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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?

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    $\begingroup$ 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 $\endgroup$
    – Henry
    Apr 26, 2017 at 14:52
  • $\begingroup$ @Henry Spot-on comments~ $\endgroup$
    – Ran Wang
    Apr 26, 2017 at 15:24
  • $\begingroup$ It is not a Gaussian distribution. You cannot get standard GRD by just scaling the variance of Gaussian. $\endgroup$
    – Ravi
    Apr 26, 2017 at 15:25
  • $\begingroup$ 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 $0.9545$ $\endgroup$
    – Henry
    Apr 26, 2017 at 15:43
  • $\begingroup$ Thanks. $\Phi(x \, \sqrt{\log(\phi)}) = f(x),$ where $\Phi(.)$ is the distribution of standard Gaussian. $\endgroup$
    – Ravi
    Apr 26, 2017 at 16:26

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

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