Central Limit Theorem: Approximations for Linear Combinations of iid Random Variables

The CLT saves the day: for 10,000 non-normal policyholders, the exact sum distribution is impossible to compute, but the sum is approximately normal regardless of the original distribution. This is why actuaries can use normal-based calculations even for skewed losses.

Let $X_1,X_2,\dots ,X_n$ be iid random variables with mean $\mu$ and finite variance ${\sigma }^2>0$. Define the sum $S_n=X_1+X_2+\cdots +X_n$. Then as $n\to \infty$:

$$\frac{S_n-n\mu }{\sigma \sqrt{n}}\stackrel{d}{\to }Z\sim N(0,1)$$

Equivalently, for the sample mean $\bar{X}=S_n/n$:

$$\frac{\bar{X}-\mu }{\sigma /\sqrt{n}}\stackrel{d}{\to }Z\sim N(0,1)$$

HIGH-FREQUENCY: The CLT applies to any distribution with finite mean and variance, exponential, Poisson, Bernoulli, uniform, or any other. The distribution does NOT need to be symmetric or continuous.

For finite (but large) $n$, the CLT gives the approximation:

Common mistakes

• Forgetting to take the square root of the variance. The z-score divides by $\sigma \sqrt{n}$, not ${\sigma }^{2}n$. In Example 1, using $\text{Var}(S)=525$ instead of $\text{SD}(S)=22.91$ gives $z=20/525=0.038$, producing a wildly wrong probability near 0.5. Trap: $\Phi (0.038)\approx 0.515$.
• Using $\mu$ instead of $n\mu$ as the mean of the sum. The mean of $S_n$ is $n\mu$, not $\mu$. In Example 1, using $\mu =3.5$ gives $z=(650-3.5)/22.91$, which is nonsensical. The sum of 180 dice has mean 630, not 3.5.
• Applying the CLT with the wrong variance for the binomial. $\text{Var}(S)=np(1-p)$, not $np$. With $n=400,p=0.1$: the correct variance is $36$, not $40$. Using 40 gives $\text{SD}=\sqrt{40}\approx 6.32$, which produces $z=10/6.32=1.582$, close to the continuity-corrected value, creating a subtle trap.

Bottom line

• CLT: $S_n\approx N(n\mu ,n{\sigma }^2)$ for large $n$, regardless of the original distribution
• Sample mean: $\bar{X}\approx N(\mu ,{\sigma }^2/n)$; standardize with $z=(\bar{x}-\mu )/(\sigma /\sqrt{n})$
• Sum z-score: $z=(s-n\mu )/(\sigma \sqrt{n})$; always divide by the standard deviation, never the variance
• Continuity correction: add or subtract 0.5 for discrete counts only (binomial, Poisson, integer sums), never a continuous mean; push the boundary 0.5 toward the region

Exam shortcut

CLT problems follow a template: (1) identify $\mu$ and ${\sigma }^2$, (2) compute $n\mu$ and $n{\sigma }^2$ (or ${\sigma }^2/n$ for the mean), (3) standardize, (4) look up $\Phi$. If the question asks about the sum, variance scales up by $n$. If about the mean, variance scales down. "Sum scales up, Mean scales down:" $\text{Var}(S_n)=n{\sigma }^2$ grows with $n$. $\text{Var}(\bar{X})={\sigma }^2/n$ shrinks. Continuity correction. "0.5 toward the region:" $P(S\le k)$ uses $k+0.5$. $P(S\ge k)$ uses $k-0.5$.

The full lesson (about 3,334 words, 22 min read) adds 3 worked examples, all 6 common mistakes, a self-check, free in the app.

Learning objectives

• 3i

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