Random variable transformation Suppose I have a scale-invariant response y and a set of predictors x, which are related as
$y^\alpha = x\beta +\epsilon$, where $\epsilon | x \sim \mathcal{N}(0, \sigma^2)$.
Here, I know that the distribution of $y^\alpha|x \sim \mathcal{N}(x\beta, \sigma^2)$.
But I also know that the distribution of $y|x$ can be obtained via variable transformation.
I am thinking about using $W = \exp(y^{\alpha})$ as my transformation in order to obtain the distribution of $y|x$, but I am not so sure where that leads me to.
Can anyone here kindly help me out? Thanks
 A: I don't think there's any fancy name for this distribution, since it's just a normally-distributed variable raised to the exponent $-\alpha$. You can still solve for the distribution though. Using $\Phi$ to denote the cdf for the standard normal distribution $\mathcal{N}(0, 1)$, we can figure out the cdf of $Y|X$:
$$\begin{align}
P(Y^\alpha < c : x) &= \Phi\left(\frac{c-x\beta}{\sigma}\right) \\
P(Y < c^{1/\alpha} : x) &= \Phi\left(\frac{c-x\beta}{\sigma}\right)
\end{align}$$
and letting $d = c^{1/\alpha}$,
$$P(Y < d : x) = \Phi\left(\frac{d^{\alpha}-x\beta}{\sigma}\right)$$
In other words, the cdf for $y | x$ is $F_{Y|X}(y) = \Phi\left(\frac{y^{\alpha}-x\beta}{\sigma}\right)$. From there, if you want the pdf then you could differentiate the cdf.

Bonus note: depending on your specific values of $\alpha$, you may eventually see some problem due to $\mathcal{N}(x\beta, \sigma^2)$ taking negative values leads to imaginary numbers in your $Y$. It might not be an issue depending on how you're applying this work. Just something to look into in case your code starts randomly blowing up later.
