1. **Problem Statement:** Given the joint pdf $$f(x,y) = \frac{2}{(1+x+y)^3}$$ for $$x>0, y>0$$ and 0 elsewhere, find: (i) Marginal densities of $$X$$ and $$Y$$ (ii) Conditional density of $$X$$ given $$Y=y$$ (iii) Conditional density of $$Y$$ given $$X=x$$ 2. **Marginal Density of $$X$$:** The marginal density $$f_X(x)$$ is found by integrating the joint pdf over all $$y$$: $$f_X(x) = \int_0^\infty f(x,y) \, dy = \int_0^\infty \frac{2}{(1+x+y)^3} \, dy$$ Let $$u = 1 + x + y \Rightarrow du = dy$$, when $$y=0, u=1+x$$ and when $$y \to \infty, u \to \infty$$. So, $$f_X(x) = 2 \int_{1+x}^\infty u^{-3} \, du = 2 \left[-\frac{1}{2u^2}\right]_{1+x}^\infty = \frac{1}{(1+x)^2}$$ for $$x>0$$. 3. **Marginal Density of $$Y$$:** Similarly, integrate over $$x$$: $$f_Y(y) = \int_0^\infty f(x,y) \, dx = \int_0^\infty \frac{2}{(1+x+y)^3} \, dx$$ Let $$v = 1 + x + y \Rightarrow dv = dx$$, when $$x=0, v=1+y$$ and when $$x \to \infty, v \to \infty$$. So, $$f_Y(y) = 2 \int_{1+y}^\infty v^{-3} \, dv = 2 \left[-\frac{1}{2v^2}\right]_{1+y}^\infty = \frac{1}{(1+y)^2}$$ for $$y>0$$. 4. **Conditional Density of $$X$$ given $$Y=y$$:** By definition, $$f_{X|Y}(x|y) = \frac{f(x,y)}{f_Y(y)} = \frac{\frac{2}{(1+x+y)^3}}{\frac{1}{(1+y)^2}} = \frac{2(1+y)^2}{(1+x+y)^3}$$ for $$x>0, y>0$$. 5. **Conditional Density of $$Y$$ given $$X=x$$:** Similarly, $$f_{Y|X}(y|x) = \frac{f(x,y)}{f_X(x)} = \frac{\frac{2}{(1+x+y)^3}}{\frac{1}{(1+x)^2}} = \frac{2(1+x)^2}{(1+x+y)^3}$$ for $$y>0, x>0$$. **Final answers:** (i) $$f_X(x) = \frac{1}{(1+x)^2}, x>0$$ and $$f_Y(y) = \frac{1}{(1+y)^2}, y>0$$ (ii) $$f_{X|Y}(x|y) = \frac{2(1+y)^2}{(1+x+y)^3}, x>0$$ (iii) $$f_{Y|X}(y|x) = \frac{2(1+x)^2}{(1+x+y)^3}, y>0$$