# How does a dependent domain in joint probability distributions affect integration limits?

$$0<x<y<C,$$ $$f(x,y)\text{ given}$$

1. If I want to find $E[X]$ do I have to integrate $x f(x)$ from $0$ to $C$ or from $0$ to $y$?
2. If I want to find $E[E[Y\mid X]]$, I integrate $E[Y\mid X]\cdot f(x)$ from $0$ to $C$, but why not from $0$ to $y$ (since $x$ is constrained by $y$)?
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One way of finding $E(X)$ would be to find $f_X(x)$ and then finding $\displaystyle\int_0^C xf_X(x)\,dx$.

Another is just to find $\displaystyle\int xf_{X,Y}(x,y)\,dx$, integrating over the set $\{(x,y) : 0<x<y<C\}$.

To integrate over that latter set, you can take either of two approaches:

First approach: First, $y$ goes from $0$ to $C$, then for each value of $y$, $x$ goes from $0$ to $y$. Thus you have $$\int_0^C\cdots\cdots\,dy$$ and inside that you have $$\int_0^y \cdots\cdots\,dx.$$

Second approach: First $x$ goes from $0$ to $C$, then for each value of $x$, $y$ goes from $x$ to $C$. Thus you have $$\int_0^C\cdots\cdots\,dx$$ and inside that you have $$\int_x^C \cdots\cdots\,dy.$$

In your second question, what you need is $E(Y\mid X=x)\cdot f_X(x)$, from $0$ to $C$.

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thanks for your detailed asnwer. in my second question i know that it goes from 0 to C, but why not from 0 to y? because X is assumed to be fixed? – Wuschelbeutel Kartoffelhuhn Jan 23 '13 at 19:20
For the same reasons that those constraints only appear in the INSIDE integrals above. – Michael Hardy Jan 23 '13 at 19:22
also in your first sentence of your answer, why do you not integrate from 0 to y(since x depends on y)? – Wuschelbeutel Kartoffelhuhn Jan 23 '13 at 19:22
Have you carefully read my answer above? I think it might make those things clear. – Michael Hardy Jan 23 '13 at 20:16
@WuschelbeutelKartoffelhuhn : $X$ can be anywhere between $0$ and $C$, so its density is non-zero only between those bounds. Given a value of $X$, $Y$ can only be between $X$ and $C$, but that doesn't alter the fact that the set of possible values of $X$ is the set of numbers between $0$ and $C$. – Michael Hardy Jan 23 '13 at 22:28