The Benini distribution is a continuous univariate distribution that is used in actuarial applications. For all $x \geqslant \sigma$ it has density function:

$$\text{Benini}(x| \alpha, \beta, \sigma) = \frac{\alpha + 2 \beta \log (x/\sigma)}{x} \cdot \exp \Bigg( - \alpha \log \Big( \frac{x}{\sigma} \Big) - \beta \log^2 \Big( \frac{x}{\sigma} \Big) \Bigg),$$

where we have the shape parameters $\alpha > 0$ and $\beta > 0$ and the scale parameter $\sigma > 0$. (The $\log^2$ notation is explained here for anyone unfamiliar with that notation.) I have plotted this density for a range of parameter values, and it seems to be strictly quasi-concave (with a single mode) in all cases, but monotonically decreasing (with a mode at its minimum value) in some cases. For the purposes of deriving a highest density region, I would like to know the general "shape" of the distribution from the parameters.

Question: Is the Benini density function always unimodal? Is it always strictly quasi-concave? Under what conditions (if any) is the density function monotonically decreasing? What is the mode of the distribution as a function of the parameters?


1 Answer 1


The density function of the Benini distribution is strictly quasi-concave, and is strictly decreasing if and only if $\beta \leqslant \alpha (1+\alpha)/2$. The mode of the distribution has the explicit form:

$$\text{Mode} = \hat{x} = \sigma \cdot \exp \bigg( \bigg\{ \frac{\sqrt{1+8\beta} - (1+2\alpha)}{4 \beta} \bigg\} \bigg),$$

where the brackets $\{ \cdots \}$ denote the ramp function. (It is easy to verify that the monotonicity cut-off $\beta = \alpha(1+\alpha)/2$ gives $\hat{x}=\sigma$, which confirms the validity of the mode formula in the monotonic case.) The mode of the distribution is not shown on the linked Wikipedia page, but it has an explicit form that is a relatively simple function of the parameters.

Proof of strict quasi-concavity: We will start by writing the log-density for the Benini distribution to show that this is a quasi-concave function. For all $x > \sigma$ we have the log-density:

$$\begin{aligned} \ell(x| \alpha, \beta, \sigma) \equiv \log \text{Benini}(x| \alpha, \beta, \sigma) &= - \alpha \log \Big( \frac{x}{\sigma} \Big) - \beta \log^2 \Big( \frac{x}{\sigma} \Big) \\[6pt] &\quad + \log \Big( \alpha + 2 \beta \log \Big( \frac{x}{\sigma} \Big) \Big) - \log (x). \end{aligned}$$

The first and second derivatives of this function are:

$$\begin{aligned} \frac{d\ell}{dx}(x| \alpha, \beta, \sigma) &= - \frac{1}{x} \Bigg[1+\alpha + 2 \beta \cdot \log \Big( \frac{x}{\sigma} \Big) - \frac{2 \beta}{\alpha + 2 \beta \log ( x/\sigma )} \Bigg], \\[6pt] \frac{d^2 \ell}{dx^2}(x| \alpha, \beta, \sigma) &= - \frac{1}{x} \Bigg[\frac{d\ell}{dx}(x| \alpha, \beta, \sigma) + \frac{2 \beta}{x} \Bigg(1 + \frac{2 \beta}{(\alpha + 2 \beta \log ( x/\sigma ))^2} \Bigg) \Bigg]. \\[6pt] \end{aligned}$$

Thus, at any critical point $\hat{x}$ we have:

$$\frac{d^2 \ell}{dx^2}(\hat{x}| \alpha, \beta, \sigma) = - \frac{1}{x} \cdot \frac{2 \beta}{x} \Bigg(1 + \frac{2 \beta}{(\alpha + 2 \beta \log ( x/\sigma ))^2} \Bigg) <0.$$

This demonstrates that every critical point is a strict local maximum, and so there must be a unique critical point that is the mode of the density. This demonstrates that the log-density is a strictly quasi-concave function, which means that the density must also be a strictly quasi-concave function. This holds for all parameterisations.

Condition for monotonicity: Since the density function is strictly quasi-concave over a support that is bounded from below, it is monotonically decreasing (starting from that bound) if and only if the first derivative at the lower bound is non-positive. Thus, the condition for the density to be monotonically decreasing (and thus have its mode at $\hat{x} = \sigma$) is:

$$\frac{d\ell}{dx}(\sigma| \alpha, \beta, \sigma) \leqslant 0.$$

This condition can be written as $1+\alpha \geqslant 2 \beta/\alpha$, which means we have:

$$\begin{matrix} \beta \leqslant \alpha (1+\alpha)/2 & & & \text{Monotonic density} & & & \hat{x} = \sigma, \\[6pt] \beta > \alpha (1+\alpha)/2 & & & \text{Non-monotonic density} & & & \hat{x} > \sigma. \\[6pt] \end{matrix}$$

This gives a nice simple parameter condition to distinguish the monotonic case from the non-monotonic case (where the density is still strictly quasi-concave). In the former case the mode is the minimum value in the support of the distribution, and in the latter case it is above this vale.

Derivation of the mode: We have already dealt with the monotonic case in the section above, so here we will deal with the non-monotonic case. In this case, the mode $\hat{x}$ is the unique critical point that solves the critical point equation. Letting $L(x) \equiv \log(x/\sigma)$ and $\hat{L} \equiv L(\hat{x})$ we have the critical point equation:

$$1+\alpha + 2 \beta \hat{L} = \frac{2 \beta}{\alpha + 2 \beta \hat{L}}.$$

Rearranging this equation gives the quadratic equation:

$$4 \beta^2 \hat{L}^2 + 2 \beta (1+2\alpha) \hat{L} + (\alpha (1+\alpha) - 2 \beta) = 0.$$

This quadratic equation has a unique solution in the support of the distribution, which is:

$$\hat{x} = \sigma \exp(\hat{L}) = \sigma \cdot \exp \Bigg( \frac{\sqrt{1+8\beta} - (1+2\alpha)}{4 \beta} \Bigg).$$

This is the mode in the case where the monotonicity condition is not met. The overall formula for the mode is obtained over all parameter values by stipulating that the exponent (the part inside the exponential) cannot be negative; this is done by applying the ramp function. This results in the explicit form shown at the top of this answer.


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