Miller Chapter 3: Basic Statistics - Study Notes

Dr. Jayanthi Sankaran

Well-Known Member
Hi Shakti and Brian,

By AIM's do you mean the 'FRM program manual' - ie. readings that one should focus on, and specific learning objectives that should be achieved with each reading?

Thanks!
Jayanthi
 

brian.field

Well-Known Member
Subscriber
From GARP's website:

The Applying Instructional Materials (AIM) Statements are made available to 2014 FRM Exam paid candidates to assist in Exam preparation. The AIM Statements are designed to serve as an additional study resource and will not in and of themselves fully prepare a candidate for the FRM Exam. They should be used as guidance and support for the readings outlined in the Study Guide.

Once you have registered for the exam, you can download a PDF of the AIMs.
 

Meredius

New Member
Hello,
I can type a couple lines. :)

1) On the above about the covariance of the sample population being different than the population's because of the n-1 denominator... what I recall is that for a population larger than 30, you start biasing the denominator as a rule of thumb to get a more unbiased (realistic?) expectation. With a very small population, the bias takes on more and more weight... kind of like using a scale to measure the weight of a table with bowling balls or marbles.... marbles should yield a more refined result but each has a more subtle weight on reality.

2) The skew and kurtosis formula's would not be asked as calculations AFAIK. It would be too tedious. It's more conceptual in understanding how it affects the peakedness or width of the distribution. I would extend that to some of the heavy derivations as well.

Happy studying. :)
 

bpdulog

Active Member
Hi David,

In page 39 of your Study Notes for Miller, Chapter 3, Basic Statistics, I am not sure I understand the following:

"In general, for (n) random variables, the number of non-trivial cross-central moments of order (m) is given by:

Equation (1)

k = (m + n - 1)! - n
m!(n - 1)!
In the case of n = 3, we have coskewness which is given by:

Equation (2)

k3 = (n + 2)(n+1)n - n
6​

How do you get equation (2) from equation (1)? Although I have seen part of equation (2) elsewhere, I don't know how to derive it!

Thanks
Jayanthi

I see n = # of random variables, but what does m stand for?
 

ShaktiRathore

Well-Known Member
Subscriber
Hi,
Yes m stands for number of non-trivial cross-central moments ,
for m=1 ,K1=[(1+ n - 1)!/1!(n - 1)!]-n=[(n)!/(n - 1)!]-n=n-n=0,
for m=2 ,K2=[(2+ n - 1)!/2!(n - 1)!]-n=[(n+1)!/2(n - 1)!]-n=n(n+1)/2 - n=n(n-1)/2
for m=3 ,K2=[(3+ n - 1)!/3!(n - 1)!]-n=[(n+2)!/6(n - 1)!]-n=(n+2)(n+1)n/6 - n
thanks
 
Last edited:

SP_SK

New Member
Jayanthi,
Its m=3 not n=3 a typo i think
k(m)=((m+n-1)!)/(m!(n-1)!)-n
k(3)=((3+n-1)!)/(3!(n-1)!)-n
k(3)=((n+2)!)/(6(n-1)!)-n
k(3)=((n+2)(n+1)n(n-1)!)/(6(n-1)!)-n [(n+2)! Can be write as (n+2)(n+1)n(n-1)!]
k(3)=((n+2)(n+1)n/6)-n
Yhanks

Hi Shakti - can you explain how are you getting k(3)=((n+2)!)/(6(n-1)!)-n after
k(3)=((3+n-1)!)/(3!(n-1)!)-n. Also what does ! stand for?
 

David Harper CFA FRM

David Harper CFA FRM
Subscriber
Hi @SP_SK The exclam (!) represents factorial and should be understood by the FRM candidate (e.g., http://mathworld.wolfram.com/Factorial.html). 13! = 13 * 12 * 11 * ... * 3 * 2 * 1. In Shakti's step above, the 3! = 3* 2 * 1 = 6. In the numerator, his (3+n+1) = (n +2), simple. In this way, his step is
  • k(3)=((3+n-1)!)/(3!(n-1)!)-n =
  • k(3)=((n+3-1)!)/((3*2*1)*(n-1)!)-n =
  • k(3)=((n+2)!)/(6*(n-1)!)-n = .. I hope that helps!
 
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