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Is there some restriction to parameters $( \alpha , \beta)$ that makes the beta distribution concave down? Bell-shaped like e.g. a normal?

For example, the cases in purple and black, but not the red green or blue cases:

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What constitutes 'bell-shaped' exactly? – gung Sep 1 '14 at 21:24
What I was hinting at is that you need a precise definition of 'bell-shaped' in order for this question to be answerable. Eg, is what you want any shape that has exactly 0 skewness & is unimodal? – gung Sep 1 '14 at 21:37
In addition to the usual comments on beta distributions, a cute result is that if $p$ is beta, then logit $p$ is unimodal. See Ch.23.5 in Mackay's book – Nick Cox Sep 1 '14 at 21:59
@gung Is there such a thing as a bimodal beta that's concave down? ... micheal - just to clarify: (i) Q looks like routine bookwork -- is this for some subject? (ii) you want concavity in the density, rather than log-concavity? If concavity in $f$, you should think about what happens at 0 and 1 with a beta, and the definition of concave, and then see what you can do with it. – Glen_b Sep 2 '14 at 0:08
The fact that the three answers you have all make different assumptions about what you mean clearly suggests that you need to make your question more clear. – Glen_b Sep 2 '14 at 6:29

In the context of a beta distribution, I would regard "bell-shaped" as meaning that there exist exactly two distinct inflection points in the interval $(0,1)$ with a global maximum in between, and $f(0) = f(1) = 0$ where $$f(x) \propto x^{a-1} (1-x)^{b-1}.$$ We clearly aren't interested in the proportionality constant in this case. The condition that $f(0) = f(1) = 0$ obviously requires $a > 1$, $b > 1$. Now via logarithmic differentiation, we obtain $$f'(x) \propto x^{a-2} (1-x)^{b-2}(a-1+(2-a-b)x),$$ so our local maximum occurs at $x = \frac{a-1}{a+b-2}.$ Then $$f''(x) \propto x^{a-3} (1-x)^{b-3} ((a-1)(a-2) + 2(a-1)(3-a-b)x + (3-a-b)(2-a-b)x^2),$$ which will have two distinct real zeroes if the discriminant of the quadratic factor is positive: $$(2(a-1)(3-a-b))^2 - 4(3-a-b)(2-a-b)(a-1)(a-2) > 0.$$ This is equivalent to $$(a-1)(b-1)(a+b-3) > 0.$$ Since we already established that $a, b > 1$, we then obtain another condition: $a + b > 3$. However, we are still not done: we must now check that the inflection points are in $(0,1)$. To do this, it suffices to check that $\displaystyle \lim_{\epsilon \to 0^+} f''(\epsilon) > 0$ and similarly $\displaystyle \lim_{\epsilon \to 0^+} f''(1-\epsilon) > 0$: the first condition requires $(a-1)(a-2) > 0$, hence $a > 2$, and similarly, the second requires $(b-1)(b-2) > 0$, or $b > 2$. (A symmetry argument would also suffice.) Therefore, the beta PDF could be said to be "bell-shaped" whenever $a > 2$ and $b > 2$.

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+1. Given edits to the question, which now suggests that "bell shaped" means unimodal with a mode in the interior of the interval $[0,1]$, the first part of the analysis provides a simple direct answer: $0\lt (a-1)/(a+b-2)\lt 1$, equivalent to both $a$ and $b$ strictly greater than $1$. – whuber Sep 2 '14 at 16:12

What we usually call a "bell-shaped" graph, is neither concave (or "concave down") nor convex (or "concave up") -it has both concave and convex parts.

For the Beta density to be (strictly) concave everywhere, its second derivative with respect to its variable must be negative. The probability density function is

$$f_X(x)=\frac{x^{\alpha-1}(1-x)^{\beta-1}} {B(\alpha,\beta)},\;\;\;\ B(\alpha,\beta) = \int_0^1t^{\alpha-1}(1-t)^{\beta-1}\,\mathrm{d}t$$

We compute that

$$f'_X(x) = f_X(x)\left(\frac {\alpha-1}{x} - \frac {\beta-1}{1-x}\right)$$


$$f''_X(x) = f'_X(x)\left(\frac {\alpha-1}{x} - \frac {\beta-1}{1-x}\right) - f_X(x)\left(\frac {\alpha-1}{x^2} + \frac {\beta-1}{(1-x)^2}\right)$$

$$\Rightarrow f''_X(x) = f_X(x)\left(\frac {\alpha-1}{x} - \frac {\beta-1}{1-x}\right)^2 - f_X(x)\left(\frac {\alpha-1}{x^2} + \frac {\beta-1}{(1-x)^2}\right)$$


$$\text{sign}\{f''_X(x)\}=\text{sign}\left\{[(\alpha-1)(1-x)-(\beta-1)x]^2- (\alpha-1)(1-x)^2-(\beta-1)x^2\right\}$$

$$=\text{sign}\left\{(\alpha-1)(\alpha-2)(1-x)^2+(\beta-1)(\beta-2)x^2- 2(\alpha-1)(\beta-1)x(1-x)\right\}$$

We see that when

$$\{1<\alpha\leq 2\}\cap\{1<\beta \leq 2\}$$

the above is certainly negative irrespective of the value of $x$ (and so the density graph will be concave for the whole of its domain).

The strict concavity result also holds if one of the parameters is equal to $1$, while the other is strictly between $1$ and $2$ (but then one endpoint of the graph won't reach zero. Still the density will be strictly concave). I don't think there is any other range of the parameters for which strict concavity holds. To play around a web-graph facility is

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hmm. I think you have an error in your calculation. I kind of suspected this one will actually turn out to be self study, which is why I held off. – Glen_b Sep 2 '14 at 4:33
To be specific, I think the term $(\beta-1)(\beta-2)x^2$ is wrong. And I think that changes the region where the curve is concave. (Of course, I could have made a mistake, but taking derivatives in R (via D) and making it a function also suggests there's a problem with your answer, since yours changes sign for cases when doing it via D doesn't.) -- I think that $1<\alpha,\beta\leq 2$ does it – Glen_b Sep 2 '14 at 4:46
(... though I should probably check that carefully. If I post an answer, I will do so.) – Glen_b Sep 2 '14 at 6:27
@Glen_b Thanks Glen, I had the sign wrong. Corrected. – Alecos Papadopoulos Sep 2 '14 at 8:05
Counter-example: when $\alpha=5, \beta=1.5$, the distribution is not concave down everywhere. – amoeba Sep 2 '14 at 10:04

I surmise that what you really mean is the density goes to zero at both the endpoints. For this, you need $\alpha > 1$ and $\beta > 1$. The closer that $\alpha$ and $\beta$ are to each other, the closer the distribution will be to symmetrical.

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