Recursive Discrete Aggregate Loss

You have 2 six-sided dice.  You roll one dice to determine the number of times you will roll the second dice.  The sum of the results of each roll of the second dice is the amount of aggregate loss.  Since the frequency and severity are discrete, for any aggregate loss amount, the number of combinations of rolls to produce such an amount is clearly countable and finite.  For example, an aggregate loss amount of 3 can be arrived at by rolling a 1 on the first dice, then rolling a 3; or rolling a 2, then rolling the combinations (1,2),(2,1); or rolling a 3 and then rolling (1,1,1) on the second dice.  The probability of experiencing an aggregate loss of 3 is:

\begin{array}{rll} \Pr(S=3) \displaystyle &=& \frac{1}{6^2} + \frac{2}{6^3} + \frac{1}{6^4} \\ \\ \displaystyle &=& \frac{49}{6^4} \end{array}

This method of calculating the probability is called the convolution method.  Now imagine the frequency and severity distributions are discrete but infinite.  To calculate \Pr(S=10) would require calculating the probability for many possible combinations.  If the discrete functions are from the (a,b,0) class, there is a recursive formula that can calculate this.  It is given by:

g_k = \displaystyle \frac{1}{1-af_0}\sum_{j=1}^k \left(a+\frac{bj}{k}\right)f_jg_{k-j}

where k is an integer, g_k = \Pr(S=n)=f_S(n), f_n = \Pr(X=n), and p_n = \Pr(N=n).  This is called the recursive method.  To start the recursion, you need to find g_0.  You can then find any g_k.  If a problem asks for F_S(3), this is equal to g_0+g_1+g_2+g_3.  You iterate through the recursion to find each g_k then add them together.

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Filed under Aggregate Models, Frequency Models, Probability, Severity Models

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