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# Credit VaR (binomial model)

#### queliujin

##### New Member
Subscriber
Hi!
On last FRM exam, there was a question like:
Portfolio of 68 bond equally weighted, each 2 million worth. 6 defaults were expected and defaults were independent.
What is 95% Credit VaR?

The answer I have found was:
6*2 - 2*68*0.04 = 6.56mil

I suppose the binomial model is used here but... could someone please explain me how does it work under this particular example?

#### ami44

##### Active Member
Subscriber
I get to a different result.

The probability p of a specific bond to default is $p=\frac{6}{68}$
The losses are binominal distributed with standard derivation $s=2 \cdot \sqrt{68 \cdot p \cdot (1-p)} = 4.68mn$
Using the approximation as normal distribution, I get for the credit var:
$$CVaR =1.65 \cdot 4.68mn = 7.72mn$$
I assume that Credit VaR is the unexpected loss only.
How did you get to your result?

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
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Hi @queliujin I agree with @ami126 who is using the normal approximation to the binomial, but it's surprising close, given that it doesn't pass the n*p > 10 and n*(1-p) > 10 test; i.e., n*p is only 6 such that the rule of thumb suggests probably too much skew ... but it still worked basically:

=BINOM.INV(n = 68 trials, p = 6/68, alpha = 0.95) = 10 defaults is the 0.95 quantile; i.e.,
• =BINOM.DIST(X = 9, n = 68, p = 6/68, cumul = TRUE) = 92.6%
• =BINOM.DIST(X = 10, n = 68, p = 6/68, cumul = TRUE) = 96.5%
So 10 defaults is the 95th quantile, and the CVaR = (10 - 6 expected) * 2 = $8.0 million; or if you like: • 0.95 quantile = 10 defaults *$2.0 mm loss = $20.00 mm quantile • Expected loss = 6 * 2 =$12.0
• CVaR = UL = quantile - EL = 20 - 12 = \$8.0
which is approximated by @ami44's 7.72 million!

I don't know why your solution finds the quantile at only 68*0.04 2.72 defaults, that's nearer to the 0.05 quantile which is the wrong side of the distribution; i.e., worst expected number of defaults at 95% confidence is about 10 defaults, not 4% of 68 which is better than the expected number of 6.

#### queliujin

##### New Member
Subscriber
Solution offered by @ami44 is clear for me whereas I am wondering about David's one.
@David Harper CFA FRM, in terms of the exam, how to calculate below using calculator only?
=BINOM.INV(n = 68 trials, p = 6/68, alpha = 0.95) = 10 defaults is the 0.95 quantile
Apologize if it is simple, I could miss somewhere... or forgot due to increasing stress before exam

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
Hi @queliujin The calculator cannot be used to find the inverse cumulative binomial, which is the Excel function =BINOM.INV(n = 68 trials, p = 6/68, alpha = 0.95) = 10 defaults is the 0.95 quantile ... It returns the smallest quantile (in this case, 10 defaults) for which the cumulative binomial dsitribution s greater than or equal to the given confidence level (in this case, 95.0%).

On the exam, for this question, it appears you would need to use @ami44 's solution as the cumulative binomial would be tedious. I was just checking the accuracy of the approximation. But, actually, I do not believe this is an actual exam question: the given solution is wrong, and an exam question is unlikely to force the need for an approximation. Just my opinion. Thanks,