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Probability random variables

  1. Oct 24, 2007 #1
    1. The problem statement, all variables and given/known data

    A random variable has a distribution function F(z) given by
    F(z) = 0 if z< -1
    F(z) = 1/2 if -1 <= z < 2
    F(z) = (1-z^{-3}) is 2 <= z

    Find the associated mean and variance.

    3. The attempt at a solution
    I drew the distribution function. I started with the associated mean (if I can figure that out the variance should follow.) I have:

    E[Z] = [tex] \sum [/tex] zp(z)

    p(z) = P[X = z]

    Therefore,
    p[X = -1] = P[X= -1] - P[X<-1]
    = F(-1) - lim F(1-1/n)
    = 1/2 - (1-2)
    = 3/2

    Sorry, if I messed up badly somewhere. The class is taught without a book and I can't seem to get anything out of my notes for this homework. Thanks.
     
    Last edited: Oct 24, 2007
  2. jcsd
  3. Oct 24, 2007 #2
    Your notation is confusing to me, but for starters 1/2 - (1-2) isn't -1/2.
     
  4. Oct 24, 2007 #3
    Sorry, for the bad arithmetic. That is fixed. The problem statement is a piecewise function. As for the rest of the notation. I don't know how else to write it. My professor said this usually isn't presented in textbooks in this manner but he thinks it is a good way to do it. So the notation may be quite strange. If there is a specific part I might be able to explain what I am doing.
     
  5. Oct 24, 2007 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    What are you adding over? The way you have written F it looks like a piecwise function and you should be integrating, not adding:
    [tex]E[Z]= \int_{-1}^\infty zF(z)dz= \int_{-1}^2 \frac{z}{2}dz+ \int_2^\infty (z- z^{-2}dz[/tex]
    Unfortunately, it looks to me like that last integral won't converge. Are you sure it wasn't [itex]F(z)= (1- z)^{-3}[/itex] if z> 2?

    If F is only defined for integer z (then you should have told us that),
    [tex]E(Z)= \sum_{-1}^\infty nF(n)= \sum_{-1}^2 \frac{n}{2}+ \sum_2^\infty (n+ n^{-2})[/tex]
    Again, there is a problem with the final sum: it doesn't converge.
     
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