I'm having a dillemma regarding the integral of a density: Suppose: $$f_{X,Y}(x,y) = \frac{1}{2}xy\times\textbf{1}_{x\in[0,2]}\times\textbf{1}_{y\in[0,x]}$$
In order to find $E[X\mid Y]$ I know I want to solve the integral: $$\int_{\mathbb{R}}x\frac{f_{X,Y}(x,y)}{f_{Y}(y)}dx$$ And for that I need to find $f_{Y}(y)$.
Since I have this condition where $y\in[0,x]$ I'm not sure how to take that into consideration where I integrate.
When looking for $f_{Y}(y)$ how should I approach? it can either be:
- $f_{Y}(y)= \int_{y}^{2}{f_{X,Y}}(x',y)dx'$
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- $f_{Y}(y)= \int_{0}^{2}{f_{X,Y}}(x',y)dx'$
If that were the other way around, I'd say with no doubt that: $$f_{X}(x)= \int_{0}^{x}{f_{X,Y}}(x,y')dy'$$
My intuition says that $x$ can run independently on the interval $[0,2]$ and is not affected by $y$ even though $y$ IS affected by $x$. Therefore I think integral number 2 is correct, but I'm not $100 \%$ about it.
In order to find $f_{Y}(y)$, you should compute it in this way: $f_{Y}(y)= \int_{\mathbb R}{f_{X,Y}}(x',y)dx'=\int_{\mathbb R}\frac{1}{2}x'y\times\textbf{1}_{x'\in[0,2]}\times\textbf{1}_{y\in[0,x']}=\\ \\ =\int_{[0,2] \cap [y,+\infty]}\frac{1}{2}x'y dx'=\frac 1 2 y \int_{[0,2] \cap [y,+\infty]}x'dx'$
and the last one depends on $y$. If $y \leq 0$ then $[0,2] \cap [y,+\infty]=[0,2]$, if $0 < y < 2$ then $[0,2] \cap [y,+\infty]=[y,2]$ and if $y > 2$ then $[0,2] \cap [y,+\infty]= \emptyset$. The case $y=2$ does not really matter.