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Finding the median of a distribution

  1. Feb 8, 2015 #1
    1. The problem statement, all variables and given/known data
    Suppose X has the Uniform (0,1) distribution. Find the median of the distribution of e^X correct to 2 decimals.

    2. Relevant equations
    F(X) = 0.5
    F(X) = ∫f(x)

    3. The attempt at a solution
    I am not entirely sure what to do here, I know to find the median you need to find when F(X) = 0.5. I assumed that e^X was just the cdf f(x) = e^X which makes F(x) still e^X.

    So then I used e^(X) = 0.5 which gives X = ln(0.5).

    However this is not correct. I am a bit confused about the first part of the question, how X being a uniform distribution would affect it.

    Any help would be appreciated

    Thanks
     
    Last edited: Feb 8, 2015
  2. jcsd
  3. Feb 8, 2015 #2

    Mark44

    Staff: Mentor

    If you think about it a bit, this can't be right, since X would have to be a negative number. If X ranges from 0 to 1, then e^X will range from ? to ?
     
  4. Feb 8, 2015 #3
    I found the answer to be e^0.5, but im not sure exactly why. Do we find the median of the uniform distribution of X (0.5) and then plug that into e^(X)? Or do I need to find the median of the distribution between 1 and e? If it's the latter, how do we know that the median is simply e^(0.5)? Are we able to say this because X is a uniform distribution?
     
  5. Feb 8, 2015 #4
    I found the answer to be e^0.5, but im not sure exactly why. Do we find the median of the uniform distribution of X (0.5) and then plug that into e^(X)? Or do I need to find the median of the distribution between 1 and e? If it's the latter, how do we know that the median is simply e^(0.5)? Are we able to say this because X is a uniform distribution?
     
  6. Feb 8, 2015 #5

    Ray Vickson

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    If ##Y = e^X##, the cumulative distribution of ##Y## is ##F(y) = P(Y \leq y) = P(e^X \leq y)##. For ##X \sim U(0,1)##, what does this give you for ##F(y)##? What is the solution of ##F(y) = 1/2##? That would give you the median.
     
  7. Feb 8, 2015 #6
    I am unsure of how to continue after
    For ##X \sim U(0,1)##, what does this give you for ##F(y)##? Am I supposed to see that X = ln(y) or am I supposed to see that Y ~ U(1,e)? Or am I missing something altogether?
     
  8. Feb 8, 2015 #7

    haruspex

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    2016 Award

    Some risk of confusion in the notation. That X = ln(Y) is self evident, so I'm assuming your question is why x' = ln(y'), x' and y' representing the medians of the respective distributions.
    As Ray wrote, for any x, the event Y < ex is the same event as X < x. Therefore FY(ex) = FX(x). Setting x=x' we get FY(ex') = 0.5.
    That is certainly not the case.
     
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