z) = E(Z^2) - [E(Z)]^2$$ 3. **Calculations:** (i) $E(X - Y) = E(X) - E(Y)$ Since $X$ and $Y$ are uniform on $[0,1]$: $$E(X) = E(Y) = \frac{1}{2}$$ Therefore: $$E(X - Y) = \frac{1}{2} - \frac{1}{2} = 0$$ (ii) $E(XY)$ Because $X$ and $Y$ are independent: $$E(XY) = E(X)E(Y) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$$ (iii) $E(X^2 + Y^2) = E(X^2) + E(Y^2)$ For uniform $[0,1]$: $$E(X^2) = E(Y^2) = \int_0^1 x^2 \, dx = \frac{1}{3}$$ Thus: $$E(X^2 + Y^2) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$$ (iv) $\mathrm{Var}(XY) = E[(XY)^2] - [E(XY)]^2$ Calculate $E[(XY)^2] = E(X^2 Y^2) = E(X^2) E(Y^2)$ (independence): $$E(X^2) = E(Y^2) = \frac{1}{3}$$ So: $$E[(XY)^2] = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$$ Recall from (ii): $$E(XY) = \frac{1}{4}$$ Therefore: $$\mathrm{Var}(XY) = \frac{1}{9} - \left(\frac{1}{4}\right)^2 = \frac{1}{9} - \frac{1}{16} = \frac{16}{144} - \frac{9}{144} = \frac{7}{144}$$ **Final answers:** (i) $0$ (ii) $\frac{1}{4}$ (iii) $\frac{2}{3}$ (iv) $\frac{7}{144}$