Risk measurement
10 November 2021
5th week
1. Probability and Statistics
• Probability distribution
• Mean
• Standard deviation
2. Law of large numbers
• Central limit theorem
3. Loss Forecasting
• Probability analysis
• Regression analysis
• Forecasting based on loss distributions
10 November 2021
Overview
• To determine expected losses, insurance actuaries apply probability and
statistical analysis to given loss situations.
• The probability of an event is simply the long-run relative frequency of the
event, given an infinite number of trials with no changes in the underlying
conditions.
• The probability of some events can be determined without experimentation e.g.:
If a “fair”coin is flipped in the air, the probability the coin will come up
”heads”is 50%, and the probability it will come up ”tails”is also 50%.
• Other probabilities, such as the probability of dying during a specified year or
the probability of being involved in an auto accident, can be estimated from
past loss data.
10 November 2021
Probability
• A convenient way of summarizing events and probabilities is through a
probability distribution.
• A probability distribution lists events that could occur and the corresponding
probability of each event’s occurrence.
• Probability distributions may be discrete, meaning that only distinct outcomes
are possible, or continuous, meaning that any outcome over a range of
outcomes could occur.
• e.g.: The number of runs scored in a baseball game is a discrete measure, as
partial runs cannot be scored. Speed and temperature are continuous
measures, as all values over the range of values can occur.
10 November 2021
Probability distribution
• Probability distributions are characterized by two important measures: central
tendency and dispersion.
• Although there are several measures of central tendency, the measure most
often employed is the mean (µ) or expected value (EV) of the distribution.
• (Other measures of central tendency are the median, which is the middle
observation in a probability distribution, and the mode, which is the
observation that occurs most often.)
• The mean or expected value is found by multiplying each outcome by the
probability of occurrence, and then summing the resulting products:
µ or EV = ∑XiPi
10 November 2021
Probability distribution
e.g.: Assume that an actuary estimates the following probabilities of various losses
for a certain risk:
Amount of Loss (Xi) Probability of Loss (Pi) XiPi
$0 X 0.30 = $0
$360 X 0.50 = $180
$600 X 0.20 = $120
∑XiPi = $300
Thus, we could say that the mean or expected loss given the probability
distribution is $300.
10 November 2021
Mean
Although the mean value indicates central tendency, it does not tell us anything
about the riskiness or dispersion of the distribution. Consider a 2nd probability-ofloss
distribution:
Amount of Loss (Xi) Probability of Loss (Pi) XiPi
$225 X 0.40 = $90
$350 X 0.60 = $210
∑XiPi = $300
This distribution also has a mean loss value of $300. However, the 1st distribution
is riskier because the range of possible outcomes is from $0 to $600. With the 2nd
distribution, the range of possible outcomes is only $125 ($350 — $225), so we
are more certain about the outcome with the 2nd distribution.
10 November 2021
Mean
• 2 standard measures of dispersion to characterize the variability or dispersion
about the mean value: the variance (σ 2 ) and the standard deviation (σ)
• The variance of a probability distribution is the sum of the squared differences
between the possible outcomes and the expected value, weighted by the
probability of the outcomes:
σ 2 = ∑ Pi(Xi − EV ) 2
• is the average squared deviation between the possible outcomes and the mean
10 November 2021
Standard deviation
Because the variance is in ”squared units”, it is necessary to take the square root of
the variance so that the central tendency and dispersion measures are in the same
units. The square root of the variance is the standard deviation. The variance and
standard deviation of the first distribution:
σ 2 = 0.30(0 − 300)2 + 0.50(360 − 300)2 + 0.20(600 − 300)2
= 27, 000 + 1, 800 + 18, 000
= 46, 800
σ = √46, 800 = 261.33
10 November 2021
Standard deviation
For the 2nd distribution, the variance and standard deviation:
σ 2 = 0.40(225 − 300)2 + 0.60(350 − 300)2
= 2, 250 + 1, 500
= 3, 750
σ = √ 3, 750 = 61.24
Thus, while the means of the two distributions are the same, the standard
deviations are significantly different.
• Higher standard deviations, relative to the mean, are associated with greater
uncertainty of loss; therefore, risk is higher.
• Lower standard deviations, relative to the mean, are associated with less
uncertainty of loss; therefore, risk is lower.
10 November 2021
Standard deviation
• The two probability distributions used in the discussion of central tendency and
dispersion are ”odd”in that only three and two possible outcomes, respectively,
could occur.
• In addition, specific probabilities corresponding to the loss levels are assigned.
In practice, estimating the frequency and severity of loss is difficult.
• Insurers can employ both actual loss data and theoretical loss distributions in
estimating losses (e.g.: the binomial, Poisson and normal distribution).
10 November 2021
Probability distribution
• Even if the characteristics of the population were known with certainty,
insurers do not insure populations.
• Rather, they select a sample from the population and insure the sample.
• Obviously, the relationship between population parameters and the
characteristics of the sample (mean and standard deviation) is important for
insurers, since actual experience may vary significantly from the population
parameters.
• The characteristics of the sampling distribution help to illustrate the law of
large numbers, the mathematical foundation of insurance.
10 November 2021
Law of large numbers
• It can be shown that the average losses for a random sample of n exposure
units will follow a normal distribution because of the Central Limit Theorem,
which states:
• If you draw random samples of n observations from any population with mean
µx and standard deviation σx, and n is sufficiently large, the distribution of
sample means will be approximately normal, with the mean of the distribution
equal to the mean of the population µx = µx and the standard error of the
sample mean σx equal to the standard deviation of the population (σx) divided
by the square root of n(σx = σx/ √ n).
• This approximation becomes increasingly accurate as the sample size, n,
increases.
10 November 2021
Law of large numbers
The central limit theorem - implications for insurers:
1. The sample distribution of means does not depend on the population
distribution, provided n is sufficiently large. Regardless of the
population distribution (bimodal, unimodal, symmetric, skewed right,
skewed left, and so on), the distribution of sample means will
approach the normal distribution as the sample size increases.
10 November 2021
Central limit theorem
10 November 2021
1. Sampling distribution versus sample size
• The normal distribution is a symmetric, bell-shaped curve.
• It is defined by the mean and standard deviation of the distribution.
• About 68% of the distribution lies within one standard deviation of the mean,
and about 95% of the distribution lies within two standard deviations of the
mean.
• The normal curve has many statistical applications (hypothesis testing,
confidence intervals, and so on) and is easy to use.
10 November 2021
1. Sampling distribution versus sample size
2. The standard error of the sample means distribution declines as the sample
size increases.
Recall that the standard error is defined as σx = σx/ √ n the standard error of
the sample mean loss distribution is equal to the standard deviation of the
population divided by the square root of the sample size.
Because the population standard deviation is independent of the sample size,
the standard error of the sampling distribution, σx, can be reduced by simply
increasing the sample size.
10 November 2021
2. Central limit theorem implication
e.g.: Assume that an insurer would like to select a sample to insure from a
population where the mean loss is $500 and the standard deviation is $350. As the
insurer increases the number of units insured (n), the standard error of the
sampling distribution σx will decline. The standard error for various sample sizes
is summarized below:
n σx
10 110.68
100 35.00
1,000 11.07
10,000 3.50
100,000 1.11
Thus, as the sample size increases, the difference between actual results and
expected results decreases. Indeed, approaches zero as n gets very large.
10 November 2021
2. Standard error of the sampling distribution versus sample size
10 November 2021
2. Standard error of the sampling distribution versus sample size
• Obviously, when an insurer increases the size of the sample insured,
underwriting risk (maximum insured losses) increases because more insured
units could suffer a loss.
• The underwriting risk for an insurer is equal to the number of units insured
multiplied by the standard error of the average loss distribution, σx.
• Recalling that σx is equal to σx/ √ n, the expression for underwriting risk: n × σx
= n × σx/ √ n = √ n × σx
• Thus, while underwriting risk increases with an increase in the sample size, it
does not increase proportionately.
• Insurance companies are in the loss business — they expect some losses will
occur. It is the deviation between actual losses and expected losses that is the
major concern. By insuring large samples, insurers reduce their objective risk.
There truly is ”safety in numbers” for insurers.
10 November 2021
2. Standard error of the sampling distribution versus sample size
• A risk manager must also identify the risks the organization faces, and then
analyze the potential frequency and severity of these loss exposures.
• Although loss history provides valuable information, there is no guarantee that
future losses will follow past loss trends.
• Risk managers can employ a number of techniques to assist in predicting loss
levels:
1. Probability analysis
2. Regression analysis
3. Forecasting based on loss distributions
10 November 2021
Loss Forecasting
• Chance of loss is the possibility that an adverse event will occur.
• The probability (P) of such an event is equal to the number of events likely to
occur (X) divided by the number of exposure units (N).
• Thus, if a vehicle fleet has 500 vehicles and on average 100 vehicles suffer
physical damage each year, the probability that a fleet vehicle will be damaged
in any given year is:
P(physical damage) = 100/500 = 0.20 or 20%
10 November 2021
1. Probability analysis
Types of events:
1. Independent events
• the occurrence does not affect the occurrence of another event
• e.g.: Assume that a business has production facilities in Prague and
Brno, and that the probability of a fire at the Prague plant is 5% and
that the probability of a fire at the Brno plant is 4%. Obviously, the
occurrence of one of these events does not influence the occurrence of
the other event. If events are independent, the probability that they will
occur together is the product of the individual probabilities. Thus, the
probability that both production facilities will be damaged by fire is:
P(fire at Prague plant) × P(fire at Brno plant) = P(fire at both plants) =
0.04 × 0.05 = 0.002 or 0.2%
10 November 2021
1. Probability analysis
2. Dependent events
• the occurrence of one event affects the occurrence of the other
• If 2 buildings are located close together, and one building catches on
fire, the probability that the other building will burn is increased.
• e.g.: Suppose that the individual probability of a fire loss at each
building is 3%. The probability that the 2nd building will have a fire
given that the 1st building has a fire, however, may be 40%. What is
the probability of two fires? This probability is a conditional
probability that is equal to the probability of the 1st event multiplied
by the probability of the 2nd event given that the 1st event has
occurred:
P(fire at one bldg) × P(fire at second plant given fire at first bldg) =
P(both burn) = 0.03 × 0.40 = 0.012 or 1.20%
10 November 2021
1. Probability analysis
3. Mutually exclusive events
• if the occurrence of one event precludes the occurrence of the 2nd
event
• e.g.: If a building is destroyed by fire, it cannot also be destroyed by
flood.
• probabilities are additive – e.g.: If the probability that a building will
be destroyed by fire is 2% and the probability that the building will be
destroyed by flood is 1%, then the probability the building will be
destroyed by either fire or flood is:
P(fire destroys bldg) + P(flood destroys bldg) = P(fire or flood
destroys bldg) = 0.02 + 0.01 = 0.03 or 3%
10 November 2021
1. Probability analysis
4. If the independent events are not mutually exclusive, then more than 1 event
could occur.
• Care must be taken not to ”double-count”when determining the
probability that at least one event will occur.
• e.g.: If the probability of minor fire damage is 4% and the probability
of minor flood damage is 3%, then the probability of at least 1 of these
events occurring is:
P(minor fire) + P(minor flood) − P(minor fire and flood) =
P(at least one event) 0.04 + 0.03 − (0.04 × 0.03) = 0.0688 or 6.88%
Assigning probabilities to individual and joint events and analyzing
the probabilities can assist the risk manager in formulating a risk
treatment plan.
10 November 2021
1. Probability analysis
• characterizes the relationship between 2 or more variables and then uses this
characterization to predict values of a variable
• 1 (the dependent) variable is hypothesized to be a function of 1 or more
independent variables.
• It is not difficult to envision relationships that would be of interest to risk
managers in which 1 variable is dependent on another variable.
• e.g.: Consider workers compensation claims. It is logical to hypothesize that
the number of workers compensation claims should be positively related to
some variable representing employment (such as the number of employees,
payroll, or hours worked).
• e.g.: We would expect the number of physical damage claims for a fleet of
vehicles to increase as the size of the fleet increases or as the number of miles
driven each year by fleet vehicles increases.
10 November 2021
2. Regression analysis
• The number of claims is plotted against payroll.
• Regression analysis provides the coordinates of the line that best fits the points
on the chart.
• This line will minimize the sum of the squared deviations of the points from
the line. Our hypothesized relationship: Number of workers compensation
claims = B0 + (B1 × P ayroll [in thousands]) where B0 is a constant and B1 is
the coefficient of the independent variable.
The coefficient of determination, R-square, ranges from 0 to 1 and measures
the model fit. An R-square value close to 1 indicates that the model does a
good company’s annual payroll in thousands of dollars and job of predicting Y
values. By substituting the estimated payroll for next year (in thousands), the
risk manager estimates that 509 workers compensation claims will occur in the
next year.
10 November 2021
2. Regression analysis
• A loss distribution is a probability distribution of losses that could occur.
• Forecasting by using loss distributions works well if losses tend to follow a
specified distribution and the sample size is large.
• Knowing the parameters that specify the loss distribution (for example, mean,
standard deviation, and frequency of occurrence) enables the risk manager to
estimate the number of events, severity, and confidence intervals.
10 November 2021
3. Forecasting based on loss distributions
• Assume that the number of weather-related property losses is normally
distributed with a mean (m) equal to 16 and standard deviation (s) equal to 3.
What is the probability that the number of weather-related property losses will
be between 16 and 22?
• Assume the number of physical damage losses for a large fleet of vehicles is
normally distributed with a mean of 400 and a standard deviation of 80. What
is the probability that: a. More than 440 losses will occur? b. Between 320 and
480 losses will occur? c. Between 460 and 520 losses will occur?
10 November 2021
Loss forecasting based on the normal distribution
Crisis matrix was designed by Klaus Winterling. The matrix is one of analytical
techniques used in risk management.
The matrix allows risks categorization by two parameters:
Probability of a risk occur at a given time - how real and probable is that the risk will
actually occurs - matrix defines three levels of probability
- low
- medium
- high
Risk effects on an SBU - what what would be the impacts of the risk on an organization
or department if the risk occurs - matrix defines three levels of effect
- negative,
- threatening
- destructive
Thank you for your attention!
10 November 2021