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Probability distribution and cumulative distribution function q's

  1. Apr 5, 2013 #1
    1. The problem statement, all variables and given/known data
    $$P_x(x)=A(1- \frac{|x|}{2}) \ \ \ |x|≤2$$
    $$P_x(x)=0 \ \ |x|>2$$

    Find A

    2. Relevant equations



    3. The attempt at a solution

    This one shouldn't be too bad but I wanted to verify that im on the right track.

    I basically have ##P_x(x)=A(1- \frac{x}{2})## when x≤2. 0 otherwise.

    So I write the integral ##\int_{-\infty}^{2} A(1-\frac{|x|}{2})dx##

    I end up with ##A (x(\frac{-x^2}{4})dx## evaluated from 2 to ##-\infty## wich will equate to ##\infty##

    The section in my text that talks about probability distribution function mentions this
    $$\int_{-\infty}^{\infty} p_x(x)dx=1$$

    So I could do ##A\int_{-\infty}^{\infty} 1-\frac{|x|}{2} dx=1##? Solving this would yield A=##\infty## which is not what im looking for I dont think. Its something to do with the absolute value of x i think...

    The form that the question writes the equation in always screws me up.

    Thanks for any help
     
  2. jcsd
  3. Apr 5, 2013 #2

    SteamKing

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    Your limits of integration do not match the domain specified for P(x). Look carefully. The ABSOLUTE VALUE of x is less than or equal to 2.
     
  4. Apr 5, 2013 #3
    How about something like this ##\int_{-2}^{2}A(1-\frac{|x|}{2}=1##

    ##\frac{1}{A}=\int_{-2}{2}(1-\frac{|x|}{2}=4##

    so ##A=\frac{1}{4}##?
     
  5. Apr 5, 2013 #4

    SteamKing

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    That looks much better.
     
  6. Apr 6, 2013 #5
    Thanks again for your help!

    I think i actually made a mistake with the absolute value. A would equal ##A=\frac{1}{2}## not ##\frac{1}{4}##

    I have another question about this problem.

    The second part is to find the CDF (cumulative distribution function) for x.

    From what I understand the pdf is ##p_x(x)## as the problem states. The CDF is ##P_x(x)##, and ##P_x(x)= \int p_x(x)##?

    Isnt that what Ive already done to find the constant A?

    So ##P_x(x)=\frac{1}{2}\int_{-2}^2(1-\frac{|x|}{2})dx##

    Isnt this what Ive already done?
     
  7. Apr 6, 2013 #6
    Ok so I did a little more digging/reading and found a few video examples too. I found this one to quite helpful Cumulative Distribution Functions : Introduction

    So to find ##P_x(x)## I did the following

    $$P_x(x)=\frac{1}{2}\int_{-2}^{x}(1-\frac{|x|}{2}) \\
    =\frac{1}{2}|_{-2}^{x} \ x-\frac{x-|x|}{4} \\
    =\frac{1}{2}[x-\frac{x^2}{2}-1]\\
    =\frac{-(x^2-4x+4)}{8}
    $$
    So my CDF is

    $$\begin{cases}0 \ \ \ \ x<-2\\ \\
    \frac{-(x^2-4x+4)}{8} \ \ \ \ -2≤x≤2\\ \\
    1 \ \ \ \ x>2 \end{cases}$$

    Does this look somewhat correct? I got a little confused when it came to values of x greater then 2 and the probability being 1. I guess I would have expected the probability to be 0 as the value is o above that upper limit...
     
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