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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. $\endgroup$ – JejeBelfort May 20 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 at 4:54
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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 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 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 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 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 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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