marginal pdf of exponential functions Hi I have a question about marginal PDFs of exponentials. Given is :
$$f(x,y)=y \text{ for } y\leqslant \ e^{-x/4} \text{ and } 0 \text{ otherwise}.$$
I have sketched the graph and got an exponential curve. Question is what parameters is the marginal PDF of $X$ and $Y$ supposed to take? My assumptions:
$$f_X(x)= \int_0^{e^{-x/4}}y \ dy$$
and $f_Y(y)= \int_0^{e^{-x/4}}y \ dx$ which doesn't make sense to me for $f(y)$ (according to the solutions given) why is the parameter in terms of $x$ and not $y$?
moreover, what would be the parameters given to the expected values?
for $\operatorname{E}(X)=\int_0^\infty x \ f(x) \ dx$ because of the asymptote.
what would it be for $\operatorname{E}(Y)= \int_0^\text{?} y \ f(y) \, dy$?
Apart from that, how would the independence be determined from the graph? I know the equation for indepedence is $f(x,y) = f(x)f(y)$ but is the correct way  of using it by plugging in x=0 and y=0 and see if they are equal to each other?
 A: It looks as if your pdf must be
$$
f(x,y) = \begin{cases} y & \text{if } x\ge0 \text{ and } 0\le y\le e^{-x/4}, \\[5pt] 0 & \text{if } x<0 \text{ or } y<0 \text{ or } y>e^{-x/4}. \end{cases}
$$
The marginal pdf for $Y$ is
$$
f_Y(y) = \int_0^\infty f_{X,Y} (x,y) \, dx.
$$
You "integrate out" $x,$ getting a function of $y.$ This is similar to a situation with a function of two discrete variables, $i$ and $j,$ both taking values in the set $\{1,2,3\}:$
$$
p_I(i) = \sum_{j=1}^3 p_{I,J}(i,j) = \overbrace{p(1,j)+p(2,j)+p(3,j)}
$$
The part under the $\overbrace{\text{overbrace}}$ is a function only  of $j$ and not of $i;$ the variable $i$ has been "summed out."
The density is $0$ when $y> e^{-x/4},$ and that is the same as $\log y > -x/4,$ or $-4\log y > x.$ And not that $\log y$ is negative since $y<1,$ so $-4\log y$ is positive.
You have
\begin{align}
f_Y(y) & = \int_0^\infty f_{X,Y}(x,y)\, dx = \int_0^\infty \left. \begin{cases} y & \text{if } x< -4\log y \\ 0 & \text{if } x>-4\log y \end{cases}  \right\} \, dx \\[10pt]
& = \int_0^{-4\log y} y \, dx = -4y\log y \quad \text{provided } 0<y<1.
\end{align}
This density is $0$ at the two endpoints $0$ and $1,$ and is not symmetric.
As for "parameters," you can't talk about those unless you choose some parametrized family of distributions that contains this one.
You also have
$$
f_X(x) = \int_0^1 f_{X,Y}(x,y)\,dy = \int_0^{e^{-x/4}} y \, dy = \frac{e^{-x/2}} 2 \quad \text{provided } x>0.
$$
This is an exponential distribution with expected value $2.$
