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Discrete Random Variable

  1. Oct 11, 2011 #1
    Let X be a discrete random variable that can assume the values -1, 0,1,2,3,4 with the probabilities 1/6, 1/12, 1/6, 1/4, 1/12, 1/4. Find the probability densities of the following random variables:

    a) Z= X^2 + 1

    h(y)= f(g^-1(y))


    Attempted Solution
    X= -1 0 1 2 3 4
    Z= 2 1 2 5 10 17

    How do i find Pr(Z)?
    The book gave an answer of
    1/12 Z=1,10
    1/4 Z= 5,17
    1/3 Z=2
    0 elsewhere.
     
  2. jcsd
  3. Oct 11, 2011 #2
    When X= -1, Z=2. So the probability of Z=2 is Pr(x=-1)= 1/6
    When X = 0, Z= 1. So the probability of Z=1 is Pr(x=0)= 1/12
    and so on

    Notice that when X= 1, Z=2 again. Thus, the probability of Z=2 is Pr(x= -1) plus Pr(x=1) which equals 1/3.
     
  4. Oct 11, 2011 #3
    Incidently, are you using Tsokos' new book? The masterpiece of typos?
     
  5. Oct 11, 2011 #4
    Ok thank you.
    Let me know if I'm thinking about this correctly.

    Z=5 only X=2 can produce this so Pr= 1/4
    Z=10 only X=3 can produce this so Pr=1/12
    Z= 17 only X=4 can produce this so Pr=1/4.
     
  6. Oct 11, 2011 #5
    Lol yes i am.
     
  7. Oct 12, 2011 #6
    Yes, this is correct.
    Do you have a test Thursday? I'm wondering if you are in my class, since this book is so new. 6:30 Tuesdays and Thursdays?
     
  8. Oct 12, 2011 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    I hope your book and your instructor do not use the terminology "probability density" in this case. Discrete random variables do not have probability densities; they have distribution functions and probability mass functions, but not densities.

    RGV
     
  9. Oct 12, 2011 #8
    No I have an exam on Friday. I take it MWF 12:55 but it may be for the same professor our exam covers Chpts. 3 and 4!
     
  10. Oct 12, 2011 #9
    Yup, must be the same guy. :)
     
  11. Oct 12, 2011 #10
    Oh he uses the term "probability density" all right. This book is a joke.
     
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