Normal Distribution

Study guides: Exam P-1, Exam MAS-I

The Normal Distribution X \sim N(\mu, \sigma^2) is a symmetric, bell-shaped continuous distribution on (-\infty, \infty) fully characterized by its mean \mu and variance \sigma^2.

f(x) = \frac{1}{\sigma\sqrt{2\pi}}\,\exp\!\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)

x \in \mathbb{R}

\text{where } \mu = \text{mean},\quad \sigma^2 = \text{variance}

• Its CDF has no closed form and is evaluated via the standard normal Z = (X - \mu)/\sigma \sim N(0,1) using \Phi tables
• By the Central Limit Theorem, sums of many independent random variables are approximately normal, making it foundational for aggregate loss models
• Symmetry gives \Phi(-z) = 1 - \Phi(z) — the identity needed for every negative z, since standard tables only list z \geq 0
• Any linear combination of independent normals is exactly normal: if X_i \sim N(\mu_i, \sigma_i^2) independently, then \sum c_i X_i \sim N\!\left(\sum c_i\mu_i,\ \sum c_i^2\sigma_i^2\right). No approximation and no CLT required — see Probabilities for Linear Combinations
• Useful percentiles: z_{0.90} = 1.282, z_{0.95} = 1.645, z_{0.975} = 1.960, z_{0.99} = 2.326
• When approximating a discrete variable (a binomial or Poisson count) by a normal, apply the continuity correction: P(X \leq k) \approx \Phi\!\left(\frac{k + 0.5 - \mu}{\sigma}\right)

Probability that Aggregate Losses Exceed a Threshold Annual aggregate losses S \sim N(\mu = 50{,}000,\, \sigma^2 = 40{,}000{,}000). Find P(S > 55{,}000).

Answer Standardize: \sigma = \sqrt{40{,}000{,}000} \approx 6{,}324.6. Z = \frac{55{,}000 - 50{,}000}{6{,}324.6} \approx 0.791 P(S > 55{,}000) = P(Z > 0.791) = 1 - \Phi(0.791) \approx 1 - 0.7855 = 0.2145 There is approximately a 21.5% probability that aggregate losses exceed $55,000.

A Difference of Two Independent Normal Losses Two independent lines of business have annual losses X \sim N(100, 400) and Y \sim N(80, 225) (in thousands). Find P(X < Y).

Answer Recast the comparison as a single normal variable D = X - Y. Means subtract; variances add, even for a difference:

\begin{align*}
E[D] &= 100 - 80 = 20 \\
\text{Var}(D) &= 400 + 225 = 625 \\
\text{SD}(D) &= 25
\end{align*}

So D \sim N(20, 625) exactly, and:

\begin{align*}
P(X < Y) &= P(D < 0) \\
         &= \Phi\!\left(\frac{0 - 20}{25}\right) \\
         &= \Phi(-0.80) \\
         &= 1 - \Phi(0.80) \\
         &= 1 - 0.7881 \\
         &= 0.2119
\end{align*}

Subtracting the variances to get 175 is the standard trap — a difference is more variable than either component, not less.