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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 $b$?

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  • $\begingroup$ What are $a$ and $b$? I think you forgot to define them. $\endgroup$ May 20, 2019 at 4:46
  • $\begingroup$ Hi, a and b are just two constants on the equations. a is on the indicator function and b is the variance of the normal random variable $ ε_𝑖 $. Hope it helps. $\endgroup$
    – Amy Zhang
    May 20, 2019 at 4:54

3 Answers 3

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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!

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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*}

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  • $\begingroup$ If I were to simulate values for Xi and plot a density of the number of defaults, how do you suggest I would do that. This is still not clear for me $\endgroup$
    – Amy Zhang
    May 24, 2019 at 7:11
  • $\begingroup$ I did not realise that was the nature of your question. I imagine it depends rather on what programming language you are using. In Mathematica you might write n = 1000; b = 0.5; a = InverseCDF[NormalDistribution[], 0.01]; k[x_] := Total[ UnitStep[a - (x + RandomVariate[NormalDistribution[0, b], n])/ Sqrt[1 + b]]]; Histogram[ k /@ RandomVariate[NormalDistribution[], 5000]] $\endgroup$
    – Ali
    May 24, 2019 at 10:24
  • $\begingroup$ Your simulation is for the case where a and b is independent. Which is not really the point of the questions, but I probably wasn't clear in the description. The main question is, now that I have a and b related to each other, I'll have 1 correlation. Let's say I choose 3 different combinations of a (and consequently b) and want to compare the densities they will generate. How could I do this? As long as I understand, my a is what I'm varying in my density function ($Pr(X+e_i<a$) $\endgroup$
    – Amy Zhang
    May 24, 2019 at 14:11
  • $\begingroup$ and because the correlation depends on b and b depends on a, which is a value I'm changing to generate a density plot, I can't see how can I have a density plot with a constant correlation among the variables. Let me know if you can understand what I mean.... $\endgroup$
    – Amy Zhang
    May 24, 2019 at 14:19
  • $\begingroup$ The code I posted is equivalent to this: n = 1000; b = 0.5; a = InverseCDF[NormalDistribution[], 0.01]*Sqrt[1 + b]; k[x_] := Total[ UnitStep[a - (x + RandomVariate[NormalDistribution[0, b], n])]]; Histogram[ k /@ RandomVariate[NormalDistribution[], 5000]]. I had used the convention that X_i=(X+\epsilon_i}/ sqrt{1+b) precisely because it means you can vary $a$ without affecting correlation and $b$ without affecting marginal probability of default. $\endgroup$
    – Ali
    May 24, 2019 at 15:02
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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*}

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