1. ## CDF's integration

how would i find the density function which generates the cdf F(x) = x^(N):

Suppose that x and y are two independent random variables, each of which is uni-
formly distributed on the unit interval [0; 1]. Find the probability that x + y =z for
any z between 0 and 2?

2. Just take the derivative and get $\displaystyle f(x) = \frac{x^{n-1}}{n}$, valid on [0, 1].

X and Y are both continuous random variables, so $\displaystyle P(X + Y = z) = 0$ for all z. It would be more meaningful to ask for the distribution of $\displaystyle X + Y$, which involves a little bit of work.

3. If I recall correctly f(z)=z on (0,1) and 2-z on (1,2)

4. Please I'm kind of desperate I've tried to do the calculations, but can't find where my mistake is !

$\displaystyle P(X+Y\leq z)=E[P(X+Y\leq z|Y)]$

And $\displaystyle P(X+Y\leq z| Y=y)=z-y$, if $\displaystyle z-y\in(0,1)$, that is $\displaystyle y\in(z-1,z)$ ; 0 otherwise.

- if z<1, there's no problem, I find zē/2 as the cdf
- if z>1, $\displaystyle \displaystyle P(X+Y\leq z)=E[(z-Y)\bold{1}_{y>z-1}]=\int_{z-1}^1 z-y ~dy=2z-z^2-\frac 12+\frac{(z-1)^2}{2}$
by differentiating, it gives the pdf on this part : 1-z, which is negative...

So what's wrong ?

5. You forgot to deal with the case where $\displaystyle z - y \in (1, 2)$ I think.

6. Well the probability would be 0, wouldn't it ?

7. Originally Posted by Moo
Well the probability would be 0, wouldn't it ?
More like 1 minus that

8. Originally Posted by theodds
More like 1 minus that
Oh yeah right, now I can see through it ! Thanks !

9. It's easier to use geometry and see the triangles,
including the ones for the complement when 1<z<2
This is a uniform distribution on the unit square.

10. Originally Posted by matheagle
It's easier to use geometry and see the triangles,
including the ones for the complement when 1<z<2
This is a uniform distribution on the unit square.
I like conditional stuff and dislike graphic stuff, hence the tortuous method

11. so its 0 or 1?