Modeling Loan Portfolio Losses with a Binomial Default Count

# Probability distribution of portfolio loss


# Probability distribution of portfolio loss












I was trying to do this exercise, and something doesn't quite add up. We need to calculate,with reference to a time horizon of one year the expected number of defaults and the probability of default equal to the expected number of defaults for a portfolio consisting of 200 loans, each with probability of default(PD) equal to 1%. It is suggested that the Poisson distribution be used. This is how I computed the expected number of defaults $E(D) = n\times PD = 200\times0.01 = 2$

Using the poisson I computed the probability of default equal to the expected number of defaults(2) as $P(X=2) = \frac{\lambda^x e^{-\lambda}}{x!}$ = $\frac{2^2 e^{-2}}{2!}$ = 0.27

What is not clear is in the next point: making use of the probability of the previous point and assuming that losses are directly proportional to the number of defaults in the amount of 20000, complete the probability distribution of portfolio loss given below:

\begin{matrix} \textbf{Loss} & \textbf{Probability} \\ 0 & 0.16 \\ 20000 & ? \\ 40000 & ? \\ 60000 & 0.22 \\ 80000 & 0.08 \\ 100000 & 0.02 \\ \end{matrix}

Can someone help me to fill this table?

If the initial value in the table(no default ) is associated with a probability of 0.16% , my answer in the first point could not be right because $P(X=0) \frac{\lambda^x e^{\lambda}}{x!} = e^{-2}= 0.135$

Then the probability of X=1 which is equal to X=2 is $P(X=1) \frac{\lambda^x e^{\lambda}}{x!} = \lambda e^{-2}= 0.27$

If we add these probabilities to those in the table, however, it goes beyond the value 1. How can I tackle this problem?

## Answer by Kurt G. (score 3, accepted)

https://quant.stackexchange.com/a/81235

- Who produced those other probabilities in that table?

- Simplicity is key. Instead of fiddling around with a Poisson distribution we take $X$ to be binomially distributed with parameters $n=200$ and $p=0.01\,.$ This produces the table \begin{align} x && Loss && Prob\\ \hline 0 && 0 && 0.13\\ 1 && 20,000 && 0.27\\ 2 && 40,000 && 0.27\\ 3 && 60,000 && 0.18\\ 4 && 80,000 && 0.09\\ 5 && 100,000 && 0.04\\ 6 && 120,000 && 0.01\\ \hline && Total && 1.00\\ \end{align} where each probability is $P(X=x)={n\choose x}p^x(1-p)^{n-x}\,.$

- If we use instead your method with Poisson, $$P(X=x)=\frac{\lambda^xe^{-\lambda}}{x!}\,,\quad\lambda=2\,,$$ we practically get the same table.