Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

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

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.