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Let the point (u, v) be chosen uniformly from the square 0<=u<=1, 0<=v<=1. Let X be the random variable that assigns to the point (u, v) the number u+v. Find the distribution function of X.

Now can I say that X=u+v and X distributed uniformly between 0 and 2 then should I find the cdf of the uniform(0,2)?

Or should I consider taking 2 integral 1 for u and 1 for v? I know the answer but I want to know the steps. :/

The answer is F(x) = 0 when x<0 xsquare/2 when 0<=x<1 -1+2x-(1/2)xsquare when 1<=x<=2 1 when x>2

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1 Answer 1

Call our random variable $W$ instead of $X$. This is because it is useful to give $x$ and $y$ their traditional geometric meanings.

We want the probability that $W\le w$. Draw the line $x+y=w$. Then $\Pr(W\le w)$ is the area of the part of the square which is "below" the line. This is clearly $0$ if $w\lt 0$, and $1$ if $w\gt 2$. Sketch several such lines, say for $w=1/2$, $w=1$, and $w=3/2$.

For $w\le 1$, the part of the square below the line is just a right triangle whose "legs" are $w$ and $w$. This is because the line meets the $x$-axis at $x=w$, and the $y$-axis at $y=w$. The area of this triangle is $w^2/2$. So if $0\le w\le 1$, then $\Pr(W\le w)=w^2/2$.

For $1\lt w\le 2$, the part of the square below the line has area $1$ minus the area of the part of the square above the line. This part is just a triangle, with easily computed area. For let us find the legs of this triangle. If $w$ is between $1$ and $2$, the line $x+y=w$ meets the line $x=1$ at $y=w-1$. So the triangle has legs $1-(w-1)=2-w$, and therefore area $(2-w)^2/2$. So if $w$ is between $1$ and $2$, then $\Pr(W\le w)=1-(2-w)^2/2$.

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Thank you a lot !! :) However I couldn't imagine the line :/ –  idobi182 Oct 29 '12 at 17:58
    
I do not know what :/ means. Fix $w$, like $w=0.6$. Then the line $x+y=0.6$ is a "diagonal" line with slope $-1$ that meets the $x$-axis at $(0.6,0)$ and the $y$-axis at $(0,0.6)$. The geometry is a bit different if, say, $w=1.3$. Drawing a couple of pictures helps a lot. –  André Nicolas Oct 29 '12 at 18:27
    
Shouldn't I consider 3 dimension ?? –  idobi182 Oct 29 '12 at 18:29
    
One can think of the problem as involving three variables, but what we have done is to fix $w$, and found an explicit formula for $\Pr(W\le w)$. So we have obtained the (cumulative) distribution of $W$, which is exactly what the question asked for, apart from calling the random variable by the name $X$. –  André Nicolas Oct 29 '12 at 18:33
    
Now I understood.Thank you for your patience :) –  idobi182 Oct 29 '12 at 18:42

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