Modeling Correlated Defaults with Latent Factors and Binary Outcomes

# Portfolio - Default Probability


# Portfolio - Default Probability












Suppose we want to identify the frequency of default on a portfolio with a 1000 loans. In the independence case, each firm’s default process follows a Bernoulli distribution with parameter $p = 0.01$.

That is, each firm has 1% probability of an independent default. This can be represented by indicator functions ${Y_i},i=1,...,1000$, where $P(Y_i = 1) = p$.

In the correlated case, all the firms have a common factor $X$ (here it could be a macro variable, for instance) and their default processes can be represented by $Y_i = I_{X_i<a}$ where $X_i = X +ε_i$, and $X ∼ N(0,1)$ and $ε_i ∼ N(0,b)$,$i = 1,...,1000$. All $ε$’s are independent from each other and from $X$.

Let $M = \sum_{i=1}^{1000}Y_i$ represent the number of defaults in your portfolio. What is the relation between $a$ and $b$ such that the marginal probability of default for each firm is still 1% and how can I calculate the default correlation $ρ(Y_i,Y_j)$ as a function of $a$ and $b$?

## Answer by demully (score 1)

https://quant.stackexchange.com/a/61120

Normals won’t cut the mustard here, for all the reasons that the Linear Probability Model fails. It can generate negative probabilities; and when it doesn’t, it is guaranteed to be heteroskedastic,

If you have evidence that some variable is relevant to default probability, then logistic regression would seem an obvious way of incorporating. This variable might be “micro”, eg company debt to EBITDA coverage, that distinguishes between companies more or less likely to default. Or could be “macro”, like eg global GDP growth, that affects the default probability of companies in general. Of course, different kinds of company might have very different (logistic) betas to these macro risk factors...

So how one might choose to model these risks is far from a closed book. But the underlying regression needs to be on the odds of default (logits or probits, usually giving very very similar results) rather than directly on the probability of default itself!

## Answer by Ali (score 0)

https://quant.stackexchange.com/a/45741

Well $\text{Pr}(X<a)=0.01$ for $a\approx -2.33$. Since $\text{Var}(X+\epsilon_{i})=1+b$ we need to scale by $\sqrt{1+b}$, so $\text{Pr}(X+\epsilon_{i}<a)=0.01$ for $a\approx -2.33\sqrt{1+b}$.

Then note that \begin{equation*} \begin{pmatrix}X_{i}\\X_{j}\end{pmatrix}= \begin{pmatrix}1 & 1 & 0\\ 1 & 0 & 1\end{pmatrix} \begin{pmatrix}X\\ \epsilon_{i} \\ \epsilon_{j}\end{pmatrix} \end{equation*} so its covariance matrix will be given by \begin{equation*} \begin{pmatrix}1 & 1 & 0\\ 1 & 0 & 1\end{pmatrix} \begin{pmatrix}1 & 0 & 0\\ 0 & b & 0 \\ 0 & 0 & b\end{pmatrix} \begin{pmatrix}1 & 1\\ 1 & 0 \\ 0 & 1\end{pmatrix} =\begin{pmatrix}1+b & 1 \\ 1 & 1 + b\end{pmatrix}. \end{equation*} After that if you want numbers I think you need to move to Mathematica or something.

Edit : Not as much as you wanted? Here's a factoid about the bivariate Normal that's kinda relevant. If \begin{equation*} f(\rho,x,y)=\frac{1}{2\pi\sqrt{1-\rho^{2}}} \exp\left\{-\frac{1}{2(1-\rho^{2})}\left(x^{2}-2\rho x y + y^{2}\right)\right\} \end{equation*} and \begin{equation*} F(\rho,s,t)=\int_{-\infty}^{s}dx \int_{-\infty}^{t}dy\, f(\rho,x,y), \end{equation*} then \begin{equation*} \frac{\partial F}{\partial \rho}(\rho,s,t)=f(\rho,s,t). \end{equation*} See e.g. Sungur (1990) Dependence Information in Parameterized Copulas, Communications in Statistics---Simulation and Computation, 19:4, 1339-1360, DOI: 10.1080/03610919008812920.

To apply this to your problem, redefine $X_{i}$ according to \begin{equation*} X_{i}=\frac{1}{\sqrt{1+b}}\left(X+\epsilon_{i}\right) \end{equation*} so that $ \begin{pmatrix} X_{i} \\ X_{j} \end{pmatrix}$ has covariance matrix $ \begin{pmatrix} 1 & \frac{1}{1+b} \\ \frac{1}{1+b} & 1 \end{pmatrix} $. Now $a = -2.33$ can remain independent of $b$, and $ \rho=\frac{1}{1+b}$.

Since total correlation $\rho=1$ implies simultaneous default with probability $0.01$, \begin{equation*} F\left(\frac{1}{1+b},a,a\right) + \int_{\frac{1}{1+b}}^{1}\frac{\partial F}{\partial \rho}(\rho,a,a) \, d\rho= 0.01. \end{equation*} and therefore \begin{align*} F\left(\frac{1}{1+b},a,a\right)&= 0.01 - \int_{\frac{1}{1+b}}^{1} f(\rho,a,a) \, d\rho \\ &= 0.01 - \int_{\frac{1}{1+b}}^{1} \frac{1}{2\pi\sqrt{1-\rho^{2}}}\exp\left\{{-\frac{a^{2}}{1+\rho}}\right\} \, d\rho. \end{align*} This quantity, also known as $\mathbb{E}\left[Y_{i}Y_{j}\right]$, is I think the one you were asking for. For small $b$, \begin{align*} \mathbb{E}\left[Y_{i}Y_{j}\right]&\approx 0.01 - \frac{1}{2\pi}\exp\left\{-\frac{a^{2}}{2}\right\}\arccos\frac{1}{1+b}\\ &\approx 0.01 - 0.01063 \arccos\frac{1}{1+b} \end{align*}

## Answer by Ali (score 0)

https://quant.stackexchange.com/a/45770

Given the structural model with correlated Gaussians that you describe, I doubt you'll get a much better answer than the one you've been given to your specific question about the two point correlation $\mathbb{E}\!\left[Y_{i}\,Y_{j}\right]$. (Love to be proved wrong! This is not my day job.)

But the approach won't scale well, and while you ask no specific question about it, you do mention a large (thousand-loan) portfolio. For that, I would suggest something more like the following.

$y_{i}$ a binary variable as above, $\boldsymbol{y}$ the vector for all loans. \begin{equation*} P(\boldsymbol{y})=\frac{1}{Z}\exp\left\{-H(\boldsymbol{y})\right\} \qquad \qquad Z=\sum_{\boldsymbol{y}}\exp\left\{-H(\boldsymbol{y})\right\}. \end{equation*} \begin{equation*} A=\begin{pmatrix} a & b & b & \cdots\\ b & a & b & \cdots\\ b & b & a & \cdots\\ \vdots & \vdots & \vdots & \ddots \end{pmatrix} \end{equation*} \begin{equation*} H(\boldsymbol{y})= -\boldsymbol{y}^{T}A\boldsymbol{y} =-(a-b)\sum_{i}y_{i} - b \left(\sum_{i}y_{i}\right)^{2} \end{equation*} Then the number of defaults has distribution \begin{equation*} P(k)=\frac{1}{Z}\binom{N}{k}\exp\left\{(a-b)k+bk^{2}\right\}. \end{equation*}

As a sanity check, consider the case with zero correlation, i.e. with $b=0$. Then \begin{equation*} P(k)=\frac{1}{Z} \binom{N}{k} \exp\left\{ak\right\} =\frac{\binom{N}{k} p^{k}(1-p)^{-k}} {\sum_{k}\binom{N}{k} p^{k}(1-p)^{-k}} =\binom{N}{k} p^{k}(1-p)^{N-k} \end{equation*} where \begin{equation*} p=\frac{1}{1+\exp\left\{- a\right\}}\qquad\qquad \exp\left\{a\right\}=\frac{p}{1-p}. \end{equation*}