Empirical Distribution

Study guide: Exam MAS-I

The Empirical Distribution of a sample x_1, \ldots, x_n is the discrete distribution that puts probability 1/n on each observation. Its CDF, the empirical distribution function F_n, is the proportion of the sample at or below x — the basic nonparametric estimate of an unknown CDF F from complete individual data.

F_n(x) = \frac{1}{n}\sum_{i=1}^{n} \mathbf{1}\{x_i \le x\}

= \frac{\#\{x_i \le x\}}{n}

• F_n is a right-continuous step function that jumps by 1/n at each observation (by m/n at a value observed m times), with F_n = 0 below x_{(1)} and F_n = 1 from x_{(n)} on.
• At each fixed x, n F_n(x) is binomial with parameters n and F(x), so F_n(x) is unbiased with \text{Var}(F_n(x)) = F(x)\,[1 - F(x)]/n, estimated by F_n(x)[1 - F_n(x)]/n. By the weak law of large numbers it is consistent: F_n(x) \to F(x) in probability.
• Empirical moments. The empirical distribution's mean is the sample mean \bar{x}; its variance divides by n, \frac{1}{n}\sum (x_i - \bar{x})^2, not by n-1 as the unbiased sample variance does.
• Plug-in estimates. Any quantity defined from F can be estimated by computing it under F_n: E[g(X)] \approx \frac{1}{n}\sum g(x_i), such as a limited expected value or a loss elimination ratio. Its percentiles feed Percentile Matching, and it is the yardstick in the Kolmogorov-Smirnov Test.
• F_n needs complete data: with censored or truncated observations the proportion at or below x is unknown, and product-limit (Kaplan–Meier) methods replace it. Kernel Density Estimation smooths the empirical distribution into a continuous density.

Empirical CDF and Moments of a Claim Sample Ten claims are reported: 200, 350, 350, 500, 500, 500, 800, 1{,}200, 1{,}200, 3{,}000. Find F_{10}(350), F_{10}(499), F_{10}(500) and F_{10}(1{,}000), and the mean and variance of the empirical distribution.

Answer Count the claims at or below each point: F_{10}(350) = 3/10 = 0.3; F_{10}(499) = 0.3 (flat between data points); F_{10}(500) = 6/10 = 0.6 (a jump of 3/10 for the three tied claims); F_{10}(1{,}000) = 0.7.

\begin{align*}
\bar{x} &= \frac{8{,}600}{10} \\
  &= 860 \\
\text{Var} &= \frac{1}{10}\sum_{i=1}^{10}(x_i - 860)^2 \\
  &= \frac{6{,}159{,}000}{10} \\
  &= 615{,}900
\end{align*}

The empirical standard deviation is \sqrt{615{,}900} = 784.8. (The unbiased sample variance would divide by 9 instead: 684{,}333.)

Plug-In Loss Elimination Ratio and a Standard Error Using the same ten claims, estimate the loss elimination ratio of a $500 deductible, and the standard error of F_{10}(500) as an estimate of F(500).

Answer Under the empirical distribution, E[X \wedge 500] is the average of the claims capped at 500:

\begin{align*}
\widehat{E}[X \wedge 500] &= \frac{200 + 350 + 350 + 500 \times 7}{10} \\
  &= 440 \\
\widehat{\text{LER}}(500) &= \frac{440}{860} \\
  &= 0.512
\end{align*}

So the deductible would eliminate about 51% of losses. For the standard error, F_{10}(500) = 0.6:

\begin{align*}
\widehat{\text{Var}}\big(F_{10}(500)\big) &= \frac{0.6 \times 0.4}{10} \\
  &= 0.024
\end{align*}

giving a standard error of \sqrt{0.024} = 0.155 — ten claims pin down F(500) only roughly.