CDF of max($X$, $Y$) - where is the mistake? $X$ and $Y$ are independent r.v.'s and we know $F_X(x)$ and $F_Y(y)$. Let $Z=max(X,Y)$. Find $F_Z(z)$.
Here's my reasoning: 
$F_Z(z)=P(Z\leq z)=P(max(X,Y)\leq z)$. 
I claim that we have 2 cases here: 
1) $max(X,Y)=X$. If $X<z$, we are guaranteed that $Y<z$, so $F_Z(z)=P(Z\leq z)=P(X<z)=F_X(z)$
2) $max(X,Y)=Y$. Similarly, $F_Z(z)=P(Z\leq z)=P(Y<z)=F_Y(z)$
Since we're interested in either case #1 or #2, 
$F_Z(z)=F_X(z)+F_Y(z)-F_X(z)*F_Y(z)$
However, it's wrong and I know it. But I would like to know where the flaw in my reasoning is. I know the answer to this problem, I just want to know at what moment my reasoning fails.
 A: I think you are fine with separating the cases, but then do not take care of them correctly. Since when you say in your case 1 that the maximum is $X$, you are "conditioning on" $X>Y$ and that changes the space over which you calculate the probabilities. 
We have two cases that either of which happens:
case 1: $X<Y<z$
case 2: $Y<X<z$.
That is 
\begin{align}
\Pr(\max\{X,Y\}<z)&=\Pr(X<Y<z)+\Pr(Y<X<z)\\
&=\int_{x=-\infty}^z\int_{y=x}^zf_Y(y)f_X(x)dydx+\int_{y=-\infty}^z\int_{x=y}^zf_Y(y)f_X(x)dxdy\\
&=\int_{y=-\infty}^z\int_{x=-\infty}^yf_Y(y)f_X(x)dxdy+\int_{y=-\infty}^z\int_{x=y}^zf_Y(y)f_X(x)dxdy\\
&=\int_{y=-\infty}^zf_Y(y)\left(\int_{x=-\infty}^yf_X(x)dxdy+\int_{x=y}^zf_X(x)dx\right)dy\\
&=\int_{y=-\infty}^zf_Y(y)\left(\int_{x=-\infty}^zf_X(x)dxdy\right)dy\\
&=F_Y(z)F_X(z).
\end{align}
A: $\max(X,Y)\le z$ means that both $X$ and $Y$ are $\le z$.
A: Your reasoning, of using the Principle of Inclusion and Exclusion, would be fine if you were dealing with minimum rather than maximum.
$\bullet \quad F_{\min \{X,Y\}}(z) ~{=~ \mathsf P(X\leq z~\cup~Y\leq z) \\=~ F_X(x)+F_Y(z)-F_X(z)\cdot F_Y(z)}\\\bullet\quad F_{\max\{X,Y\}}(z)~{=~\mathsf P(X\leq z~\cap~Y\leq z)\\=~F_X(z)\cdot F_Y(z)}$
A: Everything works until you write "either... or..." instead of "and". 
In practice, your last formula is computing the probability that $X \le z$ or $Y \le z$ but not both. Hence, for $X \le z$, you implicitly compute the probability that $Y > z$; likewise, for $Y \le z$, you consider the probability that $X > z$. 
