The normal (Gaussian) distribution plays a central role in probability and statistics. I was wondering about an analogous distribution, which I call "golden-ratio distribution" (GRD), defined as follows:
$$ f(x) = \sqrt{\frac{\log \phi}{2\pi}} \int_{-\infty}^{x} \phi^{-\frac{1}{2} t^2} \, \mathrm{dt}, $$ where $\phi = \frac{1 + \sqrt 5}{2}$ is the golden ratio.
This is the `standard' version of GRD. We can include the mean and variance also:
$$ f(x; \mu, \sigma) = \sqrt{\frac{\log \phi}{2\pi \sigma^2}} \int_{-\infty}^{x} \phi^{-\frac{1}{2} {(\frac{t-\mu}{\sigma}})^2} \, \mathrm{dt}, $$
For statistical application, the estimation would be exactly the same as it is for the Gaussian distribution; but, inferences would differ since the golden-ration distribution is more dispersed than the Gaussian.
Does GRD make sense? Has this distribution been studied before?
Best Answer
The PDF of the normal distribution with mean $0$ and variance $\sigma^2$ is given by $$\frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{1}{2} \frac{x^2}{\sigma^2}}$$
Now if we set $e^{\frac{1}{\sigma^2}}=\phi \to \sigma = \sqrt{\frac{1}{\ln(\phi)}}$.
If we plug this $\sigma$ into the PDF of the normal distribution (so that the variance is $\frac{1}{\ln(\phi)}$), the distribution would be $$\frac{1}{\sqrt{\frac{1}{\ln(\phi)}} \sqrt{2\pi}} \phi^{-\frac{1}{2}x^2} = \sqrt{\frac{\ln(\phi)}{2\pi}} \phi^{-\frac{1}{2}x^2}$$ which is exactly the PDF of your distribution. This means that your distribution is simply the normal distribution with mean $0$ and variance $\frac{1}{\ln(\phi)}$.