yes, I mean the cumulative distribution function.

if I integrate from 0 to 2 (for the first piece)
I get : (1/8)(x^2) + C        0<=x<2

if I integrate from 2 to 4 (for the second piece)
I get :  -(1/8)(x^2) + x + C    2<=x<=4

I think I am doing something wrong because the integral is greater
than one in most places (if for example let x=3) and the distribution
function (cdf) cannot be greater than one. Also, I have to a
decreasing function of x (which I know is wrong since the cdf is a
nondecreasing function of x)
I am expecting that afer x=4 the cdf=1

I think I migh be missing something I just haven't been able to see
what it is.
thanks

P.S which other info migh you need?



[EMAIL PROTECTED] (Donald Burrill) wrote in message 
news:<[EMAIL PROTECTED]>...
> On 3 Nov 2001, Gilbert wrote:
> 
> > If I have a density function defined as:
> > 
> >           f(x)=(1/4)x           0<=x<=2
> >           f(x)=-(1/4)x + 1      2<x<=4
> >           f(x)=0                 elsewhere  
> > 
> > (so the density function is a triangle of height (1/2)) 
> > 
> > how do I find the distribution function of x from here?
> > 
> > I tried integrating to get the distribution function but I think I am
> > doing something wrong.  any clues?
> 
> One presumes you mean the cumulative distribution function (cdf) ?
> 
> You have not provided enough clues for an answer (not enough for me, 
> anyway).  Integrating would seem to be appropriate.  Why do you think you 
> are "doing something wrong"?  That provides no detail whatever, and one 
> cannot base a diagnosis on zero information.
> 
>  ------------------------------------------------------------------------
>  Donald F. Burrill                                 [EMAIL PROTECTED]
>  184 Nashua Road, Bedford, NH 03110                          603-471-7128
> 
> 
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