Find the cdf given a pdf with absolute value

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Homework Help Overview

The discussion revolves around finding the cumulative distribution function (CDF) for a continuous random variable X, given its probability density function (PDF) defined as fX(x) = |x|/5 for the interval –1 ≤ x ≤ 3, and zero elsewhere.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the problem, focusing on the need to consider different cases for the CDF based on the value of x. There are attempts to compute integrals for the CDF, with some participants expressing confusion about the results yielding constants instead of a function of x.

Discussion Status

Some participants have provided guidance on how to approach the problem by suggesting the need to break it into cases based on the value of x. There is acknowledgment of the necessity for multiple integrals depending on the range of x being considered.

Contextual Notes

Participants note that the problem requires careful consideration of the intervals for x, specifically addressing the cases when -1 ≤ x ≤ 0 and x > 0. There is an emphasis on ensuring the correct setup of integrals for each case.

aquaelmo
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Homework Statement


Consider a continuous random variable X with the probability density function fX(x) = |x|/5 , – 1 ≤ x ≤ 3, zero elsewhere.
I need to find the cumulative distribution function of X, FX (x).

2. Homework Equations

The equation to find the cdf.

The Attempt at a Solution


FX(x) = ∫-1x -u/5 du + ∫-10 -u/5 du + ∫0x u/5 du

For some reason, my result is just a constant, but I can't figure out why my equation is wrong?
 
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aquaelmo said:

Homework Statement


Consider a continuous random variable X with the probability density function fX(x) = |x|/5 , – 1 ≤ x ≤ 3, zero elsewhere.
I need to find the cumulative distribution function of X, FX (x).

2. Homework Equations

The equation to find the cdf.

The Attempt at a Solution


FX(x) = ∫-1x -u/5 du + ∫-10 -u/5 du + ∫0x u/5 du

For some reason, my result is just a constant, but I can't figure out why my equation is wrong?
You have to do two cases. First take ##-1\le x \le 0## and work that. You will just need one integral. Then take ##x>0## and work that, which will take two integrals with ##x## only in the second one, etc.
 
Oh I understand, the solution will have two cases. Thank you!
 
aquaelmo said:
I believe that's what I'm doing.
For case -1 ≤ x ≤ 0, I compute the integral ∫-1x -u/5 du.
For the case x > 0, I compute the area of the first case, ∫-10 -u/5 du, then the second case, ∫0x u/5 du
No, the second case would be ##\int_{-1}^0 -\frac u 5~ du + \int_0^x \frac u 5 ~du##.
 

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