Expected Value for 2 Random Variables with Joint Probability Distribution

Question

I have trouble with determining the domain for integration in the case of having a joint pdf when one variable depends on the other. There are two examples I don't quite understand, and hopefully, you can help me out.

Example 1: $ f(x,y)=2, 0\le x\le y \le 1$ In the solution, the expected values for $X$ and $Y$ $E[X]$ and $E[Y]$ are given as: $$E[X]=\int_{0}^{1} \int_{x}^{1} 2x \,dy \,dx$$ $$E[Y]=\int_{0}^{1} \int_{y}^{0} 2y \,dx \,dy$$ In the first case, we are using $1$ and $x$ for the inner integral. In the second case, I believe we are supposed to use $x$ and $1$ rather than $0$ and $1$(based on inequality). Am I wrong? I don't see why in the first case we chose $x$ rather than $0$ for the initial value while in the second case we chose $0$ over $x$ as an initial value.

Example 2:(similar concept but applied to marginal pdf):

Let $X$ and $Y$ have a joint pdf $f(x,y)=1, x\le y\le x+1, 0\le x\le1$ Then, $$f_{X}(x)=\int_{x}^{x+1}1dy,0\le x\le1$$ $$\begin{equation} f_{Y}(y)=\begin{cases} \int_{0}^{y}1dx=y, & \text{$0\le y\le1$}.\\ \int_{y-1}^{1}1dx=2-y, & \text{$1\le y\le2$}. \end{cases} \end{equation}$$ Why does here $f_{Y}(y)$ branches into 2 parts and why can't we simply say that $f_{Y}(y)=\int_{x}^{x+1}1dx$ ? (simillar to $f_{X}(x)$) . I do get that we need to get a function in terms of $y$ but all I am doing is just looking at the given inequality.

I ask these 2 questions together because I believe the issue comes from the same misunderstanding on my part.

Answer 1

When writing (correctly) $$\mathbb E[X]=\int_{0}^{1} \int_{x}^{1} 2x \,\mathrm dy \,\mathrm dx$$ one skips intermediate steps: \begin{align} \mathbb E[X] &= \int_{\mathbb R^2} x f(x,y)\,\,\mathrm d(x,y)\tag{definition}\\ &= \int_{\mathbb R^2} x\times2\times\mathbb I_{\underbrace{\{(x,y);\,0\le x\le y \le 1\}}_{\text{support set}}}(x,y) \,\,\mathrm d(x,y)\\ &= \int_{\underbrace{(0,1)^2}_{\text{max range}}} x\times2\times\mathbb I_{\{(x,y);\,0\le x\le y \le 1\}}(x,y) \,\,\mathrm d(x,y)\\ &= \int_0^1 2x \Big\{ \int_0^1 \mathbb I_{\underbrace{\{y;\,x\le y \le 1\}}_{\text{$y$ sliced support}}}(y)\,\mathrm dy \Big\} \,\mathrm dx\\ &= \int_0^1 2x \Big\{\int_x^1 \,\mathrm dy\Big\} \,\mathrm dx \end{align} The same applies to $\mathbb E[Y]$: \begin{align} \mathbb E[Y] &= \int_{\mathbb R^2} y f(x,y)\,\,\mathrm d(x,y)\tag{definition}\\ &= \int_{\mathbb R^2} y\times2\times\mathbb I_{\underbrace{\{(x,y);\,0\le x\le y \le 1\}}_{\text{support set}}}(x,y) \,\,\mathrm d(x,y)\\ &= \int_{\underbrace{(0,1)^2}_{\text{max range}}} y\times2\times\mathbb I_{\{(x,y);\,0\le x\le y \le 1\}}(x,y) \,\,\mathrm d(x,y)\\ &= \int_0^1 2y \Big\{ \int_0^1 \mathbb I_{\underbrace{\{x;\,0\le x\le y\}}_{\text{$x$ sliced support}}}(x)\,\mathrm dx \Big\} \,\mathrm dx\\ &= \int_0^1 2y \Big\{\int_0^x \,\mathrm dx\Big\} \,\mathrm dy \end{align}