How reparameterize Beta distribution?

Question

$\newcommand{\Beta}{\operatorname{Beta}}$Consider $X \sim N(\mu,\sigma)$; I can reparameterize it by $X = \varepsilon\mu + \sigma; \varepsilon \sim N(0,I) $

But given Beta distribution $X \sim \Beta(\alpha,\beta)$; is there easy way (closed form transformation) to reparameterize $X$ with some very simple random variable (Normal, uniform )

My major goal is to do VAE such that my prior is Beta and my posterior is also Beta; so I'm thinking how to reparameterization trick for Beta. What I want to do, is instead of directly sampling from $\Beta(\alpha,\beta)$ (because couldn't do backpropagation), I want first generate $\varepsilon \sim \mathcal{Q}$ some easily sampled distribution, then apply some deterministic function that involves $\alpha, \beta$, such that after the transformation it follows $\Beta(\alpha,\beta)$.

Answer 1

There is always the obvious inverse cdf representation: $$X=F_{\alpha,\beta}^{-1}(U)$$ where $F_{\alpha,\beta}^{-1}(\cdot)$ is the inverse cdf (quantile function) of the Beta $\mathcal Be(\alpha,\beta)$ distribution.

Otherwise, the Wikipedia page lists a large collection of connections with other standard distributions, like the Gamma and the F distributions. For integer valued $\alpha$ and $\beta$, the Beta $\mathcal Be(\alpha,\beta)$ distribution is the distribution of an order statistic of a Uniform sample.

Answer 2

If you mean representing every beta-distributed random variable as some simple function of the two parameters $\alpha,\beta$ and some "standard beta" random variable, then probably it cannot be done.

One alternative to the simple standard way of parameterizing this family of distributions that has crossed my mind is as follows.

The expected value is $\mu=\dfrac\alpha{\alpha+\beta}.$

The variance is $\dfrac{\frac\alpha{\alpha+\beta} \cdot \frac\beta{\alpha+\beta}}{\alpha+\beta+1} = \dfrac{\mu(1-\mu)}{\alpha+\beta+1} = \dfrac{\mu(1-\mu)}\kappa$

where the last equality defines $\kappa.$

So we have \begin{align} \mu & = \alpha/(\alpha+\beta), \\ \kappa & = \alpha+\beta+1. \\[12pt] \alpha & = (\kappa-1)\mu, \\ \beta & = (\kappa-1)(1-\mu). \end{align} $\mu$ is the mean and $\kappa$ is the concentration. With $\mu$ fixed, $\kappa$ is proportional to the reciprocal of the variance.

Postscript: It has occurred to me that what I said in the first paragraph above is mistaken, and I've crossed it out. One can use the beta distribution with $\alpha=\beta=1,$ which is the same as the uniform distribution on $[0,1].$ If $X$ has that distribution, then $F^{-1}(X)\sim\operatorname{Beta}(\alpha,\beta),$ where $F$ is the c.d.f. of the $\operatorname{Beta}(\alpha,\beta)$ distribution.

Postpostscript: The postscript above does not represent an alternative parametrization of the family of beta distributions, since the same pair of parameters still represents the same distribution.