# CDF's integration

• Apr 10th 2011, 06:39 AM
tickman
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?
• Apr 10th 2011, 08:32 AM
theodds
Just take the derivative and get $f(x) = \frac{x^{n-1}}{n}$, valid on [0, 1].

X and Y are both continuous random variables, so $P(X + Y = z) = 0$ for all z. It would be more meaningful to ask for the distribution of $X + Y$, which involves a little bit of work.
• Apr 11th 2011, 12:48 PM
matheagle
If I recall correctly f(z)=z on (0,1) and 2-z on (1,2)
• Apr 12th 2011, 10:46 AM
Moo
Please I'm kind of desperate :D I've tried to do the calculations, but can't find where my mistake is !

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

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

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

So what's wrong ? :(
• Apr 12th 2011, 10:56 AM
theodds
You forgot to deal with the case where $z - y \in (1, 2)$ I think.
• Apr 12th 2011, 11:33 AM
Moo
Well the probability would be 0, wouldn't it ?
• Apr 12th 2011, 01:43 PM
theodds
Quote:

Originally Posted by Moo
Well the probability would be 0, wouldn't it ?

More like 1 minus that :)
• Apr 12th 2011, 01:54 PM
Moo
Quote:

Originally Posted by theodds
More like 1 minus that :)

Oh yeah right, now I can see through it ! Thanks !
• Apr 12th 2011, 03:41 PM
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.
• Apr 12th 2011, 04:00 PM
Moo
Quote:

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 (Giggle)
• Apr 13th 2011, 09:53 AM
tickman
so its 0 or 1?