# How to find the transformed density and log likelihood for this family of distributions?

Let's consider the family of transformations given by $$g_a(Y)=\begin{cases} \frac{e^{aY}-1}{a} & \text{ for } a\neq 0 \\ Y & \text{ for } a=0 \end{cases}$$ for $$Y\in\mathbb{R}$$. Analogous to the estimation of the Box-Cox parameter $$\lambda$$, the parameter $$a\in\mathbb{R}$$ can be estimated using a profile likelihood approach. Let $$Y_1,...,Y_n\in\mathbb{R}$$ be independent responses together with corresponding predictors $$\mathbf{x}_1,...,\mathbf{x}_n\in\mathbb{R}^n$$. We assume for $$a$$ that there exist $$\textbf{b}\in\mathbb{R}^p$$ and $$\sigma^2>0$$ such that $$g_a(Y_i)\sim N(\textbf{x}_i^T \textbf{b},\sigma^2)$$ for $$i=1,...,n$$.

Here is my questions:

1. Can we derive such a density function $$f_Y$$ of the untransformed observations $$Y_i$$. If so, would it be $$f_{Y_i}(\textbf{x})=\frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{1}{2}\cdot \frac{(\textbf{x}-\textbf{x}_i^T\textbf{b})^2}{\sigma^2}}$$

2. How would we find the log-likelihood function $$\ell(a,b,\sigma^2;y_1,...,y_n)$$? I know that $$\sum_{i=1}^n \log(f_{Y_i}(y_i))$$.

• Your (1) states that $Y_i$ and $g_a(Y_i)$ are identically distributed. That's clearly the case for $a=0,$ but do you believe that continues to be true for nonzero $a$? Maybe there are some typographical errors in your formulas?
– whuber
Dec 13, 2022 at 15:54

Only one calculation is needed: we have to differentiate $$g_a.$$ Everything else is just substitutions and taking logarithms.

For convenience, let $$Z=g_a(Y)$$ and write $$\mu = \mathbf x_i^\prime \beta.$$ You say $$Z$$ has a Normal distribution, which means it has a density function given by

$$f(z) = \frac{1}{\sigma\sqrt{2\pi}}\, \exp\left(-\frac{1}{2\sigma^2}(z-\mu)^2\right)$$

The transformation $$y\to g_a(y) = z$$ is everywhere increasing, differentiable, and one-to-one. For $$a\ne 0$$ its differential is

$$\mathrm d z = \mathrm d\left(\frac{e^ay-1}{a}\right) = e^{ay}\mathrm dy.$$

Consequently, $$Z$$ has a continuous distribution supported on the image of $$g_a$$ with probability element

$$h(y)\mathrm d y = f(z(y))\mathrm dz = \frac{1}{\sigma\sqrt{2\pi}}\, \exp\left(-\frac{1}{2\sigma^2}((e^{ay}-1)/a-\mu)^2\right)\, e^{ay}\mathrm dy .$$

The logarithm of the density $$h$$ is

$$\log h(y) = -\log(\sigma) - \frac{1}{2}\log (2\pi) - \frac{1}{2\sigma^2}\left(\frac{e^{ay}-1}{a}-\mu\right)^2 + ay.$$

Consequently, the log likelihood of $$n$$ independent observations according to your model is the sum of the logs of their individual likelihoods,

$$\Lambda(a,\beta,\sigma; \mathbf x, \mathbf y)) = - n\log\sigma - \frac{n}{2}\log(2\pi) - \frac{1}{2\sigma^2}\sum_{i=1}^n \left(\frac{e^{ay_i}-1}{a}-\mathbf x_i^\prime \beta \right)^2 + a\sum_{i=1}^n y_i.$$

When $$a=0$$, taking the limit of the forgoing as $$a\to 0$$ reduces each $$(e^{ay_i}-1)/a$$ to $$y_i$$ and the $$+ay_i$$ terms vanish, reproducing the usual likelihood for this Normal regression model.

• Nice one! Is it possibel to find ML-estimastes for $\hat{b}$ and $\hat{\sigma}^2$ for a fixed value of $a$? I tried to read about ML-estimates but didn't get so much from this. Dec 13, 2022 at 23:54
• And notice. Should you not include a negative sign in the exponent in both $f(z)$ and $h(y)dy$? Dec 14, 2022 at 0:12
• Good catch! I left that out of the first two equations, but the subsequent formulas involving logarithms are correct. For any fixed value of $a,$ this is an Ordinary Least Squares (OLS) problem. The ML estimate of $b$ is the OLS estimate and the ML estimate of $\sigma^2$ is $1-2/n$ times the OLS estimate of $\sigma^2.$
– whuber
Dec 14, 2022 at 14:39