1. **Problem statement:** Given continuous random variables $(X,Y)$ with joint density function $$f_{X,Y}(x,y) = k y e^{-y(1+x)}, \quad x>0, y>0,$$ and zero otherwise, where $k>0$, we need to find the constant $k$, marginal densities, check independence, and compute conditional expectation. 2. **Why must a joint pdf integrate to one?** A joint pdf $f_{X,Y}(x,y)$ represents a probability distribution over $(X,Y)$. The total probability over the entire support must be 1, so $$\int \int f_{X,Y}(x,y) \, dx \, dy = 1.$$ 3. **Why marginal densities do not determine independence?** Independence requires $$f_{X,Y}(x,y) = f_X(x) f_Y(y)$$ for all $x,y$. Marginals alone do not guarantee this factorization; joint behavior matters. 4. **Find $k$ by integrating the joint pdf over support:** $$\int_0^\infty \int_0^\infty k y e^{-y(1+x)} \, dx \, dy = 1.$$ Integrate w.r.t. $x$ first: $$\int_0^\infty e^{-y(1+x)} \, dx = e^{-y} \int_0^\infty e^{-y x} \, dx = e^{-y} \cdot \frac{1}{y}.$$ So, $$\int_0^\infty k y e^{-y(1+x)} \, dx = k y e^{-y} \cdot \frac{1}{y} = k e^{-y}.$$ Now integrate w.r.t. $y$: $$\int_0^\infty k e^{-y} \, dy = k \int_0^\infty e^{-y} \, dy = k [ -e^{-y} ]_0^\infty = k (1 - 0) = k.$$ Set equal to 1: $$k = 1.$$ 5. **Marginal density of $X$:** $$f_X(x) = \int_0^\infty f_{X,Y}(x,y) \, dy = \int_0^\infty y e^{-y(1+x)} \, dy.$$ Use substitution $t = y(1+x)$, so $y = \frac{t}{1+x}$ and $dy = \frac{dt}{1+x}$: $$f_X(x) = \int_0^\infty \frac{t}{1+x} e^{-t} \frac{dt}{1+x} = \frac{1}{(1+x)^2} \int_0^\infty t e^{-t} dt.$$ Recall $$\int_0^\infty t e^{-t} dt = 1! = 1.$$ Thus, $$f_X(x) = \frac{1}{(1+x)^2}, \quad x>0.$$ 6. **Marginal density of $Y$:** $$f_Y(y) = \int_0^\infty y e^{-y(1+x)} \, dx = y e^{-y} \int_0^\infty e^{-y x} \, dx = y e^{-y} \cdot \frac{1}{y} = e^{-y}, \quad y>0.$$ 7. **Check independence:** If independent, $$f_{X,Y}(x,y) = f_X(x) f_Y(y) = \frac{1}{(1+x)^2} e^{-y}.$$ Given joint pdf is $$f_{X,Y}(x,y) = y e^{-y(1+x)} = y e^{-y} e^{-x y}.$$ Since $$y e^{-x y} \neq \frac{1}{(1+x)^2}$$ for all $x,y$, $X$ and $Y$ are not independent. 8. **Compute conditional expectation $E[X|Y=y]$:** Conditional pdf: $$f_{X|Y}(x|y) = \frac{f_{X,Y}(x,y)}{f_Y(y)} = \frac{y e^{-y(1+x)}}{e^{-y}} = y e^{-x y}, \quad x>0.$$ This is an exponential distribution with rate $y$. Therefore, $$E[X|Y=y] = \frac{1}{y}.$$ This shows the expected value of $X$ given $Y=y$ decreases as $y$ increases. **Final answers:** - $k=1$ - $f_X(x) = \frac{1}{(1+x)^2}, x>0$ - $f_Y(y) = e^{-y}, y>0$ - $X$ and $Y$ are not independent - $E[X|Y=y] = \frac{1}{y}$ depends inversely on $y$