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How can I find the CDF and PDF of Y?

  1. Oct 6, 2015 #1
    Problem

    Let X be a uniform(0,1) random variable, and let Y=e^−X.
    Find the CDF of Y.
    Find the PDF of Y.
    Find EY.

    Relevant Equations
    5b04f0e6f9.png
    http://puu.sh/kAVJ8/0f2b1e7b22.png [Broken]


    My attempt at a solution

    If I solve for the range of y I get (1, 1/e), but because Y is not an increasing function, my second bound is smaller than my first. I am really confused as to how I would be able to solve for the CDF and PDF in this case... Any help would be greatly appreciated.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Oct 6, 2015 #2

    RUber

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  4. Oct 6, 2015 #3

    mathman

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    To get CDF:
    [itex]P(Y<y)=P(e^{-X}<y)=P(-X<lny)=P(X\ge -lny)=1+lny[/itex]
    PDF = [itex]\frac{1}{y}[/itex]
     
  5. Oct 6, 2015 #4
    This is very helpful. Thanks. However, would the CDF be 0 for y < 1 and 1 for y > 1/e? This part does not really make sense.
     
  6. Oct 6, 2015 #5

    RUber

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    Mathman's formula is pretty clear for the cdf. It is equal to 0 at y=1/e, and 1 at y=1, and increasing in between.
     
  7. Oct 11, 2015 #6

    gill1109

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    ... which is just what we expect since X certainly lies between 0 and 1, hence Y = exp(-X) lies between 1/e and 1. The density of Y is zero left of 1/e and right of 1. And in between it's the derivative of mathman's cumulative distribution function, hence 1/y.

    Check: 1/y integrates to 1 when you integrate it from 1/e to 1
     
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