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Uniform distribution- probabilities

  1. Mar 21, 2015 #1
    Hello, I am stuck at this exercise:

    1. The problem statement, all variables and given/known data

    X ~ U(0, a), a > 0 and Y = min(X; a=2).
    - Find the cumulative distribution function of Y
    -Is the variable Y continuous ?

    2. Relevant equations
    3. The attempt at a solution

    The density function for X is
    f(t)= 1/a if 0≤t≤a , 0 elsewhere
    Is it correct to write that :
    ∀t < 0, P(Y ≤ t) = 0
    ∀0 ≤ t < a/2, P(Y ≤ t) = P(X ≤ t) = t/a
    ∀t ≥ a/2, P(Y ≤ t) = 1

    And then don't know what to do
    Thanks
     
  2. jcsd
  3. Mar 21, 2015 #2

    haruspex

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    I'm not familiar with the notation min(X; a=2). What does it mean?
     
  4. Mar 21, 2015 #3

    Ray Vickson

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    You have an '##a##' in the definition of ##X## itself, and an '##a##' in the "definition" of ##Y## in terms of ##X##. Are they supposed to be the same '##a##' in both places? If so, I cannot make any sense out of what you wrote.

    On the other hand, if you really mean that ##Y = \min(X,2)##, then that would have meaning. In that case it is important to distinguish between the two cases ##0 < a \leq 2## and ##a > 2##.
     
  5. Mar 21, 2015 #4
    Oh sorry, I didn't notice that there is a mistake in the expression of Y

    It is Y=min(X; a/2)
     
  6. Mar 21, 2015 #5

    haruspex

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    In that case your CDF for Y is correct. Draw it. Is it continuous?
     
  7. Mar 21, 2015 #6
    No, it is not continuous !
    But how to find the cumulative distribution function ?
    Thanks !
     
  8. Mar 21, 2015 #7

    haruspex

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    The CDF is what you wrote in the OP. You specified P(Y<=t) for all three ranges of t.
     
  9. Mar 21, 2015 #8
    Oh, right ! I don't know why I thought that I had to find something else :eek:
    Thanks for your answers !
     
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