Random Sample

Study guide: Exam MAS-I

A Random Sample of size $n$ from a distribution with density $f(x \mid \theta)$ is a set of independent and identically distributed random variables $X_1, \ldots, X_n$ , each with that distribution; the observed values $x_1, \ldots, x_n$ are the data. Functions of the sample that involve no unknown parameter — statistics — are what estimate the population's features.

$f(x_1, \ldots, x_n \mid \theta) = \prod_{i=1}^{n} f(x_i \mid \theta)$

$\bar{X} = \frac{1}{n}\sum_{i=1}^{n} X_i$

$S^2 = \frac{1}{n-1}\sum_{i=1}^{n} \left(X_i - \bar{X}\right)^2$

$\theta$ is the unknown population parameter. Read as a function of $\theta$ with the data plugged in, the product is the likelihood behind Maximum Likelihood Estimation; matching sample moments to population ones instead is the Method of Moments.
The Sample Mean and Sample Variance estimate $\mu$ and $\sigma^2$ without bias: $E[\bar{X}] = \mu$ , $\text{Var}(\bar{X}) = \sigma^2/n$ and $E[S^2] = \sigma^2$ . The estimated standard error of $\bar{X}$ is $S/\sqrt{n}$ .
Population quantities ( $\mu$ , $\sigma^2$ , $\theta$ ) are fixed and unknown; statistics are random variables with a Sampling Distribution of their own.
Sorting the sample gives the Order Statistics $X_{(1)} \le \cdots \le X_{(n)}$ : the sample minimum, maximum, median and empirical percentiles. Their distributions come from counting how many observations fall below $x$ — a binomial $(n, F(x))$ count — so, for example, $P(X_{(n)} \le x) = F(x)^n$ .
When data are not a random sample. Claims from one catastrophe, or several years of one insured, are dependent (Clustered Data). Claims recorded only above a deductible come from a conditional distribution (Truncation), and claims capped at a policy limit are censored. Each breaks the product formula and must be modelled explicitly.

Estimating Mean and Variance from Six Claims Six claims (in $000s) are $2, 5, 6, 9, 13, 19$ . Estimate the mean, the variance and the standard error of the mean, and find the sample median.
Answer
 $\begin{align*} \bar{x} &= \frac{54}{6} \\ &= 9 \\ \sum x_i^2 &= 4 + 25 + 36 + 81 + 169 + 361 \\ &= 676 \\ s^2 &= \frac{676 - 6(9)^2}{6 - 1} \\ &= \frac{190}{5} \\ &= 38 \end{align*}$ 
So $s = 6.16$ and the standard error of $\bar{x}$ is $6.16/\sqrt{6} = 2.52$ . The sample median is the average of the third and fourth order statistics, $(6 + 9)/2 = 7.5$ .

The mean claim is estimated at $9,000, give or take about $2,500. The median of $7,500 sitting below the mean is the right skew typical of severity data.

Sample Median of Five Exponential Claims Five claims are a random sample from an exponential distribution with mean $1,000. Find the probability that the sample median exceeds the population mean.
Answer The median is $X_{(3)}$ . It exceeds 1,000 exactly when at most 2 of the 5 claims are at or below 1,000. Each claim is, independently, with probability $p = F(1{,}000) = 1 - e^{-1} = 0.6321$ , so the count is binomial $(5, p)$ with $1 - p = e^{-1}$ :
 $\begin{align*} P(X_{(3)} > 1{,}000) &= \sum_{j=0}^{2} \binom{5}{j} p^j (1-p)^{5-j} \\ &= 0.0067 + 0.0579 + 0.1989 \\ &= 0.2636 \end{align*}$ 
The sample median lands above the true mean only about 26% of the time. It estimates the population median, $1{,}000 \ln 2 = 693$ , not the mean — a right-skewed sample's median is the wrong statistic for pricing.