box-cox

**MichaelMath** · October 16th 2010, 12:51 PM

I have the BC tranformation:

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

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

How do I go about doing this?

**Guy** · October 17th 2010, 06:28 AM

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

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.

**MichaelMath** · October 19th 2010, 11:04 AM

Originally Posted by Guy

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

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?

**MichaelMath** · October 19th 2010, 11:29 AM

$f(\lambda)=y^\lambda=\frac{f(\lambda)-f(0)}{\lambda}=\lambda f(\lambda)+1$

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

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

make sense?

**Guy** · October 19th 2010, 02:38 PM

The basic argument is

$\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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