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.

Thank you for your time.


a) Correct, but I would rather go for:$$1=\int_{0}^{\infty}\int_{0}^{\infty}cxe^{-2x\left(1+y\right)}dydx=c\int_{0}^{\infty}x\int_{1}^{\infty}e^{-2xz}dzdx=c\int_{0}^{\infty}\frac{1}{2}e^{-2x}dx=\frac{c}{4}$$

b) Correct.

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 $f(x)g(y)$.

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