Actuarial Outpost Shortcut kth central moment
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 Short-Term Actuarial Math Old Exam C Forum

#1
09-24-2016, 03:24 PM
 msmnCasualty Member CAS Join Date: Feb 2015 Posts: 122
Shortcut kth central moment

I remember taking statistics course in college and my professor taught us shortcut on how to calculate kth central moments where k = 3,4,....n

Could anyone explain the shortcut to me? I really appreciate your time and help!

Thanks!
#2
09-26-2016, 01:19 AM
 Demo99 Member SOA Join Date: Jan 2013 Location: CA Posts: 237

I'm not aware of any shortcut, but I will say I haven't seen many third or higher moments on this exam. Are you talking about taking the nth derivative of the moment generating function at 0?
#3
09-26-2016, 08:25 PM
 msmnCasualty Member CAS Join Date: Feb 2015 Posts: 122

Quote:
 Originally Posted by Demo99 I'm not aware of any shortcut, but I will say I haven't seen many third or higher moments on this exam. Are you talking about taking the nth derivative of the moment generating function at 0?
Nope, I'm talking about calculating skewness and kurtosis which involve third and fourth central moment.. such that E(X_i - X_bar)^k

Common approach is to expand the term through algebra but the equation will get messy sometimes so I just want to find the alternative to save time.
#4
09-27-2016, 07:45 PM
 Z3ta Member SOA Join Date: Sep 2015 Posts: 361

The third central moment is the 3rd derivative of the natural log of the moment generating function at 0...

ie. the third central moment is $\frac{d^3}{dt^3}\left( \ln{M_X(t)} \right) |_{t=0}$

The larger central moments aren't quite as easy.

For instance the fourth central moment is $\frac{d^4}{dt^4}\left( \ln{M_X(t)} \right) |_{t=0}+3\left( \frac{d^2}{dt^2}\left( \ln{M_X(t)} \right) |_{t=0} \right)^2$ or equivalently $\frac{d^4}{dt^4}\left( \ln{M_X(t)} \right) |_{t=0}+3\sigma^4$

Is this what you were thinking of?

There are nice shortcuts for certain distributions of course (like the normal). Could you have been remembering a shortcut that only works with certain distributions?

Last edited by Z3ta; 09-27-2016 at 07:51 PM..
#5
10-01-2016, 01:44 AM
 Demo99 Member SOA Join Date: Jan 2013 Location: CA Posts: 237

I just saw something in Mahler today (only for Poisson). The second central moment of a Compound Poisson Count and Gamma Severity distribution was E[N]*E[X^2] and the 3rd central moment was E[N]*E[X^3].

Last edited by Demo99; 10-01-2016 at 01:47 PM.. Reason: correction
#6
10-01-2016, 01:46 AM
 BuddhaWilliams Member SOA Join Date: Mar 2016 Posts: 152

Quote:
 Originally Posted by Demo99 I just saw something in Mahler today. The second central moment of a Compound Poisson Count and Gamma Severity distribution was E[N]*E[X^2] and the 3rd central moment was E[N]*E[X^3].
Thats not for finding skewness/kutosis though. Thats for finding the variance of aggregate losses
#7
10-01-2016, 01:54 AM
 Demo99 Member SOA Join Date: Jan 2013 Location: CA Posts: 237

He used it for finding the skewness of the aggregate distribution. Probably won't work for just severity or counts alone.
#8
10-01-2016, 04:02 AM
 BuddhaWilliams Member SOA Join Date: Mar 2016 Posts: 152

You right
#9
10-01-2016, 10:28 AM
 Academic Actuary Member Join Date: Sep 2009 Posts: 8,967

Quote:
 Originally Posted by Demo99 He used it for finding the skewness of the aggregate distribution. Probably won't work for just severity or counts alone.
Formula based upon Poisson frequency. Does not hold in general.
#10
10-01-2016, 10:54 AM
 BuddhaWilliams Member SOA Join Date: Mar 2016 Posts: 152

You right