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Distribution of a Function of a Random Variable

  1. Nov 18, 2014 #1
    1. The problem statement, all variables and given/known data
    If X is uniformly distributed over (0,1), find the PDF of Y = |X| and Z = e^X

    Focusing on the |X| one

    2. Relevant equations
    Derivative of CDF is the PDF

    3. The attempt at a solution
    So I start by writing down the CDF of X, Fx(x):

    0 for x <0
    x for 0 ≤ x ≤ 1
    1 for x ≥ 1

    And I also know the range of Y = |X| is [0,1]
    Finding the CDF of Y:

    FY(y) = P(Y≤y)
    =
    P(|X|≤y)
    =
    P(-y ≤ X ≤ y)
    =
    Fx(y) - Fx(-y)

    So
    FY(y) = Fx(y) - Fx(-y)

    But the range of Y is [0,1] so I can get rid of the Fx(-y) and have

    FY(y) = Fx(y) = y

    So the CDF of Y is then

    0 for y < 0
    y for 0 ≤ y ≤ 1
    1 for y≥1

    To find the PDF, I can just take the derivative and have

    1 for 0 ≤ y ≤ 1
    0 otherwise

    Is this correct?
     
  2. jcsd
  3. Nov 18, 2014 #2

    Ray Vickson

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    You can simplify the argument by noting that, for X ~ Unif(0,1), we have |X| = X for all nonzero values of X; this just eliminates any need to look at the zero-probability values X < 0 or X > 1. (However, to a certain extent, the way you should do it depends on how your textbook or course notes do it.)
     
  4. Nov 18, 2014 #3
    Is the way I approached it correct though?
    Like for example, if I changed the problem statement so that X is uniformly distributed over (-5,5) and I want to find the PDF of Y= |X|, then the range of Y would be [0,5].

    The CDF of X is

    0 for x < 0
    x for -5 ≤ x ≤ 5
    1 for x > 5

    Repeating the process I did before, the CDF of Y would be

    0 for y < 0
    y for 0 ≤ y ≤ 5
    1 for y > 5

    Then the PDF would be
    1 for 0 < y < 5
    0 otherwise
     
  5. Nov 18, 2014 #4

    LCKurtz

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    Gold Member

    That is obviously wrong because the area under its graph is 5, not 1. You have the wrong distribution at the beginning for a uniform distribution on (-5,5).
     
  6. Nov 18, 2014 #5
    Oh oops, I meant to write:

    The CDF of X is

    0 for x < -5
    x for -5 ≤ x ≤ 5
    1 for x > 5

    Is that correct?

    And then I'm not quite sure I understand what you mean by "the area under its graph is 5, not 1".
    How would I find the PDF of Y?

    If I find the CDF first:
    FY(y) = P(Y≤y)
    =
    P(|X|≤y)
    =
    P(-y ≤ X ≤ y)
    =
    Fx(y) - Fx(-y)

    So
    FY(y) = Fx(y) - Fx(-y)

    But the range of Y is [0,5], so

    FY(y) = Fx(y)

    Then the pdf is (taking the derivative)

    fY(y) = fx(y)

    Which isn't 1 I guess; I'm stuck here.
     
  7. Nov 18, 2014 #6

    Ray Vickson

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  8. Nov 18, 2014 #7
    Ok so I drew a picture of X:
    j1dZEcY.png
    Looking at this chart:
    XRRRXWC.png
    The pdf of X for mine must be 1/10 for -5 ≤ x ≤ 5 and 0 otherwise right?
    And the CDF of X is
    0 for x < -5
    (x+5)/10 for -5 ≤ x < 5
    1 for x ≥5
     
  9. Nov 18, 2014 #8
    "Do you know what is meant by the area under a graph? Do you know how to compute it?"

    If the range of Y is [0,5]

    Then the area under the PDF from [0,5] needs to be 1 correct? If I drew a picture similar to the one I drew for X, that would mean the height of the rectangle is 1/5. So is the PDF of Y

    1/5 for 0 ≤ y ≤ 5
    0 otherwise
    ?
     
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