# Tag Archives: Convolution

## Ruin Theory

You walk into a casino with a certain amount of surplus money.  Every hour you spend in the casino, there is a certain probability of winning or losing some given amount of money.  A ruin theory question might ask, what is the probability that you go bankrupt in x number of hours.  The bankruptcy state is an absorbing state, so once you enter into that state, the probability of leaving that state is 0.  The solution to these types of problems usually require calculating an exhaustive list of probabilities for ruin by the convolution method.  But first, familiarity with the notation should be developed.

Definitions:

1. $\psi(u)$ — Starting with surplus $u$, this is the probability of ruin as $t$ goes to infinity with $t$ defined on $\mathbb{R}$.
2. $\bar{\psi}(u)$ —  Same as above except $t \in \mathbb{N}_+$, positive integers.
3. $\psi(u,t)$ — Probability of ruin from time 0 to time $t$ with $t \in \mathbb{R}$.
4. $\bar\psi(u,t)$ — Same as above except $t \in \mathbb{N}_+$.
The analogous survival probabilities follow the same conventions except they are denoted by $\phi$.
The following relationships are useful:
1. $\psi(u) \geq \psi(u,t) \geq \bar\psi(u,t)$
2. $\psi(u) \geq \bar\psi(u) \geq \bar\psi(u,t)$
3. $\psi(u) \geq \psi(u+k)$ for $k \geq 0$
4. $\phi(u) \leq \phi(u,t) \leq \bar\phi(u,t)$
5. $\phi(u) \leq \bar\phi(u) \leq \bar\phi(u,t)$
6. $\phi(u) \leq \phi(u+k)$ for $k \geq 0$

Filed under Probability, Ruin Theory

## 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.