Return to Answer

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

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

and

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

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

So

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

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

From the above we see that when

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

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 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 http://keisan.casio.com/exec/system/1180573226

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

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

and

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

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

So

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

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

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 http://keisan.casio.com/exec/system/1180573226

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

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

and

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

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

So

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

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

From the above we see that when

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

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).

This also holds if one but not both of the parameters is equal to $1$ (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 http://keisan.casio.com/exec/system/1180573226

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

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

and

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

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

So

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

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

From the above we see that when

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

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 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 http://keisan.casio.com/exec/system/1180573226

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

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

and

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

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

So

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

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

I conjecture that the only case where the sign of the above is negative irrespective of the value of $x$ (and so the density graph will be concave for the whole of its domain), is when $\alpha =\beta = 2$ (the case $\alpha=\beta =1$ gives us the uniform distribution). To play around a web-graph facility is http://keisan.casio.com/exec/system/1180573226

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

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

and

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

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

So

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

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

From the above we see that when

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

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).

This also holds if one but not both of the parameters is equal to $1$ (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 http://keisan.casio.com/exec/system/1180573226