If you look at a beta distribution with $\alpha=\beta=4$ it looks very similar to a Gaussian distribution. But is it? How can you prove whether a Beta(4,4) distribution is Gaussian or not?

  • 10
    $\begingroup$ Their supports are so different. $\endgroup$
    – Deep North
    Dec 8 '17 at 3:07
  • $\begingroup$ @DeepNorth - so are you suggesting Gaussian distributions are not a specific kind of Beta Distribution? $\endgroup$ Dec 8 '17 at 3:09
  • 4
    $\begingroup$ More than suggesting; if the support is different they can't be the same distribution. $\endgroup$
    – Glen_b
    Dec 8 '17 at 4:01

They are both symmetric and more or less bell shaped, but the symmetric beta (whether at 4,4 or at any other specific value) is not actually Gaussian. You can tell this even without looking at the density -- beta distributions are on (0,1) while all Gaussian distributions are on $(-\infty,\infty)$

Let's look a bit more closely at the comparison. We'll standardize the beta(4,4) so that it has mean 0 and standard deviation 1 (a standardized beta) and look at how the density compares to a standard Gaussian:

standardized beta(4,4) density and standard Gaussian density

The standardized beta(4,4) is restricted to lie between -3 and 3 (the standard Gaussian can take any value); it is also less peaked than the Gaussian and has rounder "shoulders" around 1 or so standard deviations either side of the mean. Its kurtosis is 27/11 ($\approx$2.45, vs 3 for the Gaussian).

Symmetric beta distributions with larger parameter values are closer to Gaussian.

In the limit as the parameter approaches infinity, a standardized symmetric beta approaches a standard normal distribution (example proof here).

So no specific case of the symmetric beta is Gaussian, but the limiting case of a suitably standardized beta is Gaussian. We can see this approach more easily by looking at the cdf of the beta, transformed by the quantile function of the Gaussian. On this scale the Gaussian would lie on the $y=x$ line, while the symmetric beta family would approach the $y=x$ line as the parameter got larger and larger.

In the plot below we look at the deviations from the $y=x$ line to more clearly see the approach of the beta($\alpha$,$\alpha$) to the Gaussian as $\alpha$ increases.

Plot showing quantile function of cdf standardized symmetric beta approaching linearity

  • 1
    $\begingroup$ You do not necessarily have to standardise the Beta random variable so as to have variance exactly $1$; for example a Student $t$ random variable does not have variance $1$. If instead of effectively saying that if $X_{\alpha,\alpha} \sim \text{Beta}(\alpha,\alpha)$ then $\frac{X_\alpha -\frac12}{\sqrt{1/(4(2\alpha+1))}}\xrightarrow{d}\ N(0,1)$ as $\alpha$ increases, you could have dropped the $+1$ and said $\frac{X_\alpha -\frac12}{\sqrt{1/(8\alpha)}}\xrightarrow{d}\ N(0,1)$ as $\alpha$ increases and you would have a better fit in the centre of the distribution $\endgroup$
    – Henry
    Dec 8 '17 at 13:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.