Let $X_1, X_2$ be two i.i.d. variables with density $f_1(x) = f_2(x) = 2x$ on $[0, 1]$. What is the conditional expectation of $Z=\max(X_1, X_2)$ given $X_2$? I tried to find $f_{Z|X_2}(x,y)$ but I struggle to do so. I have that the CDF is given by :
$$F_{Z|X_2}(x,y)=\Bbb P(\{\max(X_1,X_2) \leq x\} \cap \{X_2 \leq y\}) $$ $$=\Bbb P(\{X_1 \leq x\} \cap \{X_2 \leq \max(x,y)\}) = \Bbb P(X_1 \leq x) \cdot\Bbb P(X_2 \leq \max(x,y)) $$ but then I find quite annoying to derivate this to obtain $f_{Z|X_2}(x,y)$. Is there a way out ?
Just as an alternative to the previous answer (it is always good to have multiple approaches):
$$\mathbb E(\max(X_1,X_2)|X_2)=\mathbb E(X_1\cdot\mathbb 1_{\{X_1>X_2\}}|X_2)+X_2\,\mathbb E(1_{\{X_1\leq X_2\}}|X_2).$$
Both integrals can be solved easily and the result is again $\frac13(X_2^3+2)$.
EDIT: To solve the above, one can use the following somewhat canonical approach to put a fixed value for the condition. Just remember $\{X_2=y\}=\{\omega\in\Omega:X_2(\omega)=y\}$. This event is measurable and you can apply the formulas for the conditional expectation with respect to an event. This works because these events generate the sigma-algebra $\sigma(X_2)$.
It is written like this (I just plug in information from the condition to simplify):
$$\mathbb E(X_1\cdot\mathbb 1_{\{X_1>X_2\}}|X_2=y)=\mathbb E(X_1\cdot 1_{\{X_1>y\}})=\int_{y}^12x^2\,dx=\frac23(1-y^3)=:g(y).$$ By theory, the conditional expectation with respect to another r.v. is a measurable function of $X_2$, therefore $\mathbb E(X_1\cdot\mathbb 1_{\{X_1>X_2\}}|X_2)=g(X_2).$
The same holds for the other integral $$\mathbb E(1_{\{X_1\leq X_2\}}|X_2=y)=\mathbb E(1_{\{X_1\leq y\}})=\int_0^y2x\,dx=y^2.$$ You can use the cdf here, but it is not given. So you would have to compute the integral anyways. Therefore $\mathbb E(1_{\{X_1\leq X_2\}}|X_2)=\mathbb P(X_1\leq X_2|X_2)=X_2^2$.
Personally, I do not like to put random variables into the delimiters of integrals and other functions, but it is a usual notation.