1. ## box-cox

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?

2. 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.

3. 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?

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

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

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

make sense?

5. 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
$