I'm kind of new here so plz tell me how I can make my questions clearer?
Let $X$ and $Y$ be two i.i.d $U[0,2]$ random variables and $Z = \max(X,Y)$. What is the expected value of $Z$?
The answer is $\frac{4}{3}$ but I absolutely have zero clue about how to start to even get there, can anyone please help a poor lost soul?
A good way to handle problems you are a bit confused on is to go back to the cumulative distribution functions. In this case, the CDF for a uniform random variate on $[0,2]$ is simply $P(X \leq x) = \frac{x}{2}$ for $0\leq x \leq 2$ (and zero for $x<0$ and one for $x>2$).
Now we would llike to find the CDF for $\max(X,Y)$. So, what is the probability that said maximum is less than some value $m$? Well, $X$ and $Y$ are independent, so
$$P(X\leq m \wedge Y \leq m) = P (X\leq m ) \cdot P (Y\leq m ) \\ P(\max(X,Y) \leq m) = (\frac12 m)( \frac12 m)= \frac{m^2}{4} $$ So we have a random variate $M$ whose CDF is $\frac14 m^2$; its probability density function is the derivative of the CDF, so $$f(m) = \frac{m}{2}$$ Finally, the expected value of $M$ is $$ \int_0^2 m\,f(m)\,dm = \int_0^2 \frac{m^2}{2}\,dm = \frac{2^3}{6} = \frac43 $$