troubleshooter
Active Member
Yes. About a week to go before the results are published. It's already making me nervous... I think it's 50/50 for me...
Level 2: Post what your remember here...
- Thread starter troubleshooter
- Start date
Hmmm its a long wait now for me...I also in same boat with 50/50 chances. Come the result day and we know what are actually the results.BTW best of luck to me and then everybody waiting for the results
Suzanne Evans
Well-Known Member
Good luck to all!
Suzanne
Suzanne
Hi David,
i passed May Level I exam. Thanks for your good materials. I think ìll start for L2 right away. I just went through the L II exam questions and think they would really be good preparation questions for the exam. My question is, if its possible to summarize all shown questions from the last 14 pages with the 4 answers (when given) and the correct answers as well, because they are not really very neatly arranged. To collect all of the questions and answers from the 14 pages is a little hardSo best case would be a 80 questions block (if 80 questions are mentioned) with 4 answers for each question and 80 marked correct answers. Would that be possible?
i passed May Level I exam. Thanks for your good materials. I think ìll start for L2 right away. I just went through the L II exam questions and think they would really be good preparation questions for the exam. My question is, if its possible to summarize all shown questions from the last 14 pages with the 4 answers (when given) and the correct answers as well, because they are not really very neatly arranged. To collect all of the questions and answers from the 14 pages is a little hardSo best case would be a 80 questions block (if 80 questions are mentioned) with 4 answers for each question and 80 marked correct answers. Would that be possible?
Suzanne Evans
Well-Known Member
Hi Tom,
Are you referring to the mock exam?
I'm not sure I understand which questions you are referencing.
Thanks,
Suzanne
troubleshooter
Active Member
I would think David would take the exam feedback in this forum under consideration while preparing his sample questions...
Suzanne Evans
Well-Known Member
Hi Tom,
I honestly don't know where that would be helpful as the questions everyone quoted could vary and don't give the actual question on the exam. One word misplaced could make a big difference. David definitely takes all feedback when considering future products as well as creation of his pq's.
Thanks,
Suzanne
vallerano
New Member
jdg123 - I am aware of no definition for "liquidity duration" in any of the assignments. Admittedly, I've only read them a few dozen times, so maybe i missed itbut even Google doesn't give me solid references. I'd appreciate any legitimate reference ... I can see that it sounds like an average days to liquidate or something ... (it sounds to me like an abuse of "duration" frankly)
After a long search, the reference for "liquidity" duration is: chapter 17 of Modern Investment Management: An Equilibrium Approach, by Rosengarten and Zangari. The authors also use the term in quotation marks...
Any chance you can let us know what the book said? (I suppose I could look too!). Glad you found it. Thanks.
vallerano
New Member
Sure. I am quite surprised that this half page on a vague concept is tested regularly (concept appeared in previous L2 exam as well, just found it in the L2 feedback for Nov 2011
)
So the purpose of LD is to come up with an estimate for the time (in days) needed to sell an equity portfolio. There is also an assumption for a daily limit for volume that will not disturb the market. The formula is: LD= Number of shares held/(Daily volume limit * Daily volume of security). On portfolio level, the LDs for each security should be weighted by the shares' weight in the portfolio. Can only be used for bonds if traded volume information is available, otherwise the LD will be a result of "discussions with the portfolio managers".
)So the purpose of LD is to come up with an estimate for the time (in days) needed to sell an equity portfolio. There is also an assumption for a daily limit for volume that will not disturb the market. The formula is: LD= Number of shares held/(Daily volume limit * Daily volume of security). On portfolio level, the LDs for each security should be weighted by the shares' weight in the portfolio. Can only be used for bonds if traded volume information is available, otherwise the LD will be a result of "discussions with the portfolio managers".
David Harper, CFA, FRM, CIPM
Hi David,
I am not able to understand an equation from Qn 8 of GARP 2013 Sample PQs
1+ r = (1-pi) * (1+y) + pi*R
How come this translated to
1+r = (1-pi) * (1+y) - (1-pi) *FV/MV and how does it solve further?
If you already have this kind of a qn posted somewhere on website, please guide me to that page?
Hi David,
I am not able to understand an equation from Qn 8 of GARP 2013 Sample PQs
1+ r = (1-pi) * (1+y) + pi*R
How come this translated to
1+r = (1-pi) * (1+y) - (1-pi) *FV/MV and how does it solve further?
If you already have this kind of a qn posted somewhere on website, please guide me to that page?
Hi AlokS,
I actually do not understand the algebra in the 2nd step of the answer to question 8 in the Sample PQ. I am going to submit it to GARP, as the last term appears to be a typo, or at least, like you I do not understand it: (1-pi) *FV/MV.
Their correct answer follows instead from the first step: 1+r = (1-π) * (1+y) + π*R
This is the idea that, if you charge nothing extra for the risky option on the right hand side (i.e., "risk-neutral" in this context), you should get the same return from a riskless asset (1+r, on the left-hand side) as you would expect, ex ante, to get from an risky bond, where expected is the weighted average of the two outcomes: no default, earns you yield of (1+y) with probability of (1-π) and recovery R with probability of π (right-hand side).
Since, 1+r = (1-π) * (1+y) + π*R, if R=0, then π*R drops out and:
(1+r)/(1+y) = (1-π), such that
π = 1 - (1+r)/(1+y), with the yield in this case given by 1,000,000 face/800,000 - 1 = 25.0% with annual compounding,
such that π = 1 - (1+5%)/(1+25%) = 16.0%. I hope that explains,
I actually do not understand the algebra in the 2nd step of the answer to question 8 in the Sample PQ. I am going to submit it to GARP, as the last term appears to be a typo, or at least, like you I do not understand it: (1-pi) *FV/MV.
Their correct answer follows instead from the first step: 1+r = (1-π) * (1+y) + π*R
This is the idea that, if you charge nothing extra for the risky option on the right hand side (i.e., "risk-neutral" in this context), you should get the same return from a riskless asset (1+r, on the left-hand side) as you would expect, ex ante, to get from an risky bond, where expected is the weighted average of the two outcomes: no default, earns you yield of (1+y) with probability of (1-π) and recovery R with probability of π (right-hand side).
Since, 1+r = (1-π) * (1+y) + π*R, if R=0, then π*R drops out and:
(1+r)/(1+y) = (1-π), such that
π = 1 - (1+r)/(1+y), with the yield in this case given by 1,000,000 face/800,000 - 1 = 25.0% with annual compounding,
such that π = 1 - (1+5%)/(1+25%) = 16.0%. I hope that explains,
Hi David,
it seems like GARP is trying to confuse people during the exam with questions about stuff you though easy while learning. But then in the exam it turns out that the questions are very different from the ones you were working on while preparing. One specific question on Expected shortfall. We learned like ES 95% is just the average of tail losses above the 95% VaR. What if they question you 95.5% or 96.5% ES like the question below, which i took from the section "Post what you can remember here"... ?
5) Confidence VAR
94 a
95 b
96 c
97 d
98 e
99 f
What is the Expected shortfall at 95.5% confidence
I guess the answer is Answer: Average of c,d,e,and f
it seems like GARP is trying to confuse people during the exam with questions about stuff you though easy while learning. But then in the exam it turns out that the questions are very different from the ones you were working on while preparing. One specific question on Expected shortfall. We learned like ES 95% is just the average of tail losses above the 95% VaR. What if they question you 95.5% or 96.5% ES like the question below, which i took from the section "Post what you can remember here"... ?
5) Confidence VAR
94 a
95 b
96 c
97 d
98 e
99 f
What is the Expected shortfall at 95.5% confidence
I guess the answer is Answer: Average of c,d,e,and f
Hi Tom, That is an interesting variation on ES. I agree it would be unexpected, although personally I don't think it's sneaky: it biases way in favor of candidates with a robust understanding of ES as a conditional mean.
If we reduce the sample just to keep it simple, say n = 100 losses and they happen to be ordered, such that the worst 3 losses are conveniently (absolute values):
100, 99, 98
The easy case is a 97% ES, because that is the (conditional) average of the worst 3 losses = (100+99+98)/3 = 99
But let's explicate this average, which is really given by [(100*1%) + (99*1%) + (99*1%)] / 3% = 99
i.e., the weighted average of the three outcomes, (100* prob 1%) + (99* prob 1%) + (99* prob 1%), which equals 2.97 which is the expected outcome for this tail, but in the context of the overall parent distribution, which is then translated into its own probability distribution (i..e., this is the meaning of "conditional") by dividing by 3%, the sum of the probabilities.
Now we can handle any ES.
For example, what is the 97.5% ES? It is the conditional mean of the 2.5% tail:
[(100*1%) + (99*1%) + (98*0.5%)] / 2.5% = 99.20; i.e., in terms of the parent distribution, 99 remains with a f(x) =1.0%, but we only "need half of it" in order to "carve out" our 2.5% child distribution. I hope that helps,
If we reduce the sample just to keep it simple, say n = 100 losses and they happen to be ordered, such that the worst 3 losses are conveniently (absolute values):
100, 99, 98
The easy case is a 97% ES, because that is the (conditional) average of the worst 3 losses = (100+99+98)/3 = 99
But let's explicate this average, which is really given by [(100*1%) + (99*1%) + (99*1%)] / 3% = 99
i.e., the weighted average of the three outcomes, (100* prob 1%) + (99* prob 1%) + (99* prob 1%), which equals 2.97 which is the expected outcome for this tail, but in the context of the overall parent distribution, which is then translated into its own probability distribution (i..e., this is the meaning of "conditional") by dividing by 3%, the sum of the probabilities.
Now we can handle any ES.
For example, what is the 97.5% ES? It is the conditional mean of the 2.5% tail:
[(100*1%) + (99*1%) + (98*0.5%)] / 2.5% = 99.20; i.e., in terms of the parent distribution, 99 remains with a f(x) =1.0%, but we only "need half of it" in order to "carve out" our 2.5% child distribution. I hope that helps,
Hi David, thanks a lot. I think it works now Although i think you meant 98 x 0.5% ?
That means if the question above was talking about n = 1oo ==> ES (95.5) = (100*1%+99*1%+98*1%+97*1%+96*0.5%)/4.5% = 98.22
Could the ask the same question if n would be 99 ? Or would that be too complicated ?
Thanks a lot again.
Although i think you meant 98 x 0.5% ? That means if the question above was talking about n = 1oo ==> ES (95.5) = (100*1%+99*1%+98*1%+97*1%+96*0.5%)/4.5% = 98.22
Could the ask the same question if n would be 99 ? Or would that be too complicated ?
Thanks a lot again.
Mark W
Active Member
Hi Tom,
GARP 'could' ask any question (that's their prerogative within (estimatable but not-known) limits), but I would find it unlikely that they would make it unecessarily tedious or off-topic...after all, there are only 3 mins per question.
Thanks - Mark
GARP 'could' ask any question (that's their prerogative within (estimatable but not-known) limits), but I would find it unlikely that they would make it unecessarily tedious or off-topic...after all, there are only 3 mins per question.
Thanks - Mark
Hi Tom77, Thank you for the correction, fixed above (see? you do understand it quite well enough to spot an error!)
In regard to ES(95.5), yes, I get exactly 98.22 also (but not %, unless the loss amounts are %)!
I agree with Mark: the original cited example from the Nov Part 2 exam was unexpected due to its difficulty, yet still GARP is very unlikely to get too adventurous. Although I do think questions like these might inform an exam strategy: any FRM exam is going to contain some very difficult questions (i.e., that a majority of candidates will get wrong). Since all questions have the same weight, it's not worth getting bogged down time-wise on any one question. And the questions are not grouped or sequenced such that easy questions appear earliest, so I think your first goal is to finish the exam even if you must skip the most difficult questions on first pass (I actually think it might be better to skip-and-save-for-last the most difficult questions due to the possibility the certain questions are potential time traps, but i think that is a style more than a recommendation). Thanks,
In regard to ES(95.5), yes, I get exactly 98.22 also (but not %, unless the loss amounts are %)!
I agree with Mark: the original cited example from the Nov Part 2 exam was unexpected due to its difficulty, yet still GARP is very unlikely to get too adventurous. Although I do think questions like these might inform an exam strategy: any FRM exam is going to contain some very difficult questions (i.e., that a majority of candidates will get wrong). Since all questions have the same weight, it's not worth getting bogged down time-wise on any one question. And the questions are not grouped or sequenced such that easy questions appear earliest, so I think your first goal is to finish the exam even if you must skip the most difficult questions on first pass (I actually think it might be better to skip-and-save-for-last the most difficult questions due to the possibility the certain questions are potential time traps, but i think that is a style more than a recommendation). Thanks,
Thanks a lot David for clarifying. I think i'll try that strategy.
Sorry for one last question about this topic: What would be the ES when we would have the same setup like the above example (n=100, ES 95.5%=98.22) but with only 4 given worst losses like 99, 98, 97, 96. That would be the example i gave first:
5) Confidence VAR
94 a
95 b
96 c
97 d
98 e
99 f
What is the Expected shortfall at 95.5% confidence
I guess the answer is Answer: Average of c,d,e,and f
According to our logic above it should be : ES (95.5) = (99*1%+98*1%+97*1%+96*1.5%)/4.5% = 97.33 or would it then be just the average of 97.5.
Sorry for one last question about this topic: What would be the ES when we would have the same setup like the above example (n=100, ES 95.5%=98.22) but with only 4 given worst losses like 99, 98, 97, 96. That would be the example i gave first:
5) Confidence VAR
94 a
95 b
96 c
97 d
98 e
99 f
What is the Expected shortfall at 95.5% confidence
I guess the answer is Answer: Average of c,d,e,and f
According to our logic above it should be : ES (95.5) = (99*1%+98*1%+97*1%+96*1.5%)/4.5% = 97.33 or would it then be just the average of 97.5.
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