coskewness and cokurtosis
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masterwu2014
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Hi David:
I saw in your video that as per (2013) exam curriculum (co-skew and co-kurtosis) were not high exam relevance testing material. But you mentioned that it is still yet to be seen if they will be of more higher exam relevance in the future. Do you think it will be of much higher exam relevance in this coming 2014 may level 1 exam? pls advise. thank you.
-mw2014.
I saw in your video that as per (2013) exam curriculum (co-skew and co-kurtosis) were not high exam relevance testing material. But you mentioned that it is still yet to be seen if they will be of more higher exam relevance in the future. Do you think it will be of much higher exam relevance in this coming 2014 may level 1 exam? pls advise. thank you.
-mw2014.
Hi @masterwu2014
Right, I saw no mention in the Nov FRM de-brief so I would continue to characterize co-skew and co-kurtosis with low (to very low) exam relevance (I don't think I've even seen a question about them; I think they are hard to query). But you never know, GARP apparently decided to retain the AIM "... and interpret the concepts of coskewness and cokurtosis," so some exam reference is possible. I hope that helps, thanks,
Right, I saw no mention in the Nov FRM de-brief so I would continue to characterize co-skew and co-kurtosis with low (to very low) exam relevance (I don't think I've even seen a question about them; I think they are hard to query). But you never know, GARP apparently decided to retain the AIM "... and interpret the concepts of coskewness and cokurtosis," so some exam reference is possible. I hope that helps, thanks,
Hi @David Harper CFA FRM CIPM, I am having problems deriving the Skew and Kurtosis for question 5 found behind the study notes of Miller, Chap3 Basic Statistics. How is the 3rd & 4th central moment derived?
Hi @tosuhn Source Q&A is here @ https://forum.bionicturtle.com/threads/p1-t2-307-skew-and-kurtosis-miller.6825/#post-23328
The nth moment about the mean (i.e., nth central moment) is given by E [(X - μ)^n] such that the 3rd central moment = E [(X - μ)^3] and fourth central moment = E [(X - μ)^4]. As Miller shows, these are standardized by dividing by, respectively, σ^3 and σ^4 to return skew and kurtosis.
The nth moment about the mean (i.e., nth central moment) is given by E [(X - μ)^n] such that the 3rd central moment = E [(X - μ)^3] and fourth central moment = E [(X - μ)^4]. As Miller shows, these are standardized by dividing by, respectively, σ^3 and σ^4 to return skew and kurtosis.
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Hi @David Harper CFA FRM CIPM thanks for your reply. The assumption made is it due to the question that it is a bond and thus we used the bernoulli/ binomial distribution where there are only 2 instance (i.e. default and no default) and thus we use mean=np and variance=npq?
Hi @tosuhn Yes, that is exactly correct! The same function--i.e., for variance, skew, kurtosis--applies, but we just happen to have here a Bernoulli, which seems to cause confusion only because it is deceptively simple. A variance is given by E[(X- μ)^2], which is a 2nd central moment such that n = 2, and therefore when p = 5% = 0.05, we can find the variance with:
given mean = 0*95% + 1*5% = 0.05,
E[(X- μ)^2] = 95%*(0 - 0.05)^2 + 5%*(1 - 0.05)^2 = 0.04750, which of course is given straightaway by n*p*(1-p) = 1*5%*95% = 0.04750
So, it's funny, i think the hardest part is just noticing that it's a very simple discrete distribution:
given mean = 0*95% + 1*5% = 0.05,
E[(X- μ)^2] = 95%*(0 - 0.05)^2 + 5%*(1 - 0.05)^2 = 0.04750, which of course is given straightaway by n*p*(1-p) = 1*5%*95% = 0.04750
So, it's funny, i think the hardest part is just noticing that it's a very simple discrete distribution:
- 0 with prob of 95%
- 1 with prob of 5%.
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