Normal Approximation

Study guides: Exam P-1, Exam FAM

The Normal Approximation replaces the distribution of a sum or average of many independent and identically distributed random variables by a normal distribution with the same mean and variance, as the Central Limit Theorem justifies; probabilities are then read from the standard normal CDF \Phi.

P(S_n \le s) \approx \Phi\!\left(\frac{s - n\mu}{\sigma\sqrt{n}}\right)

P(\bar{X} \le x) \approx \Phi\!\left(\frac{x - \mu}{\sigma/\sqrt{n}}\right)

• S_n = X_1 + \cdots + X_n and \bar{X} = S_n/n is the sample mean; \mu and \sigma^2 are the mean and variance of one X_i. The recipe: identify n, compute E[S_n] = n\mu and \text{Var}(S_n) = n\sigma^2 (Moments for Linear Combinations), standardise, look up \Phi.
• Continuity correction. When S_n is integer-valued — a claim count, a binomial, a Poisson — widen the event by half a unit first: P(S_n \le k) \approx \Phi\!\left(\frac{k + 0.5 - n\mu}{\sigma\sqrt{n}}\right) and P(S_n \ge k) \approx 1 - \Phi\!\left(\frac{k - 0.5 - n\mu}{\sigma\sqrt{n}}\right). Never apply it to a continuous sum.
• A binomial(n, p) is approximately N\big(np,\ np(1-p)\big) and a Poisson(\lambda) approximately N(\lambda, \lambda) for large n or \lambda — each is a sum of i.i.d. pieces.
• If the summands are themselves normal, nothing is approximate: the sum is exactly normal (Probabilities for Linear Combinations).
• How large n must be depends on the summands: for a binomial, the usual rule of thumb is that np and n(1-p) are both at least 5.
• Inverted, it sets a fund: the total exceeded with probability \alpha is about n\mu + z_{1-\alpha}\,\sigma\sqrt{n}, where z_{1-\alpha} is the standard normal percentile.

Sizing a Claim Fund at the 95th Percentile A plan covers 400 independent members. Each member's annual claim has mean $250 and standard deviation $1,000. How large must the fund be for a 95% chance of covering total claims?

Answer

\begin{align*}
E[S] &= 400(250) \\
     &= 100{,}000 \\
\text{SD}(S) &= 1{,}000\sqrt{400} \\
             &= 20{,}000 \\
\text{Fund} &= 100{,}000 + 1.645(20{,}000) \\
            &= 132{,}900
\end{align*}

The fund is $132,900, or $332.25 per member: a loading of $82.25 (33%) over the expected claim, bought by the uncertainty in a 400-member pool.

Claim Count with a Continuity Correction Each of 1,000 independent policies has a claim with probability 0.05. Approximate the probability of at least 60 claims.

Answer N is binomial with E[N] = 50 and \text{Var}(N) = 1{,}000(0.05)(0.95) = 47.5, so \text{SD}(N) = 6.892. N is integer-valued, so P(N \ge 60) becomes P(N > 59.5):

\begin{align*}
P(N \ge 60) &\approx 1 - \Phi\!\left(\frac{59.5 - 50}{6.892}\right) \\
            &= 1 - \Phi(1.38) \\
            &= 1 - 0.9162 \\
            &= 0.0838
\end{align*}

The exact binomial answer is 0.0867. Standardising 60 without the correction gives 1 - \Phi(1.45) = 0.0735, noticeably further off.

How Many Policies for a Stable Loss Ratio Each policy's loss has coefficient of variation \sigma/\mu = 2. How many i.i.d. policies are needed so that total losses exceed 110% of expected with probability at most 5%?

Answer

\begin{align*}
P(S_n > 1.1n\mu) &\approx P\!\left(Z > \frac{0.1n\mu}{\sigma\sqrt{n}}\right) \\
                 &= P\!\left(Z > \frac{0.1\sqrt{n}}{2}\right)
\end{align*}

This is at most 0.05 when 0.05\sqrt{n} \ge 1.645:

\begin{align*}
\sqrt{n} &\ge 32.9 \\
n &\ge 1{,}082.4
\end{align*}

At least 1{,}083 policies. The answer scales with \text{CV}^2: a line with twice the CV needs four times the book.