Convolution

Study guides: Exam FAM, Exam ASTAM

Convolution is the operation that gives the distribution of a sum of independent random variables; in the aggregate loss model S = X_1 + \cdots + X_N it builds the distribution of k claims, f_X^{*k}, out of the severity distribution and then mixes over the claim count N.

f_X^{*k}(x) = \sum_{y=0}^{x} f_X^{*(k-1)}(x-y)\, f_X(y)

F_X^{*k}(x) = \sum_{y=0}^{x} F_X^{*(k-1)}(x-y)\, f_X(y)

f_S(x) = \sum_{k=0}^{\infty} \Pr(N=k)\, f_X^{*k}(x)

• The recursion starts from f_X^{*0}(0) = 1 and f_X^{*0}(x) = 0 for x > 0 (zero claims total zero), with f_X^{*1} = f_X. Severity is on the integers 0, 1, 2, \ldots in some monetary unit; a continuous severity is first discretized onto a span (by rounding or local moment matching).
• f_X^{*k} is the distribution of X_1 + \cdots + X_k for independent, identically distributed claims, which is why the model assumes the X_i are i.i.d. and independent of N (frequency-severity structure).
• If f_X(0) = 0, then f_X^{*k}(x) = 0 for k > x, so f_S(x) needs only the terms k = 0, \ldots, x. If f_X(0) > 0 (a deductible applied per loss) every k contributes, and the sum is finite only when N is bounded.
• For (a,b,0) and (a,b,1) frequencies the Panjer recursion gives the same probabilities more cheaply; for large portfolios the normal approximation replaces both.

Aggregate Claims From a Small Book A policy has at most two claims a year: \Pr(N=0)=0.5, \Pr(N=1)=0.3, \Pr(N=2)=0.2. Each claim is 1 or 2 (in \$1{,}000s) with probabilities 0.6 and 0.4, independent of N and of each other. Find \Pr(S \le 2) and the stop-loss premium E[(S-2)_+].

Answer Two claims convolve to

\begin{align*}
f_X^{*2}(2) &= 0.6^2 = 0.36 \\
f_X^{*2}(3) &= 2(0.6)(0.4) = 0.48 \\
f_X^{*2}(4) &= 0.4^2 = 0.16
\end{align*}

Mixing over N:

\begin{align*}
f_S(0) &= 0.5 \\
f_S(1) &= 0.3(0.6) = 0.18 \\
f_S(2) &= 0.3(0.4) + 0.2(0.36) = 0.192 \\
f_S(3) &= 0.2(0.48) = 0.096 \\
f_S(4) &= 0.2(0.16) = 0.032
\end{align*}

\begin{align*}
\Pr(S \le 2) &= 0.5 + 0.18 + 0.192 \\
&= 0.872 \\[4pt]
E[(S-2)_+] &= 1(0.096) + 2(0.032) \\
&= 0.16
\end{align*}

Aggregate claims stay at or below \$2{,}000 with probability 0.872, and a stop-loss cover attaching at \$2{,}000 costs \$160 in expectation.

Convolution With a Per-Loss Deductible The number of losses is binomial with m = 3, q = 0.4. Ground-up losses are 1, 2 or 3 (in \$1{,}000s) with probabilities 0.25, 0.40, 0.35. A deductible of 1 applies to each loss. Let S be aggregate payments. Calculate \Pr(S \le 2).

Answer Per loss, Y = (X-1)_+ takes values 0, 1, 2 with probabilities 0.25, 0.40, 0.35 — a mass at zero, so every k contributes. Binomial weights are 0.216, 0.432, 0.288, 0.064.

\begin{align*}
f_Y^{*2}(0) &= 0.0625 \\
f_Y^{*2}(1) &= 2(0.25)(0.40) = 0.2 \\
f_Y^{*2}(2) &= 2(0.25)(0.35) + 0.40^2 = 0.335 \\
f_Y^{*3}(0) &= 0.015625 \\
f_Y^{*3}(1) &= 3(0.25)^2(0.40) = 0.075 \\
f_Y^{*3}(2) &= 3(0.25)^2(0.35) + 3(0.25)(0.40)^2 = 0.185625
\end{align*}

\begin{align*}
f_S(0) &= 0.216 + 0.432(0.25) + 0.288(0.0625) + 0.064(0.015625) \\
&= 0.343 \\
f_S(1) &= 0.432(0.40) + 0.288(0.2) + 0.064(0.075) \\
&= 0.2352 \\
f_S(2) &= 0.432(0.35) + 0.288(0.335) + 0.064(0.185625) \\
&= 0.25956 \\[4pt]
\Pr(S \le 2) &= 0.343 + 0.2352 + 0.25956 \\
&= 0.83776
\end{align*}

Check: payments are binomial with q = 0.4(0.75) = 0.3, so \Pr(S=0) = 0.7^3 = 0.343. Aggregate payments are at most \$2{,}000 with probability about 0.838.

How it appears on the exam

• FAM (calculate probabilities using the convolution method): a frequency with two or three possible values and a severity on two or three points; asked for \Pr(S = x), \Pr(S \le x), or a stop-loss premium E[(S-d)_+] built from the first few f_S(x).
• ASTAM (derive probability and distribution functions ... with (a,b,0) or (a,b,1) frequency): the same calculation with a per-loss deductible or a zero-modified frequency, often with the instruction to use convolution for small x and the Panjer recursion as the alternative.
• Pairs with coverage modifications (the deductible creates f_Y(0) > 0) and with the stop-loss premium identity E[(S-d)_+] = E[S] - d + \sum_{x<d}(d-x)f_S(x).

Exam traps

• Dropping the k=0 term. f_X^{*0}(0) = 1, so f_S(0) always includes \Pr(N=0); leaving it out gives probabilities that sum to less than one.
• Assuming f_S(0) = \Pr(N=0) when f_X(0) > 0. With a per-loss deductible, f_S(0) = \sum_k \Pr(N=k) f_X(0)^k — in the second example 0.343, not 0.216.
• Multiplying CDFs. F_X^{*2}(x) is not F_X(x)^2; that is the distribution of the maximum of two claims, not their sum.
• Forgetting the order count. f_X^{*2}(3) = 2(0.6)(0.4), not (0.6)(0.4): 3 = 1+2 = 2+1.