Find $E[Y\mid X=x]$

Question

Let $(X,Y)$ $$f(x,y)=\frac{x^2}{4}\hspace{1cm} 0\leq y\leq x\leq 2$$ Find $E[Y\mid X=x]$

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I know $$\dfrac{f(x,y)}{f_X(x)}=\frac{x^2/4}{x^3/4}=\frac{1}{x}$$ My question is about the limits of integration,

I don't know if they are $[0,2]$ $$E[Y\mid X=x]=\int_0^2y\cdot\frac{1}{x}\,dy$$

or $[0,x]$

$$E[Y\mid X=x]=\int_0^x y\cdot\frac{1}{x}\,dy$$

Answer 1

What you calculated is $$ \frac{f(x,y)}{f_X(x)}=\frac{1}{x},\quad \color{red}{0\le y\le x\le 2} $$

Hence for $x>0$, $$ E(Y|X=x)=\int_0^2 \frac{yf(x,y)}{f_X(x)}dy=\int_0^x\frac{y}{x}dy=\frac{x}{2}. $$ If $x<0$, then $f(x,y)=0$, and thus $$ E(Y|X=x)=0. $$