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Any feedback on FRM Level 2 2010 exam

stefanminchev

New Member
Guys, just came from FRM level 2 in Sofia, Bulgaria. I was the only one attendee.

I think that GARP had an error in one of their questions. It was:
You have 15 bonds, equal probability of default of 3%. Independent of each other. What is the probability of exactly one bond defaulting.
The answers were:
a) 2.9%
b) 3%
c) around 28%
d) around 67%

In my opinion none of them is correct.

Any thoughts on this one?
 

David Harper CFA FRM

David Harper CFA FRM
Staff member
Subscriber
Hi fat tails,

No opinions are required: none is correct.
Binomial PDF (i.i.d.), where P[k=1 | n=15,p=3%] = BINOMIST(k=1,n=15, p = 3%, false) = 29.4%

Thanks, David
 

stefanminchev

New Member
Thanks for replying David.
Actually, I assume, it is my mistake.
The third answer was most probably 29.4 instead of 28%, as I wrote it my first comment.

Sorry about the confusion.
 
I remembered there is a choice of 29.4% there.


This is to apply the formula nCr*p^r(1-p)^(n-r)... and n = 15 r =1. Plug in the formula it will be there.
 

Laughlin1

New Member
Hi David

On this question, if we cannot use excel in the exam, can you remind us how this can be worked out long-hand.

The below (using methodology from Jorion CD-ROM question 23 in Practice Credit Risk Exam) gets the answer of 29.4%, but this is using exponent of 1/12 for monthly, which you did not assume in your BINOMDIST calculation.

=1-(1-.03)^(1/12) = 0.002535 or (p)
=15*16/2 =120
=(p * 120)^1 * (1 - p)^14
=29.4%

Thanks,
Maurice
 

David Harper CFA FRM

David Harper CFA FRM
Staff member
Subscriber
HI Maurice,

I am not sure of the logic behind this Q23 (e.g., the first appears to maybe assume a monthly PD = 3% and solve for an annual PD ....) but please note, your number is slightly different than the answer: correct is 29.3776% versus 29.3585%. I can't quite understand the idea in your displayed calc from the CD-ROM

You have to apply Daniel's equation, which is the formula for a binomial PMF:
http://en.wikipedia.org/wiki/Binomial_distribution
(minor note: see how i lazily above called it a PDF. Wrong, it is a discrete function, so binomial p.m.f. is correct. pdf is correct for continuous normal but technically pmf is correct for discrete binomial)

the only way i know to do this is break the formula into its two parts:

1. 3%^1*(1-3%)^14 = 1.96%; i.e., the Prob of 1 default followed by 14 non-defaults. But now that is specific *permutation* (sequence) of 1 default followed by 15 survivals.

2. Since we don't care about the order (sequence), we need to mutiply by the number of combinations (http://en.wikipedia.org/wiki/Combination)--i.e., how many different ways can we choose 1 from 15; "n choose k" is here "15 choose 1"--which is above denoted by Daniel as "nCr" and = 15!/(1!14!)
... but with a bit of practice, you'll see you don't need to multiply this out. You can visualize the answer to this term will be 15. If 1 = survive and 0 = default,

011111111111111
101111111111111
110111111111111
....etc
... fifteen arrangements that include 1 default (15) each with a prob = 1.96%, so 15*1.96%

hope that helps, of course it's much easier with the luxury of time...

David
 
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