2. Lower bound of the shape factor

For a random variable f with mean mf , the following characteristics are defined:


$$\bullet \text{ Kurtosis (K), } K \equiv \frac{\text{CM}[\mathfrak{4}]}{\text{CM}[\mathfrak{2}]},$$

CM½ � <sup>4</sup> <sup>∗</sup>CM½ � <sup>2</sup> • Shape Factor (SF), SF � <sup>S</sup> K <sup>2</sup> <sup>¼</sup> <sup>2</sup> . CM½ � <sup>3</sup>

We can prove by Hölder inequality (https://en.wikipedia.org/wiki/Hölder's\_ine quality) that.

SF ≥ 1 :

$$\int \left(f - m\_f\right)^3 d\mu \le \int \left|f - m\_f\right|^3 d\mu = \int \left|f - m\_f\right|^2 \left|f - m\_f\right|^1 d\mu \tag{1}$$

$$\leq \left( \int \left| f - m\_f \right|^4 d\mu \right)^{\frac{1}{2}} \left( \int \left| f - m\_f \right|^2 d\mu \right)^{\frac{1}{2}}.\tag{2}$$

<sup>A</sup> better inequality <sup>K</sup> <sup>≥</sup> <sup>S</sup><sup>2</sup> <sup>þ</sup> <sup>1</sup> is proved in [5–7]. But by Hölder inequality we can ½ �∗ACM½ � <sup>2</sup> also know that ACM <sup>4</sup> <sup>2</sup> <sup>¼</sup> <sup>1</sup> iff <sup>f</sup> is constant: if <sup>f</sup> is not constant, the shape factor ACM½ � <sup>3</sup> must be larger than the lower bound 1.

The contribution to SF > 1 plausibly comes from two parts: Eq. (1) due to symmetry, the more symmetric the distribution, the larger the contribution to SF, or conversely, the smaller the SF, the more asymmetric the distribution; and Eq. (2) due to ACM convexity or steepness, the steeper the PDF, the smaller the SF.

This property of the shape factor identified our exceptional perils as possessing very steep and asymmetric PDF whose SF are small.

#### 2.1 Are there better definitions of shape factor?

To measure the steepness or the convexity, we can get similar inequality to Eq. (2) by Hölder inequality for absolute moment:

$$\frac{AM[4] \ast AM[2]}{AM[3]} \ge 1 \text{ and } \frac{AM[3] \ast AM[1]}{AM[2]} \ge 1.$$

From absolute central moment define:

$$SF1 \equiv \frac{A \text{CM}[4] \* A \text{CM}[2]}{A \text{CM}[3]} \ge \mathbf{1} \text{ and } SF2 \equiv \frac{A \text{CM}[\mathfrak{J}] \* A \text{CM}[\mathbf{1}]}{A \text{CM}[2]} \ge \mathbf{1}.$$

For nonnegative random variables such as the reinsurance contract loss distribution, use the following inequality for moment:

$$\frac{M[4] \ast M[2]}{M[3]^2} \ge 1 \text{ and } \frac{M[3] \ast M[1]}{M[2]^2} \ge 1.$$

From another application of Hölder inequality, we get yet other measures of convexity from absolute central moment:

$$A \text{CM}[r] \le A \text{CM}[s]^\sharp, where \ 0 \le r \le s,$$

ACM½ �<sup>r</sup> ACM½ �<sup>s</sup> SF3½�� <sup>r</sup> <sup>r</sup> <sup>≤</sup>1, where <sup>0</sup> , <sup>r</sup> , <sup>1</sup> and SF3½s� � <sup>s</sup> <sup>≥</sup>1, where <sup>s</sup> > 1: ACM½ � <sup>1</sup> ACM½ � <sup>1</sup> Similar definition from absolute moment:

$$AM[r] \le AM[s]^\sharp, where \ 0 \le r \le s\_r$$

AM½ �<sup>r</sup> AM½ �<sup>s</sup> SF4½�� <sup>r</sup> <sup>r</sup> <sup>≤</sup> <sup>1</sup>, where <sup>0</sup> , <sup>r</sup> , <sup>1</sup> and SF4½s� � <sup>s</sup> <sup>≥</sup>1, where <sup>s</sup> > 1: AM½ � <sup>1</sup> AM½ � <sup>1</sup>

Checking against NormalDistribution½μ; σ�, we see their minimum based on ½ �∗AM½ � <sup>2</sup> AM½ � <sup>3</sup> <sup>∗</sup>AM½ � <sup>1</sup> AM½ � <sup>2</sup> absolute moment: AM <sup>4</sup> <sup>2</sup> , <sup>2</sup> , and SF4 2½ �¼ 2, are all 1, but that by AM½ � <sup>3</sup> AM½ � <sup>2</sup> AM½ � <sup>1</sup> absolute central moment are not: min SF1 = 1.1781, min SF2 = 1.27324, min SF3 [2] =1.5708. Moreover, the convex index SF1, SF2, and SF3 out of absolute central moment are shift invariant besides the scale transformation invariant of the random variable, so they are preferred over the ones based on absolute moment.

½ �∗M½ � <sup>2</sup> ½ �∗M½ � <sup>1</sup> The only case in favor of <sup>M</sup> <sup>4</sup> and <sup>M</sup> <sup>3</sup> <sup>2</sup> <sup>2</sup> is when the numerical calculation <sup>M</sup>½ � <sup>3</sup> <sup>M</sup>½ � <sup>2</sup> error with extreme parameters arrive at negative kurtosis, then the calculated SF are meaningless (An example of BesselK function inaccuracy brings about negative kurtosis for generalized hyperbolic distribution can be found in [8]).

Even though both SF and SF1 are invariant under linear transformation of the distribution, and both measure the convexity, SF ≥ SF1 can additionally measure the asymmetry, combining these two into one quantity. Since most distributions in reinsurance are not symmetric, SF is preferred over SF1. That only SF measured both asymmetry and convexity, while the others cannot, can be seen from Figure 1, �xn <sup>1</sup>þ<sup>n</sup> for the case of exponential distribution family with PDF <sup>e</sup> nx� , x<sup>∈</sup> <sup>ð</sup>0; <sup>∞</sup>Þ, n <sup>&</sup>gt; 0, which is WeibullDistribution½n; 1� or GammaDistribution½1; 1; n; 0�, where only SF has a nontrivial interior global minimum.
