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Statistics: variable transformation proof?

  1. Mar 3, 2014 #1
    1. The problem statement, all variables and given/known data
    Ok this might be a stupid question, but:

    https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-frc3/t31/q77/s720x720/10001118_10202561443653973_1625797585_o.jpg
    Why is this the case? I think for all of this to be right, then the assumption of ##Y=u(X) \Leftrightarrow y=u(x)##. But how do I prove this?

    3. The attempt at a solution

    An attempt: OK, assume ##X## has a probability distribution given by ##f(x)##, and ##Y=2X## has a probability distribution of ##g(x)##. Then if ##Y = u(X)##, with w being the inverse one-on-one function of u, ##P(X=x) = P(w(Y)=x) = P(Y=u(x))=P(Y=y=u(x))##.

    This is as far as I got, but now I am confused. What to do?
     
  2. jcsd
  3. Mar 3, 2014 #2

    micromass

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    You're almost there. Note by definition that

    [tex]g(y) = P\{Y = y\}[/tex]

    and you must prove that

    [tex]g(y) = f(w(y))[/tex]

    which is by definition

    [tex]f(w(y)) = P\{X = w(y)\}[/tex]

    So you must somehow find why

    [tex]P\{Y = y\} = P\{X=w(y)\}[/tex]
     
  4. Mar 3, 2014 #3
    OK, I see. Thanks for the help :)
     
  5. Mar 3, 2014 #4

    Ray Vickson

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    You already got the answer below, but just to clarify: the one-to-one assumption is crucial. If you had a non-monotone function, such as ##u(x) = x^2##, and if the range of ##X## included both positive and negative values, then the mapping ##Y = u(X)## might not be 1:1, and so you might need a more complicated evaluation of ##P(Y = y)##.
     
    Last edited: Mar 3, 2014
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