Severity Discretization

Study guide: Exam ASTAM

Severity Discretization replaces a continuous severity distribution with an arithmetic one, a discrete severity distribution that puts all of its probability on the multiples $0, h, 2h, \ldots$ of a span $h$ . The aggregate claims distribution can then be computed exactly for the discretized severity, by convolution or by the recursive formula. Loss Models gives two methods: the method of rounding, and local moment matching, which keeps both the probability and the mean of every span.

$f_0 = F_X\!\left(\tfrac{h}{2}\right)$

$f_j = F_X\!\left(jh + \tfrac{h}{2}\right) - F_X\!\left(jh - \tfrac{h}{2}\right)$

$\begin{aligned} m_0^k + m_1^k &= F_X\big((k+1)h\big) - F_X(kh) \\ kh\,m_0^k + (k+1)h\,m_1^k &= \int_{kh}^{(k+1)h} x\,dF_X(x) \end{aligned}$

$f_0 = 1 - \frac{E[X \wedge h]}{h}$

$f_j = \frac{2E[X \wedge jh] - E[X \wedge (j-1)h] - E[X \wedge (j+1)h]}{h}$

Symbols: $h$ is the span, and $f_j$ is the probability that the discretized severity places on $jh$ , for $j = 1, 2, \ldots$ in the second and last blocks. The first two blocks are the method of rounding. In local moment matching, $m_0^k$ and $m_1^k$ are the masses placed at the two ends of the span $[kh, (k+1)h]$ , and the point $jh$ collects $f_j = m_0^j + m_1^{j-1}$ from the two spans that meet there. $E[X \wedge u]$ is the limited expected value.
Why the recursion needs it: The recursive formula sums over the claim amounts $y = 1, 2, \ldots, x$ , so it works only when the severity lives on a lattice. With a continuous severity, the same recursion becomes an integral equation that rarely has a closed-form solution. Discretizing first turns it back into a finite sum, and a smaller span buys accuracy at the cost of more terms.
Rounding: Each point $jh$ collects the probability within half a span on either side of it, so the discretized CDF agrees with $F_X$ at the midpoints $jh + h/2$ . The method does not keep the mean, because every loss is moved to its nearest grid point.
Local moment matching: The two span equations in the third block keep each span's probability and its contribution to the mean. As a result, the discretized severity has the same mean as $X$ and the same limited expected value at every multiple of $h$ . Solving the span equations gives the closed forms in the last two blocks.

Rounding an Exponential Severity Claim amounts are exponential with mean 1,000. Discretize the severity with a span of $h = 500$ by the method of rounding, and find the probabilities at 0, 500 and 1,000.
Answer With $F_X(x) = 1 - e^{-x/1000}$ , each grid point collects the probability within 250 of it.
 $\begin{align*} f_0 &= F_X(250) \\ &= 1 - e^{-0.25} \\ &= 0.2212 \\ f_1 &= F_X(750) - F_X(250) \\ &= e^{-0.25} - e^{-0.75} \\ &= 0.3064 \\ f_2 &= F_X(1{,}250) - F_X(750) \\ &= e^{-0.75} - e^{-1.25} \\ &= 0.1859 \end{align*}$ 
The discretized severity puts probability 0.2212 on a claim of zero, although no real claim is zero. Summing $jh\,f_j$ over every point gives a mean of 989.66 rather than 1,000. Rounding the claims below 250 down to zero removes 26.50 from the mean, and rounding the larger claims gives back only 16.16 of it.

Local Moment Matching on the First Two Spans For the same severity and span, find $f_0$ and $f_1$ by local moment matching. Check them against the closed forms.
Answer On $[0, 500]$ , the probability is $F_X(500) = 0.393469$ . The integral is $\int_0^{500} x\,dF_X(x) = E[X \wedge 500] - 500\,S_X(500) = 393.469 - 303.265 = 90.204$ . The mass at 0 contributes nothing to it, so the span equations give
 $\begin{align*} m_1^0 &= \frac{90.204}{500} \\ &= 0.180408 \\ m_0^0 &= 0.393469 - 0.180408 \\ &= 0.213061 \end{align*}$ 
On $[500, 1{,}000]$ , the probability is $0.238651$ and the integral is $174.037$ . Solving $m_0^1 + m_1^1 = 0.238651$ and $500\,m_0^1 + 1{,}000\,m_1^1 = 174.037$ gives $m_1^1 = 0.109423$ and $m_0^1 = 0.129228$ . The point 500 collects mass from both spans.
 $\begin{align*} f_0 &= m_0^0 \\ &= 0.2131 \\ f_1 &= m_1^0 + m_0^1 \\ &= 0.180408 + 0.129228 \\ &= 0.3096 \end{align*}$ 
The closed forms agree: $f_0 = 1 - 393.469/500 = 0.2131$ , and $f_1 = [2(393.469) - 0 - 632.121]/500 = 0.3096$ . Unlike rounding, this discretization keeps the mean at exactly 1,000.

How it appears on the exam

ASTAM asks candidates to derive the discretized version of a continuous distribution by the method of rounding and by local moment matching. A question gives the severity and the span and asks for the first few probabilities.

Aggregate claims: The discretized severity then feeds the recursive formula or a convolution, so the same question often continues to an aggregate probability or a stop-loss premium.

Closed forms: For local moment matching, the limited-expected-value form is usually quicker than solving the span equations, because the distribution tables give $E[X \wedge u]$ for every distribution they list.

Exam traps

The first point: Rounding gives $f_0 = F_X(h/2)$ , because the first interval is only half a span wide. Using $F_X(h)$ double-counts the probability between $h/2$ and $h$ .

A mass at zero: The discretized severity has $f_0 > 0$ even when every real claim is positive. The recursion then needs the factor $1/(1 - a f_0)$ and the starting value $P_N(f_0)$ rather than $\Pr(N = 0)$ .

Both spans: In local moment matching, an interior point receives mass from the span on each side of it. Taking only $m_0^j$ or only $m_1^{j-1}$ for $f_j$ leaves probabilities that sum to less than 1.

Units of the span: The recursion runs in units of $h$ . After discretizing with $h = 500$ , an aggregate value of $x = 3$ means 1,500, so a deductible or limit must be converted to spans first.