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

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.

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2
try to write $\max{(X,Y)}\le z$ in a nice way and use independence. – n.c. Jan 16 at 11:20
@n.c. This is the nicest that I get: $F_Z(z)=P(\{Z\leq z\})=P(\{\omega:Z(\omega)\leq z\}) =P(\{\omega:\max (X, Y)(\omega)\leq z\}) =P(\{\omega:X(\omega)\leq z\ \lor Y(\omega)\leq z\})$ I'm not sure about the last step. – Kasper Jan 16 at 11:21

3
your last equality is not true. it should be a $\wedge$. Otherwise the sets are not equal. Then just apply independence and identically distributed – n.c. Jan 16 at 11:32

## 1 Answer

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 $$F_Z(z)=F(z)^n$$.

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thanks for this excellent answer ! +1 – Kasper Jan 16 at 13:34
@Kasper You are welcome – n.c. Jan 16 at 13:35
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