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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, voldemort Apr 16 '14 at 22:48

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Dilip Sarwate, Did, mookid, egreg, voldemort
If this question can be reworded to fit the rules in the help center, please edit the question.

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 '14 at 22:32
up vote 0 down vote accepted

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 $z\ge 2$.

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