PDF of a $\ Z = 1 - max(X, Y) $?

Question

Given that $\ X $ and $\ Y $ are uniform random variables, independent of each other over the range $\ [0, 1] $, how to find the PDF of Z, where $\ Z = 1 - max(X, Y) $.

I could find the PDF of $\ max(X, Y) $ as $\ 2z $ ->$\ [0, 1] $, and $\ 0 $ otherwise, by the usual way, but no clue how to start this.

One thing I could make out is $\ Z $ exists over the same range of $\ [0, 1] $

Thanks.

Answer 1

$$\mathbb P(Z\geq z)=\mathbb P(\max(X,Y)\leq 1-z)=\mathbb P(X\leq 1-z, Y\leq 1-z)=\mathbb P(X\leq 1-z)\mathbb P(Y\leq 1-z)=\left(\int_0^{1-z}1_{[0,1]}(x)dx\right)^2.$$ Therefore, $$f_Z(z)=\frac{d}{dz}\mathbb P\{Z\leq z\}=2\cdot 1_{[0,1]}(1-z)\int_0^{1-z}1_{[0,1]}(x)dx=2\cdot 1_{[0,1]}(z)\int_0^{1-z}1_{[0,1]}(x)dx=2\cdot 1_{[0,1]}(z)(1-z)$$

Answer 2

I go on with your pdf of $V$, which is $f_V(v)=2v \ \textbf I_{[0,1]}$, where $V=\max (X,Y)$. I use $V$ here, since it is not $Z$ yet.

The cdf is $F_V(v)=v^2 \ \textbf I_{[0,1]}$. Next we re-arrange the inequality:

$P(1-V\leq Z)=P(V\geq 1-Z)=1-P(V\leq 1-Z)=1-(1-z)^2$. Thus the cdf Z is

$$F_Z(z)=\begin{cases} 0, \ z< 0 \\ 1-(1-z)^2, \ 0\leq z<1 \\1, \ z \geq 1 \end{cases}$$