Any feedback on FRM Level 2 2010 exam

[stefanminchev]

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]

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]

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.

 

[David Harper CFA FRM]

okay, no problem, thanks b/c otherwise i would report it to GARP so they could review ....David

 

[frm_daniel]

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]

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]

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

 

[Laughlin1]

Hi David

A (tardy) thank you for your detailed and clear reply from 25 May.

Maurice