# finding the cdf of the max of a joint distribuion [closed]

Random variables X and Y have a distribution given by the following pdf: fx,y (x,y) = 1, 0 < x < 1 , 0 < y < 2x ; and o otherwise

find the cdf of Z = max(x,y)

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## closed as off-topic by Dilip Sarwate, Did, mookid, egreg, voldemortApr 16 at 22:48

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This has very much the aspect of a homework assignment. But even if it's not homework, you should add your thoughts about the problem. –  egreg Apr 16 at 22:32

Draw a picture of the part of the plane on which the joint density function "lives." This is the triangle with corners $(0,0)$, $(1,0)$, and $(1,2)$. The argument below cannot be understood without the picture.
We have $F_Z(z)=\Pr(Z\le z)$. It is a little easier to look for $\Pr(Z\gt z)$.
First suppose $z\ge 1$. Then $Z\gt z$ precisely if $Y\gt z$. This probability is the area of the part of our triangle that is above the line $y=z$. The "base" of that triangle has size $1-\frac{z}{2}$. So by similarity its area is $\frac{\left(1-\frac{z}{2}\right)^2}{1^2}$ times the area of the full triangle. Thus if $1\le z\lt 2$ then $$F_Z(z)=1- \left(1-\frac{z}{2}\right)^2.$$
Now we find $F_Z(z)$ for $0\lt z\lt 1$. Again, $\Pr(Z\gt z)$ is easier to get at. Look at the picture. There is a term $\left(1-\frac{z}{2}\right)^2$, obtained like earlier. Another way we can have $Z\gt z$ is if $Y\le z$ but $X\gt z$. That happens if $(X,Y)$ lands in the rectangle of height $z$ with base running from $(z,0)$ to $1,0)$. This base has length $1-z$. It follows that if $0\le z\lt 1$ then $$F_Z(x)=1-\left(1-\frac{z}{2}\right)^2-z(1-z).$$ For completeness, note that $F_Z(z)=0$ if $z\le 0$, and $F_Z(z)=1$ if $z\ge 2$.