Show that the distrubtion function $F_Z$ of $Z$ suffices $F_Z(z)=F(z)^2$

Question

Let $X$ and $Y$ be two stochastically independent, equally distributed random variables with distribution function $F$. Define $Z = \max (X, Y)$.

1) Show that the distrubtion function $F_Z$ of $Z$ suffices $F_Z(z)=F(z)^2$.

I got this:

$F_Z(z)=P(\{Z\leq z\})=P(\{\max (X, Y) \leq z\})=...$

A hint would be appreciated.

Answer 1

Just for completeness:

$$\{\omega\in\Omega:\max{\{X(\omega),Y(\omega)\}}\le z\}=\{\omega\in\Omega:X(\omega)\le z\wedge Y(\omega)\le z\}$$

$"\supset"$: If both, $X(\omega)$ and $Y(\omega)$ are smaller or equal $z$, so is the maximum of both.

$"\subset"$:If the $\max{\{X(\omega),Y(\omega)\}}$ is smaller or equal $z$, then by definition both, $X(\omega)$ $\textbf{and}$ $Y(\omega)$ have to be smaller or equal $z$.

Therefore

$$F_Z(z)=P[Z\le z]=P[\max{\{X,Y\}}\le z]=P[X\le z\wedge Y\le z]=P[X\le z]P[Y\le z]=P[X\le z]^2=F(z)^2$$ where we have used independence and identically distributed in the two last equalities.

Note that in general, if you have $n\in \mathbb{N}$, $X_1,\dots,X_n$ iid. random variables with distribution function $F(z)$, then the random variable $Z:=\max{\{X_1,\dots,X_n\}}$ has the distribution function $$F_Z(z)=F(z)^n$$.