Help with the Continuous Distribution Function (CDF)?

#1
Hey, new to these forums, I was wondering if you guy could help me with a question about the CDF I have. Basically I'm wondering if this CDF:

G(t)=
{0 , t<0
t^2 , 0<=t<1
t , 1<=t<2
1 , t>=2 }

The same as this one:

G(t)=
{0 , t<0
t^2+t , 0<=t<2
1 , t>=2 }
?

Thanks in advance :)
 

Dason

Ambassador to the humans
#2
Neither of those is a CDF. And they aren't equivalent. So I guess the answer is "no" any way you look at it.
 
#3
Sorry, it's just an example, I know it doesn't actually qualify as a CDF per se, I was just wondering if the idea worked... I'm try to figure out a question using the "CDF technique" where I have f(x,y), 0<x<1, 0<y<1, and I'm trying to find g(t). So I separated it into different parts:

t<=0: G(t)=P[X+Y<t]=0
t>=2: G(t)=P[X+Y<t]=1
0<t<1: G(t)=?
1<=t<2: G(t)=?

but then when I take the derivative of G(t) I'll get two functions (one from 0 to 1, one from 1 to 2) for my g(t). Any advice?

Thanks!