Statistics

Study guides: Exam PA, Exam MAS-I

Statistics is the discipline of collecting, summarizing, and drawing inferences about a population from sample data. A statistic is any quantity computed from a sample; the most fundamental are the sample mean $\bar{X}$ and sample variance $S^2$ , which estimate the population mean $\mu$ and variance $\sigma^2$ .

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

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

Dividing the sum of squared deviations by $n-1$ (rather than $n$ ) makes $S^2$ an unbiased estimator of $\sigma^2$
Every statistic has its own Sampling Distribution, which quantifies how it varies from sample to sample
Descriptive statistics summarize data (mean, variance, quantiles); inferential statistics generalize to the population via estimation (e.g. Maximum Likelihood Estimation) and Hypothesis Testing
Statistics builds on Probability: a probability model for the data is assumed, then its parameters are estimated and tested
Actuaries use statistics to model claim frequency and severity, set premiums, and quantify reserve uncertainty

Estimating the Mean and Variance of Claim Sizes Five reported claims (in $ $1{,}000$ s) are $4, 6, 7, 9, 9$ . Estimate the mean and variance of the claim-size distribution.

Answer $\bar{X} = \frac{4+6+7+9+9}{5} = \frac{35}{5} = 7$ Sum of squared deviations: $(4-7)^2 + (6-7)^2 + (7-7)^2 + (9-7)^2 + (9-7)^2 = 9 + 1 + 0 + 4 + 4 = 18$ . $S^2 = \frac{18}{5-1} = 4.5$ The estimated mean claim is $ $7{,}000$ with sample variance $4.5$ (in $\$1{,}000^2$ units).

Standard Error of the Sample Mean Using the sample above ( $n = 5$ , $S^2 = 4.5$ ), estimate the standard error of $\bar{X}$ .

Answer The standard error estimates the standard deviation of the Sampling Distribution of $\bar{X}$ : $\text{SE}(\bar{X}) = \frac{S}{\sqrt{n}} = \frac{\sqrt{4.5}}{\sqrt{5}} = \frac{2.121}{2.236} \approx 0.949$ So the mean claim estimate of $ $7{,}000$ carries a standard error of about $ $949$ .