Let the point $(u, v)$ be chosen uniformly from the square $0\leq u\leq 1$, $0\leq v\leq 1$. 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 $$F(x) = \left\{\begin{align}0,&\quad\text{when}\quad x<0\\ \tfrac{x^2}{2},&\quad\text{when}\quad0\leq x<1\\ -1+2x-\tfrac{x^2}{2}&\quad\text{when}\quad 1\leq x\leq 2\\ 1,&\quad\text{when}\quad x>2\\ \end{align}\right.$$


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 $\Pr(W\le w)=1-(2-w)^2/2$.

  • $\begingroup$ Thank you a lot !! :) However I couldn't imagine the line :/ $\endgroup$ – idobi182 Oct 29 '12 at 17:58
  • $\begingroup$ 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. $\endgroup$ – André Nicolas Oct 29 '12 at 18:27
  • $\begingroup$ Shouldn't I consider 3 dimension ?? $\endgroup$ – idobi182 Oct 29 '12 at 18:29
  • $\begingroup$ 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 $X$. $\endgroup$ – André Nicolas Oct 29 '12 at 18:33
  • $\begingroup$ Now I understood.Thank you for your patience :) $\endgroup$ – idobi182 Oct 29 '12 at 18:42

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