Given the joint density of $X$ and $Y$,
$$f_{X,Y}(x,y)=\begin{cases}e^{-y}&\text{for }0\le x<y<\infty\\0&\text{otherwise}\end{cases},$$
(a) find the covariance and correlation of $X$ and $Y$;
(b) find $E[X\mid Y=y]$ and $E[Y\mid X=x]$; and
(c) find $E[X]$ and $\mathrm{Var}[X]$.
I know that
$$\mathrm{Cov}[X,Y]=E[XY]-E[X]E[Y],$$
$$\mathrm{Corr}[X,Y]=\frac{\mathrm{Cov}[X,Y]}{\sqrt{\mathrm{Var}[X]\mathrm{Var}[Y]}},\text{ and}$$
$$E[XY]=\int_0^\infty\int_0^yxye^{-y}\,\mathrm dx\,\mathrm dy=3,$$
so theoretically, I should be able to compute the covariance by first finding the marginal densities for $X$ and $Y$, then computing the expectations of $X$ and $Y$. Alternatively, since
$$E[X]=E[E[X\mid Y=y]],$$
I can accomplish the same thing by finding the conditional densities and their expectations.
However, the ordering of the questions has me wondering:
(1) Can I compute the covariance and correlation directly knowing just the joint PDF?
As for finding the (necessary?) expectations, I've run into a complication...
The marginal density for $Y$ is
$$f_Y(y)=\int_0^ye^{-y}\,\mathrm dx=xe^{-y}\bigg|_0^y=\begin{cases}ye^{-y}&\text{for }y\ge0\\0&\text{otherwise}\end{cases}$$
so the conditional density of $X$ given $Y=y$ is
$$f_{X\mid Y}(x\mid y)=\begin{cases}\frac{e^{-y}}{ye^{-y}}=\frac1y&\text{for }0\le x<y<\infty\\0&\text{otherwise}\end{cases}$$
Then the conditional expectation of $X$ given $Y=y$ is
$$E[X\mid Y=y]=\int_0^y\frac xy\,\mathrm dx=\frac{x^2}{2y}\bigg|_0^y=\frac y2$$
and so
$$\boxed{E[X]=E[E[X\mid Y=y]]=E\left[\frac y2\right]=\frac y2}$$
Similarly, the marginal density for $X$ is
$$f_X(x)=\int_x^\infty e^{-y}\,\mathrm dy=-e^{-y}\bigg|_x^\infty=\begin{cases}e^{-x}&\text{for }x\ge0\\0&\text{otherwise}\end{cases}$$
so the conditional density of $Y$ given $X=x$ is
$$f_{Y\mid X}(y\mid x)=\begin{cases}\frac{e^{-y}}{e^{-x}}=e^{x-y}&\text{for }0\le x<y<\infty\\0&\text{otherwise}\end{cases}$$
and the expectation of $Y$ given $X=x$ would be
$$E[Y\mid X=x]=\int_x^\infty ye^{x-y}\,\mathrm dy=-(y+1)e^{x-y}\bigg|_x^\infty=x+1$$
and so the expectation of $Y$ is
$$\boxed{E[Y]=E[E[Y\mid X=x]]=E[x+1]=x+1}$$
The complication is that I get different expectations for $X$ and $Y$ when trying to verify their values using the corresponding marginal densities:
$$\boxed{E[X]=\int_0^\infty xe^{-x}\,\mathrm dx=1\\E[Y]=\int_0^\infty y^2e^{-y}\,\mathrm dy=2}$$
I thought that perhaps the problem lies with the supports of $f_X$ and $f_Y$. We have $0\le x<y<\infty$ to start with, so the marginal densities could instead be
$$f_X(x)=\begin{cases}e^{-x}&\text{for }0\le x<y\\0&\text{otherwise}\end{cases}$$
$$f_Y(y)=\begin{cases}ye^{-y}&\text{for }x\le y\\0&\text{otherwise}\end{cases}$$
But even then,
$$\boxed{E[X]=\int_0^yxe^{-x}\,\mathrm dx=1-(y+1)e^{-y}\neq\frac y2\\E[Y]=\int_x^\infty y^2e^{-y}\,\mathrm dy=(x^2+2x+2)e^{-x}\neq x+1}$$
(2) Which expectations are correct? Why the discrepancy between methods?
Thanks in large part to the commenters for getting me back on the right track. Posting the key points of my solution here for completeness.
$$E[XY]=\int_0^\infty\int_0^yxye^{-y}\,\mathrm dx\,\mathrm dy=3$$
$$E[X]=\int_0^\infty\int_0^yxe^{-y}\,\mathrm dx\,\mathrm dy=1$$
$$E[Y]=\int_0^\infty\int_0^yye^{-y}\,\mathrm dx\,\mathrm dy=2$$
$$E[X^2]=\int_0^\infty\int_0^yx^2e^{-y}\,\mathrm dx\,\mathrm dy=2$$
$$E[Y^2]=\int_0^\infty\int_0^yy^2e^{-y}\,\mathrm dx\,\mathrm dy=6$$
$$\boxed{\mathrm{Cov}[X,Y]=E[XY]-E[X]E[Y]=1}$$
$$\boxed{\mathrm{Corr}[X,Y]=\frac{E[XY]-E[X]E[Y]}{\sqrt{(E[X^2]-E[X]^2)(E[Y^2]-E[Y]^2)}}=\frac1{\sqrt2}}$$
$$\boxed{E[X\mid Y=y]=\int_0^y\frac xy\,\mathrm dx=\frac y2}$$
$$\boxed{E[Y\mid X=x]=\int_x^\infty y^2e^{x-y}\,\mathrm dy=x+1}$$
$$\begin{cases} E[X]=E[E[X\mid Y]]=E\left[\frac Y2\right]=\frac{E[Y]}2\\[1ex] E[Y]=E[E[Y\mid X]]=E[X+1]=E[X]+1 \end{cases}\implies\boxed{E[X]=1}$$
$$E[X^2\mid Y=y]=\int_0^y\frac{x^2}y\,\mathrm dx=\frac{y^2}3$$
$$\mathrm{Var}[X\mid Y]=E[X^2\mid Y]-E[X\mid Y]^2=\frac{Y^2}3-\left(\frac Y2\right)^2=\frac{Y^2}{12}$$
$$E[\mathrm{Var}[X\mid Y]]=E\left[\frac{Y^2}{12}\right]=\frac{E[Y^2]}{12}=\frac12$$
$$\mathrm{Var}[E[X\mid Y]]=\mathrm{Var}\left[\frac Y2\right]=\frac{\mathrm{Var}[Y]}4=\frac{E[Y^2]-E[Y]}4=\frac12$$
$$\boxed{\mathrm{Var}[X]=E[\mathrm{Var}[X\mid Y]]+\mathrm{Var}[E[X\mid Y]]=1}$$