# Calculation marginal density functions given a multi-variable PDF

Let $$c$$ be a positive constant and $$f(x,y)= \begin{cases} cxe^{-2x(1+y)}&\text{if }x>0\,y>0 \\0&\text{otherwise. } \end{cases}$$ a) Which value of $$c$$ makes $$f$$ a PDF on $$\mathbb{R}^2$$?

b) Let $$(X,Y)$$ have joint probability density function $$f$$ given above, with the right constant $$c$$ from part (a). Find the marginal density functions $$f_{X}$$ of $$X$$ and $$f_{Y}$$ of $$Y$$.

c) Are $$X$$ and $$Y$$ independent?

Here is my attempt at the problem:

a) We need to find $$c$$ such that \begin{align*} 1 = \int_{\mathbb{R}}\int_{\mathbb{R}}f(x,y)\,dx\,dy & =c\int_{0}^{\infty}\int_{0}^{\infty} xe^{-2x(1+y)}\,dx\,dx \\ & = c\int^{\infty}_{0}\left[ \frac{-xe^{-2x(1+y)}}{2(1+y)}\Bigg|^{\infty}_{0}+\int_{0}^{\infty}\frac{e^{-2x(1+y)}}{2(1+y)}\,dx \right]\,dy \\ & =\frac{c}{4}\int_{0}^{\infty}\frac{1}{(1+y)^2}\,dy \\ & = \frac{c}{4}, \end{align*} so we must have $$c=4.$$

b) We can find the marginal density function of $$X$$ as follows \begin{align*} f_{X}(x) =4\int^{\infty}_{0}xe^{-2x(1+y)}\,dy & = 4xe^{-2x}\int^{\infty}_{0}e^{-2xy}\,dy \\ & = 2e^{-2x}. \end{align*}

Then we get $$f_X(x)= \begin{cases} 2e^{-2x}&\text{if }x>0 \\0&\text{otherwise. } \end{cases}$$ And for $$f_Y,$$ we have \begin{align*} f_{Y}(y) =4\int^{\infty}_{0}xe^{-2x(1+y)}\,dx &= 4\left[ \frac{xe^{-2x(1+y)}}{-2(1+y)}\Bigg|^{\infty}_{0}+\int^{\infty}_{0}\frac{e^{-2x(1+y)}}{2(1+y)}\,dx \right] \\ & = \frac{1}{(1+y)^2}. \end{align*} So that $$f_Y(y)= \begin{cases} (1+y)^{-2}&\text{if }y>0 \\0&\text{otherwise. } \end{cases}$$

c) It follows from the above calculations that $$X$$ and $$Y$$ are not independent since $$f_Xf_Y\ne f_{XY}.$$

Is any of my work correct? Any feedback is much appreciated, and if you think I should add more details to my calculations, please point it out and I will edit my work accordingly.

a) Correct, but I would rather go for:$$1=\int_{0}^{\infty}\int_{0}^{\infty}cxe^{-2x\left(1+y\right)}dydx=c\int_{0}^{\infty}x\int_{1}^{\infty}e^{-2xz}dzdx=c\int_{0}^{\infty}\frac{1}{2}e^{-2x}dx=\frac{c}{4}$$
c) Correct, but actually the non-independence can also be concluded directly from the fact that $$f_{X,Y}(x,y)$$ cannot be written as a product of the form $$f(x)g(y)$$.