Tag Archives: Geometric

Frequency with Respect to Exposure and Coverage Modifications

In a portfolio of risks, there are two types of modifications which can influence the frequency distribution of payments.

  1. Exposure Modification (not in syllabus) — increasing or decreasing the number of risks or time periods of coverage in the portfolio
  2. Coverage Modification — applying limits, deductible or adjusting for inflation in each individual risk
EXPOSURE MODIFICATION
If there is an exposure modification, you would adjust the frequency distribution by scaling the appropriate parameter to reflect the change in exposure.  The following list provides the appropriate parameter to adjust for each distribution:
  1. Poisson:  \lambda
  2. Negative Binomial:  r
    (Geometric is a Negative Binomial with r = 1)
  3. Binomial:  m
    (Only valid if the new value remains an integer)
Example 1:  You own a portfolio of 10 risks.  You model the frequency of claims with a negative binomial having parameters r = 2 and \beta = 0.5.  The number of risks in your portfolio increases to 15.  What are the parameters for the new distribution?
Answer:  The frequency distribution now has parameters r = 3 and \beta = 0.5  Note that since the mean and variance are r\beta and r\beta(1+\beta) respectively, the new mean and variance are multiplied by 1.5.
COVERAGE MODIFICATION
Coverage modifications shift, censor, or scale the individual risks, usually in the presence of a deductible or claim limit, and they change the conditions that trigger a payment.  For example, adding a deductible d is considered a coverage modification and this changes the condition for payment because any losses below the deductible do not qualify for payment.  If the risk is represented by random variable X, then adding a deductible would change the random variable to (X-d)_+. Scaling in the presence of a deductible or claim limit also affects the frequency distribution.  The following lists parameters affected by coverage modifications:
  1. Poisson:  \lambda
  2. Negative Binomial:  \beta
  3. Binomial:  q
  4. Geometric:  \beta
These parameters are scaled by the probability that a payment occurs.
Example 2:  The frequency of loss is modeled as a Poisson distribution with parameter \lambda = 5.  A deductible is imposed so that only 80% of losses result in payments.  What is the new distribution?
Answer:  It is Poisson with \lambda = 0.8(5) = 4.
Example 3:  The frequency of payment N is modeled as a negative binomial with parameters r = 3 and \beta = 0.5.  Losses X are pareto distributed with parameters \alpha = 2 and \theta = 100.  The deductible is changed from d=10 to d=15.  What are the new parameters in the frequency distribution?
Answer:  Firstly, N is the frequency of payment.  So it reflects the current deductible.  If you wanted the distribution of N without deductible, you would divide \beta by \Pr{(X>10)}.  Now to find the distribution of N with the deductible of 15, you multiply \beta by Pr(X>15).  To summarize:
\begin{array}{rll} \beta_{new} &=& \displaystyle \beta_{old}\times\frac{\Pr(X>15)}{\Pr(X>10)} \\ \\ &=& \displaystyle 0.5\times \frac{0.75614367}{0.82644628} \\ \\ &=& 0.45746692 \end{array}
Advertisements

Leave a comment

Filed under Coverage Modifications, Deductibles, Frequency Models, Limits

The Poisson Gamma Mixture Pattern

Suppose a random variable N has a frequency distribution that is Poisson with parameter \lambda. Suppose the parameter \lambda is also a random variable and it has a gamma distribution with parameters \alpha and \theta. Then N is equivalent to a negative binomial with parameters r = \alpha and \beta = \theta.

Note that

  1. When \alpha =1, the gamma distribution is equivalent to an exponential distribution.
  2. This also means the negative binomial has parameter r=1 which is equivalent to a geometric distribution.
Pop Quiz!
You own a space mining company and have sent several exploration bots to scout possible mineral rich asteroids.  Each bot discovers pockets of valuable resources on different asteroids at a rate of \lambda per year.  The parameter \lambda varies by bot according to an exponential distribution with parameter \theta = 3.
  1. What is the expected number of discoveries per year for a bot chosen at random?
    Answer:  3
  2. What is the variance?
    Answer:  12

Leave a comment

Filed under Frequency Models

Frequency Models

Frequency models count the number of times an event occurs.

  1. The number of customers to arrive each hour.
  2. The number of coins lucky Tom finds on his way home from school.
  3. How many scientists a Tyrannosaur eats on a certain day.
  4. Etc.
This is in contrast to a severity model which measures the magnitude of an event.
  1. How much a customer spends.
  2. The value of a coin that lucky Tom finds.
  3. The number of calories each scientist provides.
  4. Etc.
The following distributions are used to model event frequency.  For notation, p_n means Pr(N=n).

Poisson:

\begin{array}{lr}\displaystyle p_n = e^{-\lambda} \frac{\lambda^n}{n!} & \lambda > 0 \end{array}
Properties:
  1. Parameter is \lambda.
  2. Mean is \lambda.
  3. Variance is \lambda.
  4. If N_1, N_2, ..., N_i are Poisson with parameters \lambda_1, \lambda_2, ..., \lambda_i, then N = N_1 + N_2 + ... + N_i is Poisson with parameter \lambda = \lambda_1 + \lambda_2 + ... + \lambda_i.

Negative Binomial:

\begin{array}{lr} \displaystyle p_n = {{n+r-1}\choose{n}}\left(\frac{1}{1+\beta}\right)^r\left(\frac{\beta}{1+\beta}\right)^n & \beta>0, r>0 \end{array}
Properties:
  1. Parameters are r and \beta.
  2. Mean is r\beta.
  3. Variance is r\beta\left(1+\beta\right).
  4. Variance is always greater than the mean.
  5. Is equal to a Geometric distribution when r=1.
  6. If N_1, N_2, ..., N_i are negative binomial with parameters \beta_1 = \beta_2 = ... = \beta_i and r_1, r_2, ..., r_i, then the sum N = N_1 + N_2 + ... + N_i is negative binomial and has parameters \beta = \beta_1 and r = r_1+r_2+...+r_i.  Note: \beta‘s must be the same.

Geometric:

\begin{array}{lr} \displaystyle p_n = \frac{\beta^n}{\left(1+\beta\right)^{n+1}} & \beta>0 \end{array}
Properties:
  1. Parameter is \beta.
  2. Mean is \beta.
  3. Variance is \beta\left(1+\beta\right).
  4. If N_1, N_2, ..., N_i are geometric with parameter \beta, then the sum N = N_1+N_2+...+N_i is negative binomial with parameters \beta and r = i.

Binomial:

\displaystyle p_n = {{m} \choose {n}}q^n\left(1-q\right)^{m-n}
where m is a positive integer, 0<q<1.
Properties:
  1. Parameters are m and q.
  2. Mean is mq.
  3. Variance is mq\left(1-q\right).
  4. Variance is always less than mean.
  5. If N_1, N_2, ..., N_i is binomial with parameters q and m_1, m_2, ..., m_i, then the sum N=N_1+N_2+...+N_i is binomial with parameters q and m = m_1+m_2+...+m_i.

The (a,b,0) recursion:

These distributions can be reparameterized into a recursive formula with parameters a and b.  When reparameterized, they all have the same recursive format.
\displaystyle p_k = \left(a+ \frac{b}{k}\right)p_{k-1}
It is more common to write
\displaystyle \frac{p_k}{p_k-1} = a+\frac{b}{k}
The parameters a and b are different for each distribution.
  1. Poisson:
    a = 0 and b =\lambda.
  2. Negative Binomial:
    \displaystyle a = \frac{\beta}{1+\beta} and \displaystyle b = \left(r-1\right)\frac{\beta}{1+\beta}.
  3. Geometric:
    \displaystyle a = \frac{\beta}{1+\beta} and \displaystyle b = 0.
  4. Binomial:
    \displaystyle a = -\frac{q}{1-q} and \displaystyle b = \left(m+1\right)\frac{q}{1-q}.
Pop Quiz!
  1. A frequency distribution has a = 0.8 and b = 1.2.  What distribution is this?
    Answer: Negative Binomial because both parameters are positive. 
  2. A frequency distribution has mean 1 and variance 0.5.  What distribution is this?
    Answer: Binomial because the variance is less than the mean. 

Leave a comment

Filed under Frequency Models, Probability