# I m not understanding CDF

1. Nov 22, 2004

### semidevil

I"m not understanding CDF...

My textbook doesnt seem to give enough information on it.

I"m not understanding what do to when involved with CDFs in general.

So if I want to convert a pdf to cdf, given a function
fy(Y) =

0 for y <0
2y for 0 <= y <= 1/2
6 - 6y for 1/2 < y <= 1
0 for y > 1

so the book does this:
for 0 <= y <= 1/2, it integrates (0 dt from -infinite to 0) + the integeral of (2dt from 0 to y).

then, for 1/2 < y <= 1, it does the integeral of 0dt from -infinite to 0,+ the integeral of 2dt from 0 to 1/2dt, + the integeral of 6 - 6t from 1/2 to y.

answer is 6y - 3y^2 - 2.

and y > 1, = 1.

so we get the new cdf.

the book doesnt give any explanation of what it did. So what just happened here? how did they do all that just by looking at the function? why did they integrate from - infinite of 0, and then from 0 to y, and then 0 to 1/2.

well, you get my point...I dont know how to approach those problems..

2. Nov 23, 2004

### matt grime

Given a PDF, f, the associated CDF is usually F(k):= P(X<k), which is exactly the integral from -inf to k of f. They split F into several parts in this example because the PDF is defined in several parts.