We say that a Random Variable B has a Binomial Distribution, that is $B$ ~ $Binomial(n,p)$ if for $k \in \{ 0 , 1, ... , n \} $ : $$P(B=k) = \binom{n}{k} p^k (1p)^{ni)}$$ It is not immediately clear from this definition that this defines a discrete probability distribution, in order to show this we need to check that: $$ \sum_{i=0}^{n} \binom{n}{k} p^i (1p)^{ (ni)} = 1$$ Identity 1 In order to show this result, we make use of the binomial theorem which states that: $$ (x + y)^n = \sum_{i=0}^{n} \binom{n}{k} x^i y^{(ni)}$$ If we take $x = p$, and $y = (1p)$, then: $$ (x+y)^n = (p + (1p))^n = (1)^n = 1 $$ Allowing us to show that: $$ 1 = 1^n = (p + (1p))^n = \sum_{i=0}^{n} \binom{n}{k} p^i (1p)^{(ni)} $$ Identity 2 What other results can we derive just by manipulating the binomial theorem? If we consider the binomial expansion of $(1 + 1)^n$ then we get the following useful result: $$ 2^n = (1 + 1)^n = \sum_{i=0}^{n} \binom{n}{k} 1^i 1^{(ni)} = \sum_{i=0}^{n} \binom{n}{k} $$ I remember being pretty surprised the first time I saw this result, but there is actually another way of thinking about it which is more intuitive. We note that $$ \binom{n}{k} $$ can also be though of as the number of possible ways of making a subgroup of size $k$ from a collection of $n$ distinct objects. If we consider the power set, $\mathcal{P} (S)$, which is the collection of all possible subsets of a set. Clearly: $$\sum_{i=0}^{n} \binom{n}{k} =  \mathcal{P} (S)  $$ Therefore, if we can derive a formula for the cardinality of the power set, we know that the left hand side is equal to this, and we will be able to prove our result. To show this, let us consider the ordered set of elements of $S$: $$\{ S_1 , S_2 , ... S_n \} $$ Then we can count the number of subsets of $S$ by noting that we can form a distinct subset of $S$ by specifying a string of length $n$, made up of $0$s and $1$s, where a $1$ in the $ith$ column denotes that $s_i$ is included in the subset. Clearly this is another way of enumerating all the possible subsets of $S$, but when counted in this manner, we see that there are $2^n$ elements in the power set. Use in General Insurance We can use the binomial distribution to build a very simple loss model. Let's assume we have $n$ annual contracts, each of which either has a single loss, or no losses, further lets assume that the severity of each loss if it occurs is a fixed amount $1$. If the probability of a given contract having a loss is $p$, then the total loss amount, $X$, for the year is defined by a binomial distribution. $X$ ~ $Bin(n,p)$. Once we have specified the Gross Loss Distribution, we can very easily calculate the expected losses to various reinsurance contracts. For example, let's suppose that $n = 10$, and $p = 0.1$, and we have an Aggregate XoL contract of $8$ xs $2$, then the expected loss to the contract is: $$ \sum_{i=3}^{n} \binom{n}{k} p^i (1p)^{(ni)} (i2)$$ This formula can be directly calculated, and in this case is equal to $0.085$. It would be nice to always be able to directly calculate the expected loss to a reinsurance programme without Monte Carlo simulation, so why do we not use the binomial distribution more? To see why let's look at the underlying Gross Loss Distribution? Since we are dealing with a binomial distribution, we have simple closed forms for the central moments of the distribution, let's examine those: $$\mathbf{E} [ X ] = np$$ $$Var [ X ] = np(1p)$$ Given $(1p) <= 1$, we can see that $Var [ X ] < \mathbf{E} [ X ]$. The reason that this is an issue is that empirically, we know that this relationship does not usually hold for a portfolio of insurance losses. In addition to this, we also have the artificial restriction that the severity has to be the same for each loss. This is why we are much more likely to use a Compound Poisson Distribution instead. 
AuthorI work as a pricing actuary at a reinsurer in London. Categories
All
Archives
May 2020

Leave a Reply.