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What conditions must satisfy the mean and variance of a Beta distribution so that the parameters $\alpha,\beta$ are not both less than 1?

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  • $\begingroup$ Please explain what you mean by a "distribution to be convex." You probably need to restrict the convexity to the support of the distribution function (for otherwise no distribution function is convex). But are you referring to its density function, cumulative density, or perhaps something else? $\endgroup$ – whuber Mar 20 '15 at 17:33
  • $\begingroup$ @whuber See edit. $\endgroup$ – becko Mar 20 '15 at 18:13
  • $\begingroup$ Thank you for the clarification. I have taken the liberty of correcting your definition by inserting the word "segment". Otherwise (in the form originally stated) your definition would not hold for any distribution. $\endgroup$ – whuber Mar 20 '15 at 18:21
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The parameters of a $\text{Beta}(\alpha,\beta)$ distribution with mean $0\lt m\lt 1$ and variance $0\lt v\lt m(1-m)$ are

$$\alpha = m\frac{m(1-m)- v}{v},\quad \beta = (1-m)\frac{m(1-m)-v}{v}.$$

Figure 1

This shaded contour plot of $\alpha$ has contours ranging from $0$ (at the top of the colored region) to $1$ (along the bottom). The plot of $\beta$ is its mirror image.

If they are not both less than $1$, then algebra shows us that

$$v \le m\max\left(\frac{m(1-m)}{1+m}, \frac{(1-m)^2}{2-m}\right).$$

Figure

The valid set of all possible means and variances of Beta distributions is contained beneath the gray curve. Within that set, those where one or both of $\alpha$ and $\beta$ are $1$ or greater is shown in darker blue. These tend to have lower variances on the whole.

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The mean of the Beta distribution is

$$\mu = \frac {\alpha}{\alpha + \beta}$$

We want to see whether restricting the permissible range of $\mu$, will guarantee that we will have either $\{\alpha \geq 1, \beta >0\}$, OR $\{\alpha >0, \beta \geq 1\}$.

Treating the mean as a function of the parameters we obtain

$$\frac {\partial \mu}{\partial \alpha} > 0, \;\; \frac {\partial \mu}{\partial \beta} <0$$

So it is monotonically increasing in $\alpha$ and monotonically decreasing in $\beta$.

So

$$\min \mu = 1/(1+\beta)\implies \{\alpha \geq 1, \beta >0\} \tag {1}$$

but the situation $\{\alpha >0, \beta \geq 1\}$ permits all possible values of $\mu$ (in $(0,1)$).

In other words by restricting the mean to lie in the interval $[1/(1+\beta),\, 1)$ we can guarantee that we will have $\{\alpha \geq 1, \beta >0\}$. But there is no restriction on the mean that will guarantee us that $\{\alpha >0, \beta \geq 1\}$.

So we should turn to the variance, which is

$$\sigma^2 = \frac {\alpha \beta}{(\alpha + \beta)^2(\alpha + \beta +1)}$$

It is not difficult to determine that no restriction on the range of the variance can guarantee that we will have $\{\alpha >0, \beta \geq 1\}$.

So :

If we impose the restriction $\mu \geq 1/(1+\beta)$, then we will certainly have $\{\alpha \geq 1, \beta >0\}$, i.e. "not both parameters smaller than unity".

But this in a sense is a partial result, since there is also the other way in which "not both parameters are smaller than unity". In other words, this approach imposes the additional restriction that only $\beta$ is allowed to be smaller than unity. It is therefore incompatible with Beta distributions for which we want to be able to have $\alpha \leq 1$ (and $\beta >1$).

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  • $\begingroup$ It's unclear what your answer is. Could you more specifically describe the solution set? $\endgroup$ – whuber Mar 20 '15 at 22:17
  • $\begingroup$ @whuber Did some reworking. $\endgroup$ – Alecos Papadopoulos Mar 20 '15 at 23:53
  • $\begingroup$ I'm afraid I'm still not seeing it. In fact, looking at the last paragraph, you seem to say the solution set is empty ("incompatible"). A clear answer would provide usable criteria to determine exactly when any ordered pair $(\mu,\sigma^2)$ is or is not the mean and variance of a Beta distribution whose parameters are not both less than one. Those criteria would be in terms of $\mu$ and $\sigma^2$ alone. $\endgroup$ – whuber Mar 21 '15 at 0:00
  • $\begingroup$ @whuber Oh, I, on the other hand, see it very clearly, and it is a case of "directly enforce what you would want to indirectly impose". Let me see if I can salvage this in some way. $\endgroup$ – Alecos Papadopoulos Mar 21 '15 at 0:09

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