# Golden-Ratio Distribution - analogous to Normal distribution

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?

• It is a Gaussian distribution, though with a variance not equal to $1$. These are common. What you would need to do is motivate a variance of $\frac{1}{\log \phi} \approx 4.78$ as being in some sense special. Since your $\sigma^2$ is then not the variance, you would have to find some useful meaning for it Apr 26, 2017 at 14:52
• @Henry Spot-on comments~ Apr 26, 2017 at 15:24
• It is not a Gaussian distribution. You cannot get standard GRD by just scaling the variance of Gaussian.
– Ravi
Apr 26, 2017 at 15:25
• So what is the probability your "standard GRD" is less than $\frac{2}{\sqrt{\log \phi}}$? I would be surprised if it was far away from $0.9545$ Apr 26, 2017 at 15:43
• Thanks. $\Phi(x \, \sqrt{\log(\phi)}) = f(x),$ where $\Phi(.)$ is the distribution of standard Gaussian.
– Ravi
Apr 26, 2017 at 16:26

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 $$\frac{1}{\ln(\phi)}$$.