# Collection of binary random variables

A portfolio of 13 different companies have loans. Company $i$ defaults on their loan with probability $p_i$ and survives with probability $q_i=1-p_i$. Let $Y_i=1$ denote default.

Question: How could I get to a reasonable guess, preferably using probability theory, on what is the probability of default for the whole portfolio?

I am pondering about the risk on a loan to the entire portfolio. I am free to make assumptions and be creative, for example not find an exact solution but bound the probability by some interval. I made three scenarios, each with different assumptions:

• I) If one defaults, all other defaults.
• II) The firms are independent
• III) They are positively/negatively correlated in the sense that if one falls, it is a higher/lower probability of the other falling.

# I)

Pr(portfolio.defaults) = Pr(someone defaults) = 1 - Pr(no one defaults). Here my thinking stops, can I

# II)

Here it is like a binomial with $n=13$ although the probabilities $p$ are not the same but are changing. I could make an upper bound and say it is equal to $Bin(n=13, p=\max\{p_i\})$ but surely there is a distribution for this? Maybe I could use the PDF of a Bin directly via ${\displaystyle \Pr(Y_i=k)={\binom {n}{k}}p_i^{k}(1-p_i)^{n-k}}$

# I) and II)

We have that Pr(portfolio.default) = by assummption I = Pr(someone defaults) = 1 - Pr(no one defaults). Since they are independent by assumption II we have $$Pr(\text{portfolio.default}) = 1 - (q_1 \times ... \times q_{13})$$

# III)

First, to ease notation let $N$ be the total number of defaults, i.e. $$N = \sum_{i=1}^{13} Y_i.$$

Concerning the first scenario: in order to be able to compute $\mathbb P(N=0)$ you need to make some assumptions about the joint distribution of the vector $(Y_1, Y_2, \dots, Y_{13})$. In general you can only say that $$\mathbb P(N=0)=\mathbb P(Y_1=0, \dots, Y_{13}=0)$$ and to get an expression for the right hand side you need the joint distribution.

For II: This formula is only useful in theory but not really applicable in practice (especially if your portfolio consists of more than 13 companies) $$\mathbb{P}(N=n) = \sum_{I \subseteq \{1,\dots,13\}\\ \quad |I|=n} \prod_{i \in I}p_i \cdot \prod_{j \in \{1,\dots,13\}\setminus I}(1-p_j).$$ So you consider a sum for all subsets of size $n$ and use the independence to get the product of the $p_i$.

One classic approach is to use the Poisson approximation in this case. There it is assumed $N$ has a Poisson distribution with parameter $\lambda=p_1+p_2+\dots+p_{13}$, which simplifies the calculation of $\mathbb P (N=n)$ and then you bound the error.

For the third scenario it is difficult to say anything other than $$\mathbb P(N=0)=\mathbb P(Y_1=0, \dots, Y_{13}=0),$$ since knowing all pairwise correlation does not describe the joint distribution (this is only the case for joint normal distributions). So you would have to be more specific.

• Great! Sometimes one assumes the value of the firm $V_i$ follows a lognormal distribution (although it is probably not trivial to estimate $\mu$ and $\sigma$) and for a given low value $d$ of $V_i$ it is assumed to be in default. Thereby $Pr(V_i < d) = p_i$ so the lognormal distribution is mapped to a binary $Y_i$. Could I use the lognormal, and then compute the joint distribution (which would involve 13 integrals?). – jacob Dec 19 '16 at 8:38
• yeah you could, but in the end you will always need a dependence structure between the different $Y_i$ (resp. $V_i$) in order to be able to calculate joint probabilities. Probably it is easiest if you assume independence – Cettt Dec 19 '16 at 8:40
• I have worked a bit with the Poission distrubtion, but can you provide an argument or link as to why this would be a useful approximation? Note that $p_i \approx 0 \implies \lambda < 1$ – jacob Dec 19 '16 at 8:41
• you can find a bit in these lecute notes fam.tuwien.ac.at/~schmock/notes/ExtensionsCreditRiskPlus.pdf – Cettt Dec 19 '16 at 8:44