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Finding CDF for Continuous RVs problem

  1. Mar 29, 2012 #1
    1. The problem statement, all variables and given/known data
    I'm given the pdf and asked to find F(y)/ cdf. I've calculated it many times, but I'm not getting the right numbers. the pdf is

    f(y)= .5, .............-2≤y≤0
    .75-.25y,.....1≤y≤3
    0,................elsewhere

    so that means

    f(y)= 0,............y< -2
    0.5, ..........-2≤y≤0
    0,................0<y<1
    .75-.25y,.....1≤y≤3
    0,................ y>3

    right?

    So, my CDF is

    F(y)= c1,................................ y<-2
    .5y +c2,......................... -2≤y≤0
    c3, ...................................0<y<1
    .75y- .125y^2+ c4,........... 1≤y≤3
    c5,......................................y>3

    so, I found the constants, but I'm not getting c5=1.

    0=c1=.5y+c2 at y=-2
    so 1=c2


    .5y+1=c3 at y=0
    so 1=c3


    1=c3=.75y- .125y^2+ c4 at y=1
    1= .75- .125+ c4
    1-.625= .375=c4


    .75y-.125y^2+ .375= c5 at y=3
    .75*3 -.125(9) +.375= c5
    2.25 - 1.125 + .375= 1.5=c5≠1


    I don't know where I'm going wrong. Could you show me?? -_- Much thanks.


    I hope this is legible....
     
  2. jcsd
  3. Mar 29, 2012 #2
    Assuming you've given the right problem statement, I think the question is just flawed. Take a step back from calculus for a bit and consider you are just accumulating area. The function is zero before y = -2, so at that point the accumulation is zero. You then accumulate a rectangle of length 2 (from -2 to 0) and height .5. The area accumulated it 2*.5 = 1. You've already gotten 1 area... so the fact you will be accumulating more is a problem.
     
  4. Mar 29, 2012 #3
    @RoshanBBQ,

    Yeah, I looked at it from a geometric pov and it came out to be more than 1, but I've worked this problem many times, with a calculator and without... being careful with the numbers and such... I've just e-mailed my professor, so we shall see D:.... Thanks for your quick response!
     
  5. Mar 29, 2012 #4

    Ray Vickson

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    Homework Helper

    For any z > -2 the CDF is [itex]F(z) = \int_{-2}^z f(y) \, dy,[/itex] so there is no need to introduce arbitrary constants of integration. However, your f(y) integrates to 3/2 > 1, so is not a legitimate probability density function.

    RGV
     
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