How Do You Convert a PDF to a CDF Using Integration?

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To convert a probability density function (PDF) to a cumulative distribution function (CDF), one must integrate the PDF over its defined intervals. The given function has different expressions for specific ranges of y, requiring piecewise integration. For the interval 0 <= y <= 1/2, the integral results in y^2, while for 1/2 < y <= 1, the integration yields 6y - 3y^2 - 2. The integration from negative infinity to zero accounts for the PDF's behavior outside the defined intervals, ensuring the CDF is properly defined. Understanding the piecewise nature of the PDF is crucial for correctly calculating the CDF.
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I"m not understanding CDF...

My textbook doesn't 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).

answer is y^2.

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 doesn't 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 don't know how to approach those problems..
 
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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.
 
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