Find Distribution Function of X: f(x) = x/2, 1/2, (3-x)/2

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Homework Help Overview

The discussion revolves around finding the cumulative distribution function (CDF) for a given piecewise probability density function (PDF) of a random variable X. The PDF is defined in three segments across different intervals, and participants are exploring how to integrate these segments correctly to derive the CDF.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to integrate the PDF over specified intervals to find the CDF, questioning the limits of integration and the constants that arise from the piecewise nature of the function. There is also confusion regarding specific terms in the resulting expressions, such as the origin of certain constants.

Discussion Status

Some participants have provided insights into the integration process and the cumulative nature of the CDF. There is an acknowledgment of the need to sum the probabilities from previous intervals, but uncertainty remains regarding the specific steps and calculations for each segment.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the amount of guidance they can seek. The discussion reflects a mix of understanding and confusion about the integration process and the properties of cumulative distribution functions.

kuahji
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Find the distribution function of the random variable X whose probability density is given by

f(x)= x/2 for x < x \leq1
1/2 for 1< x \leq2
(3-x)/2 for 2 < x \leq3
0 elsewhere

Ok, so I know have to take the integral of each piece, but I'm having a hard time figuring out the limits of the integrals.

The book has the answer
F(x)= 0 for x \leq0
x^2/4 for 0 < x \leq1
1/4(2x-1) 1 < x \leq2
1/4(6x-x^2-5) for 2 < x < 3
1 for x\geq3

Which is about what I got when I took the "indefinite" integrals of each piece (again, not sure how to go about setting the limits), but for example in this section 1/4(6x-x^2-5) for 2 < x < 3 where is the -5 coming from? Any help would be appreciated.
 
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I suppose you mean cumulative distribution function? The cumulative distribution function is given as F(x)=\int_{-\infty}^x f(t)dt. Because it's a piecewise function, the constants are coming out of summing all the parts of the density less than the current piece you're integrating over. For the part you appear to have questions with, if you were to find F(x) where 2<x<3, you would find F(3) (definite integral) and then add it to \int_{2}^x f(t)dt

(Post isn't updating quickly for me so I don't know if this turned out right with my rampant typos)
 
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Right, that much was in the text, but I'm still a bit confused.

x^2/4 for 0 < x \leq1 I believe from the original step I just take the integral from 0 to x.

But for the next step, I'm not sure what to do. I mean I know its suppose to add up to one, but figuring out the second and consecutive steps is where I'm having the problem.
 
The cumulative distribution function keeps track of the total probability found between -\infty and x. In a sense it is a "running total". If you want to find F(x) for 1<x<2, you have F(1)=1/4, then F(x)= \int_{-\infty}^x f(t)dt = \int_{-infty}^{1}f(t)dt+\int_{1}^{x}f(t)dt = \frac{1}{4}+\int_{1}^{x}\frac{1}{2}dt = \frac{1}{4}+\frac{x}{2}-\frac{1}{2} = \frac{1}{4}(2x-1)

Edit: fixing typos
 
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Ok, I see how it works now. Thanks for the help, it was really appreciated. ^_^
 

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