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Math Help - box-cox

  1. #1
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    box-cox

    I have the BC tranformation:

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

    I need to show that this give a continous famly of transformations in \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 y^{(\lambda)} is continuous in \lambda. For \lambda \ne 0 this trivial. To show continuity at \lambda = 0 it suffices to show that

    \displaystyle<br />
\lim_{\lambda \to 0} y^{(\lambda)} = \log y<br />

    which is easy. The simpliest way of going about this is to define f(\lambda) = y^\lambda and write 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 y^{(\lambda)} is continuous in \lambda. For \lambda \ne 0 this trivial. To show continuity at \lambda = 0 it suffices to show that

    \displaystyle<br />
\lim_{\lambda \to 0} y^{(\lambda)} = \log y<br />

    which is easy. The simpliest way of going about this is to define f(\lambda) = y^\lambda and write 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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    f(\lambda)=y^\lambda=\frac{f(\lambda)-f(0)}{\lambda}=\lambda f(\lambda)+1

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

    f'(0)= log(y)

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

    \displaystyle<br />
\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<br />
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