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Thread: box-cox

  1. #1
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    box-cox

    I have the BC tranformation:

    $\displaystyle y^{(\lambda)}= \left\{ \begin{array}{11}
    \frac{y^\lambda - 1}{\lambda} & \mbox{ \lambda \neq 0$}\\
    log_e (y) & \mbox{ \lambda = 0$}\\
    \end{array} $

    I need to show that this give a continous famly of transformations in $\displaystyle \lambda$.

    How do I go about doing this?
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  2. #2
    Guy
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    I think the question is just asking for you to show that, for fixed y, the function $\displaystyle y^{(\lambda)}$ is continuous in $\displaystyle \lambda$. For $\displaystyle \lambda \ne 0 $ this trivial. To show continuity at $\displaystyle \lambda = 0$ it suffices to show that

    $\displaystyle \displaystyle
    \lim_{\lambda \to 0} y^{(\lambda)} = \log y
    $

    which is easy. The simpliest way of going about this is to define $\displaystyle f(\lambda) = y^\lambda$ and write $\displaystyle y^{(\lambda)} = \frac{f(\lambda) - f(0)}{\lambda - 0}$ which is a form that we all recognize.
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  3. #3
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    Quote Originally Posted by Guy View Post
    I think the question is just asking for you to show that, for fixed y, the function $\displaystyle y^{(\lambda)}$ is continuous in $\displaystyle \lambda$. For $\displaystyle \lambda \ne 0 $ this trivial. To show continuity at $\displaystyle \lambda = 0$ it suffices to show that

    $\displaystyle \displaystyle
    \lim_{\lambda \to 0} y^{(\lambda)} = \log y
    $

    which is easy. The simpliest way of going about this is to define $\displaystyle f(\lambda) = y^\lambda$ and write $\displaystyle y^{(\lambda)} = \frac{f(\lambda) - f(0)}{\lambda - 0}$ which is a form that we all recognize.
    am retarded, can't remember calculus. Can somebody dig me out of the dirt?
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  4. #4
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    $\displaystyle f(\lambda)=y^\lambda=\frac{f(\lambda)-f(0)}{\lambda}=\lambda f(\lambda)+1$

    $\displaystyle f'(\lambda)=y^\lambda log(y)$

    $\displaystyle f'(0)= log(y)$

    make sense?
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  5. #5
    Guy
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    The basic argument is

    $\displaystyle \displaystyle
    \lim_{\lambda \to 0} y^{(\lambda)} = \lim_{\lambda \to 0} \frac{f(\lambda) - f(0)}{\lambda - 0} = \frac{d}{d\lambda} y^\lambda \big|_{\lambda = 0} = ... = \log y
    $
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