What are the properties of the “unfolded” gamma distribution generalization of a normal distribution?

In a prior post, I developed an "unfolded" gamma distribution generalization of a normal distribution as an example of how to relate a gamma distribution to a normal distribution. This yielded

$$\text{ND}(x;\mu,\sigma^2,a) = \dfrac{a e^{-2^{-\frac{a}{2}} \left(\frac{1}{\sigma }\right)^a \left| x-\mu\right| ^a}}{2 \sqrt{2} \sigma \Gamma \left(\frac{1}{a}\right)} \,,$$

where the mean is $\mu$, the variance is $\sigma^2$, and the shape is $a>0$, where $a=2$ for an ordinary normal distribution.

This appears to be a different distribution from the generalized error distribution

$$\text{GED}(x;\mu,\alpha,\beta)=\frac{\beta}{2\alpha\Gamma(1/\beta)} \; e^{-(|x-\mu|/\alpha)^\beta}\,,$$

where $\mu$ is the location, $\alpha>0$ is the scale, and $\beta>0$ is the shape and where $\beta=2$ yields a normal distribution. It includes the Laplace distribution when $\beta=1$. As $\beta\rightarrow\infty$, the density converges pointwise to a uniform density on $(\mu-\alpha,\mu+\alpha)$.

Questions Are these distributions the same? If so, how does one convert between them? If not, and $\text{ND}(x;\mu,\sigma^2,a)$ is a different distribution, what are its other properties, for example, does it reduce to a distribution other than a normal distribution?

• Assuming the domain is $x\in\mathbb R$, comparing the parts of the two densities that depend on $x$ shows immediately that the role played by $\mu$ is the same in both (it's the location parameter) and $a=\beta$ (the shape parameter) and $\sigma=\alpha$ (the scale parameter). – whuber Mar 5 '18 at 13:53
• @whuber Close, but when one lets $\beta=a=1$, then $\sqrt{2} \sigma=\alpha$. There is a bit more to this, will explore further. – Carl Mar 5 '18 at 18:47
• I'm having trouble reading the "ND" expression. If it means the argument of the exponential is $$-2^{-a/2} \sigma^{-a}|x-\mu|^a=-\left(\frac{|x-\mu|}{\sigma\sqrt{2}}\right)^a,$$ then the scale factor must be $\sigma\sqrt{2}$ and everything works out. – whuber Mar 5 '18 at 19:32

Indeed, my notation was bad. Better notation is now used on the post this came from,

The generalized gamma distribution

$$\text{GD}\left(x;\alpha ,\beta ,\gamma ,\mu \right)=\begin{array}{cc} & \begin{cases} \dfrac{\gamma e^{-\left(\dfrac{x-\mu }{\beta }\right)^{\gamma }} \left(\dfrac{x-\mu }{\beta }\right)^{\alpha \gamma -1}}{\beta \,\Gamma (\alpha )} & x>\mu \\ 0 & \text{other} \\ \end{cases} \\ \end{array}\,,$$

implies that

\begin{align} \text{GND}(x;\mu,\alpha,\beta) &= \frac{1}{2}\text{GD}\left(x;\frac{1}{\beta},\alpha,\beta,\mu \right)+\frac{1}{2}\text{GD}\left(-x;\frac{1}{\beta},\alpha,\beta,\mu \right)\\ &= \frac{\beta e^{-\left(\dfrac{\left|x-\mu\right|}{\alpha }\right)^{\mathrm{\Large{\beta}}}}}{2 \alpha \Gamma \left(\dfrac{1}{\beta }\right)}\\ \end{align} \,,

is a generalization of the normal distribution, where $\mu$ is the location, $\alpha>0$ is the scale, and $\beta>0$ is the shape and where $\beta=2$ yields a normal distribution. It includes the Laplace distribution when $\beta=1$. As $\beta\rightarrow\infty$, the density converges pointwise to a uniform density on $(\mu-\alpha,\mu+\alpha)$.

See the original post for the logic behind this.

The result from Whuber's comment in an answer:

It seems like both expressions can be related to the function below

$$f(x;c_1,c_2,c_3) = c_1 e^{-c_2 \vert x - \mu \vert^{c_3}}$$

GED: $\frac{\beta}{2\alpha\Gamma(1/\beta)} \; e^{-(|x-\mu|/\alpha)^\beta}$

$$\begin{array}{rcl} c_1 &=& \frac{\beta}{2\alpha\Gamma(1/\beta)} \\ c_2 &=& \left(\frac{1}{\alpha} \right)^\beta \\ c_3 &=& \beta \end{array}$$

ND: $\dfrac{a e^{\left(-2^{-\frac{a}{2}} \left(\frac{1}{\sigma }\right)^a \left| x-\mu\right| ^a\right)}}{2 \sqrt{2} \sigma \Gamma \left(\frac{1}{a}\right)}$

$$\begin{array}{rcl} c_1 &=& \frac{a}{2 \sqrt{2} \sigma\Gamma(1/a)} \\ c_2 &=& 2^{-\frac{a}{2}}\left( \frac{1}{\sigma} \right)^a \\ c_3 &=& a \end{array}$$

and you can say

$$\begin{array}{rcl} \mu &=& \mu \\ {\alpha} &=& \sqrt{2}{\sigma} \\ \beta &=& a \end{array}$$