Law of Large Numbers

The Law of Large Numbers (LLN) states that the sample mean \bar{X}_n of n independent and identically distributed Random Variables with finite mean \mu converges to \mu as the sample size grows. It is the formal reason risk pooling works: the average loss per policy becomes predictable as an insurer writes more independent policies.

\bar{X}_n = \frac{1}{n}\sum_{i=1}^{n} X_i \;\xrightarrow[n\to\infty]{}\; \mu

• Weak LLN: for any tolerance \epsilon > 0, \ P\!\left(\,|\bar{X}_n - \mu| > \epsilon\,\right) \to 0 as n \to \infty — the sample mean is increasingly likely to land near \mu.
• It requires the X_i to be i.i.d. with finite mean \mu; the standard proof also assumes finite variance \sigma^2 and applies Chebyshev's inequality to \bar{X}_n, which has \text{Var}(\bar{X}_n) = \sigma^2/n.
• Distinguish it from the Central Limit Theorem: the LLN says where the sample mean settles (at \mu), while the CLT describes the shape of its fluctuations around \mu at scale 1/\sqrt{n}.
• It underpins Credibility Theory and the Expected Value premium principle — charging \mu per risk is sustainable only because averages stabilize over a large book of business.

Stability of Average Loss Across a Portfolio Each policy's annual loss is i.i.d. with mean \mu = 500 and standard deviation \sigma = 2{,}000. For a book of n = 10{,}000 policies, bound the probability that the average loss per policy differs from $500 by more than $100.

Answer The sample mean has variance \text{Var}(\bar{X}_n) = \sigma^2/n. Apply Chebyshev's inequality:

\begin{align*}
\text{Var}(\bar{X}_n) &= \frac{2{,}000^2}{10{,}000} \\
                      &= 400 \\
P\!\left(|\bar{X}_n - 500| > 100\right) &\leq \frac{\text{Var}(\bar{X}_n)}{100^2} \\
                      &= \frac{400}{10{,}000} \\
                      &= 0.04
\end{align*}

So there is at least a 96% chance the average loss per policy lands within $100 of $500 — the per-policy average is far more predictable than any single loss.

Sample Size Needed for a Target Precision Using the same per-policy loss (\mu = 500, \sigma = 2{,}000), how many policies are needed so that the average loss is within $50 of \mu with at least 95% probability (via Chebyshev)?

Answer Require the Chebyshev bound on the miss probability to be at most 0.05:

\begin{align*}
\frac{\sigma^2 / n}{50^2} &\leq 0.05 \\
n &\geq \frac{\sigma^2}{0.05 \cdot 50^2} \\
  &= \frac{2{,}000^2}{0.05 \cdot 2{,}500} \\
  &= \frac{4{,}000{,}000}{125} \\
  &= 32{,}000
\end{align*}

A book of at least 32,000 independent policies guarantees the target precision under the (conservative) Chebyshev bound.