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I will take only the BoxCox's transformation for $\lambda \neq 0$ so it's:

$Y_i=\displaystyle\frac{(X_i^{\lambda}-1)}{\lambda}$

and its inverse:

$X_i=e^{ \displaystyle ln(Y_i \lambda+1)/\lambda}$

Assuming that transformation will normalize the random variable $X_i$ then $Y_i\sim N(u_y,\sigma^2_y)$ the distribution for $X_i$'s distribution can be get as:

$f_X(x_i)=\phi_y\displaystyle\left({\displaystyle\frac{{x_i^\lambda}-1}{\lambda}}\right) x_i^{\lambda-1}$ where $\phi_y$ is $y$'s distribution

But when I try to get its support, that is the set where $f_{X}(x_i)>0$ and it integrates 1, I come across with this trouble below:

$-\infty < y_i < \infty$, that is the support of $Y_i$

$-\infty< y_i \lambda<\infty$

$-\infty< y_i \lambda+1<\infty$

the trouble comes when I'd apply the $ln$ function (it's clean that I'm using limits but I preferred make the notation easier):

$ln(-\infty) < ln(y_i\lambda+1)< ln(\infty)$

Note that $ln(-\infty)$ is not real valuated.

Is there any way to get rid of this issue or should I ignore that and accept it's a impossible-to-solve trouble?

Davi Américo
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    The support of Y is the positive half line. Also $X_i=(Y_i \lambda+1)^{1/\lambda}$ – Glen_b Oct 08 '21 at 05:19
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    @Glen I don't follow, because this post explicitly posits that $Y$ has a Normal distribution. What one can show is that this is a mathematically impossible assumption, for then when $Y\lambda+1\le 0$ the $1/\lambda$ power usually is undefined. However, if the chance of this event is tiny, we might consider truncating $Y$ to the interval $(-1/\lambda,\infty).$ See https://stats.stackexchange.com/a/541931/919 for details. – whuber Oct 08 '21 at 13:36
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    Quite right; I muddled symbols there, the constraint is on X rather than Y; given that we don't know which $\lambda$ we may end up choosing, the restriction is often given that way so that the transformation will be defined for all values of $\lambda$. – Glen_b Oct 08 '21 at 15:18
  • I'd like to know about inverse distribution when it is normal because I wanna make a classifier boxcox based. – Davi Américo Oct 08 '21 at 23:01

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