1. **Problem Statement:** Given the joint probability density function (pdf) for $(X_1, X_2)$ with $-1 \leq \alpha \leq 1$: $$f(x_1, x_2) = \begin{cases} [1 - \alpha(1 - 2e^{-x_1})(1 - 2e^{-x_2})] e^{-(x_1 + x_2)}, & x_1 \geq 0, x_2 \geq 0 \\ 0, & \text{elsewhere} \end{cases}$$ Find: (i) The marginal distribution of $X_1$. (ii) The marginal distribution of $X_2$. (b) Show that $X_1$ and $X_2$ are independent if and only if $\alpha = 0$. 2. **Formula and Important Rules:** - The marginal pdf of $X_1$ is found by integrating the joint pdf over all $x_2$: $$f_{X_1}(x_1) = \int_0^\infty f(x_1, x_2) \, dx_2$$ - Similarly, the marginal pdf of $X_2$ is: $$f_{X_2}(x_2) = \int_0^\infty f(x_1, x_2) \, dx_1$$ - Two random variables are independent if and only if their joint pdf factors as the product of their marginals: $$f(x_1, x_2) = f_{X_1}(x_1) f_{X_2}(x_2)$$ 3. **Find the marginal distribution of $X_1$:** $$f_{X_1}(x_1) = \int_0^\infty [1 - \alpha(1 - 2e^{-x_1})(1 - 2e^{-x_2})] e^{-(x_1 + x_2)} \, dx_2$$ Rewrite the integral: $$= e^{-x_1} \int_0^\infty [1 - \alpha(1 - 2e^{-x_1})(1 - 2e^{-x_2})] e^{-x_2} \, dx_2$$ Distribute inside the integral: $$= e^{-x_1} \left[ \int_0^\infty e^{-x_2} \, dx_2 - \alpha(1 - 2e^{-x_1}) \int_0^\infty (1 - 2e^{-x_2}) e^{-x_2} \, dx_2 \right]$$ Calculate the integrals: - $$\int_0^\infty e^{-x_2} \, dx_2 = 1$$ - $$\int_0^\infty (1 - 2e^{-x_2}) e^{-x_2} \, dx_2 = \int_0^\infty e^{-x_2} \, dx_2 - 2 \int_0^\infty e^{-2x_2} \, dx_2 = 1 - 2 \times \frac{1}{2} = 1 - 1 = 0$$ So the marginal pdf simplifies to: $$f_{X_1}(x_1) = e^{-x_1} [1 - \alpha(1 - 2e^{-x_1}) \times 0] = e^{-x_1}$$ for $x_1 \geq 0$. 4. **Find the marginal distribution of $X_2$:** By symmetry, the marginal pdf of $X_2$ is: $$f_{X_2}(x_2) = \int_0^\infty f(x_1, x_2) \, dx_1 = e^{-x_2}$$ for $x_2 \geq 0$. 5. **Show independence iff $\alpha = 0$:** - If $\alpha = 0$, then $$f(x_1, x_2) = e^{-(x_1 + x_2)} = f_{X_1}(x_1) f_{X_2}(x_2)$$ so $X_1$ and $X_2$ are independent. - If $\alpha \neq 0$, the joint pdf contains the term: $$1 - \alpha(1 - 2e^{-x_1})(1 - 2e^{-x_2})$$ which does not factor into a product of functions of $x_1$ and $x_2$ separately. Therefore, $X_1$ and $X_2$ are independent if and only if $\alpha = 0$. **Final answers:** - Marginal pdf of $X_1$: $$f_{X_1}(x_1) = e^{-x_1}, \quad x_1 \geq 0$$ - Marginal pdf of $X_2$: $$f_{X_2}(x_2) = e^{-x_2}, \quad x_2 \geq 0$$ - Independence holds iff $$\alpha = 0$$