The joint probability of default simply explained

It is essential to understand the risk of simultaneous default by several entities, particularly when it comes to credit derivatives such as basket credit default swaps. The joint probability of default and its correlation are key to understanding this risk.

 

The correlation between two random variables X and Y is given by the formula :

 

correlation(X, Y) = covariance(X, Y) / (σ_X * σ_Y)

Where:

- covariance(X, Y) is the covariance between X and Y.

- sigma_X and sigma_Y are the standard deviations of X and Y, respectively.

 

For default events, we often deal with indicator variables 1_A and 1_B which take the value 1 if default occurs and 0 otherwise. The covariance between these indicator variables can be defined as:

 

covariance(1_A, 1_B) = E[1_A * 1_B] - E[1_A] * E[1_B]

 

Here:

- E[1_A] = p(A) is the probability of default of entity A.

- E[1_B] = p(B) is the probability of default of entity B.

- E[1_A * 1_B] = p(A and B) is the joint default probability of A and B.

 

For a default event A, where x_i can be 0 or 1:

E[1_A] = 1 * p(A) + 0 * (1 - p(A)) = p(A)

 

Variance is the expectation of the squared deviation from the mean: Var(1_A) = E[(1_A - p(A))^2]

 

Since (1_A) can be 0 or 1, we can expand this as: Var(1_A) = p(A) * (1 - p(A))^2 + (1 - p(A)) * (0 - p(A))^2

 

Simplifying this, we get:

 

Var(1_A) = p(A) * (1 - p(A))^2 + (1 - p(A)) * p(A)^2= Var(1_A) = p(A) * (1 - p(A))

 

Using these definitions, the correlation of default can be expressed as:

 

corr(A, B) = [p(A and B) - p(A) * p(B)] / √[(p(A) * (1 - p(A)) * p(B) * (1 - p(B))]

 

And the joint default probability: p(A and B) = correlation(A, B) * √[(p(A) * (1 - p(A)) * p(B) * (1 - p(B))]+ p(A) * p(B)

 

Consider an equal investment of $5 million in two bonds (B1 and B2) with the following parameters:

 

* Probability of default for both bonds p(A) = p(B) = 0.05 (5%).

* Default correlation correlation(A, B) = 0.3.

* Recovery rate = 0.

 

The joint default probability =0,3*√[(0,05*0,95)*(0,05*0,95)]+

0.05*0.05=0,1675

 

The expected loss is p(A and B) = 0,1675* 10 M= $167,500

 

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