What's new

Credit Risk: General Questions, question 213.1

Thread starter #1
In the Credit Risk: General Questions, number 213.1, I’m struggling with David’s calculation for the expected loss. I understand the calculation for the probability of zero defaults of 98.9846%, but I don’t understand the probability of default of 1.0119% of a 1mm loss and I similarly don’t understand the .0034% probability of default on 2 bonds. Any help on this is greatly appreciated.
 
#2
Hi Chris,

It's nothing but the bond values weighted by binomial probability of default in the 3 bond portfolio to calculate EL, nCr * (prob. of default)^r* (prob. of no default)^n-r.

here is the calculation for clarity -> (1-0.003396)^3*(0 mn.)+3*(0.003396)*(1-0.003396)^2 * ($1mn.)+3*(0.003396)^2*(1-0.003396)*($2mn)+(0.003396)^3 *($3mn.)
-> 98.9846%*0 + 1.0119%*$1.0 million + 0.0034%*$2.0 million + ~0%*$3.0 million = $10,188

Hope this is clear.
Thanks
Aaditya
 
#4
In the Credit Risk: General Questions, number 213.1, I’m struggling with David’s calculation for the expected loss. I understand the calculation for the probability of zero defaults of 98.9846%, but I don’t understand the probability of default of 1.0119% of a 1mm loss and I similarly don’t understand the .0034% probability of default on 2 bonds. Any help on this is greatly appreciated.
Hi Chris,

Can you please show me the question in full.

Thanks
 
Top