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The Beta distribution has the PDF:

$$f\left(x\right)=\frac{x^{\alpha-1}\left(1-x\right)^{\beta-1}}{\mathrm{B}\left(\alpha,\beta\right)}$$

for $0<x<1$, and $f(x)=0$ otherwise. The parameters $\alpha,\beta$ are positive real numbers.

The mean and variance are given by:

$$\mu=\frac{\alpha}{\alpha+\beta},\quad\sigma^{2}=\frac{\alpha\beta}{\left(\alpha+\beta\right)^{2}\left(1+\alpha+\beta\right)}$$

which can be inverted to give $\alpha,\beta$ in terms of the mean and the variance as $\alpha=\lambda\mu$ and $\beta=\lambda\left(1-\mu\right)$, where

$$\lambda=\frac{\mu\left(1-\mu\right)}{\sigma^{2}}-1$$

Now I want to impose the condition that $\alpha,\beta \ge 1$. What does this imply for the mean and the variance? That is, is there a simple condition on $\mu,\sigma^2$ that is equivalent to $\alpha,\beta \ge 1$?

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I was messing up the algebra, but I think I got it now:

$$\left[\mu\left(1-\mu\right)-\sigma^{2}\right]\min\left(\mu,1-\mu\right)\ge\sigma^{2}$$

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  • $\begingroup$ That has $\sigma^2$ on both sides. I think you can rewrite it as $\sigma^{2}\le \frac{\mu\left(1-\mu\right)\min\left(\mu,1-\mu\right)}{1+\min\left(\mu,1-\mu\right)}=\frac{\mu\left(1-\mu\right)}{1+\frac1{\min\left(\mu,1-\mu\right)}}\le \frac{\mu\left(1-\mu\right)}{3}$ and compare that to the looser $\sigma^{2}\le \mu\left(1-\mu\right)$ when you allow smaller $\alpha, \beta$ $\endgroup$
    – Henry
    Commented May 27, 2023 at 15:15

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