# Integration to find pdf/cdf

• Mar 11th 2010, 01:07 PM
jpquinn91
Integration to find pdf/cdf
I have a pdf:

0 x<=4
kx 4<x<=9
0 x>9

I found k=2/65

So I thought the cdf would be (x^2)/32 but apparently not

• Mar 11th 2010, 01:11 PM
Moo
Hello,

For $t<4$, the cdf is 0.

For $t>9$, the cdf is 1.

For $t\in[4,9]$, the cdf is $\int_4^t kx ~dx$

Compute that again :)
• Mar 11th 2010, 01:21 PM
jpquinn91
ok thanks i've got that now. So for a cdf where

0<x<4 & 4<x<8 have different functions, what limits would i use?
• Mar 11th 2010, 01:48 PM
jpquinn91
would the limits just be 0 - t and 4 - t?
• Mar 12th 2010, 02:49 PM
jpquinn91
can someone please help me as I am almost sure that the limits for 0 to 4 and 4 to 8 would be either 0 to t and 4 to t or 0 to t and t to 8 but I dont know which 1, thanks
• Mar 12th 2010, 03:50 PM
matheagle
from 0 to 4, there is no probability
I don't know why you keep writing 8, but the probabilities exist between 4 and 9.

Hence F(x)=0 for any x less than 4, since f(t)=0 in that region..........

$F(x)=\int_{-\infty}^xf(t)dt=\int_{-\infty}^x0dt=0$

If 4<x<9, then $F(x)=\int_{-\infty}^xf(t)dt=\int_{-\infty}^40dt+\int_4^x ktdt$

and after 9, we have

$F(x)=\int_{-\infty}^xf(t)dt=\int_{-\infty}^40dt+\int_4^9 ktdt+\int_9^x 0dt=0+1+0=1$