In a compound Binomial distribution the skew(S) = B^3 * E (N-E(N)^3) according to the chapter on Risk Models. How could I derive the E(N-E(N)^3) -> is it the third derivative of a moment generating funcion of binomial distribution evaluated at 0? Is there any quicker and simpler way of deriving Skew (S) for compound binomial? It is not given in the tables as it is in case of compound Poisson.
Do you mean the negative binomial? No, the third derivative of the MGF at t=0 will give you E(X^3), in general. However, the third derivative of the cumulant generating function will indeed give you Skew(X) - though I should add I'm not sure if it exists for the distribution in question. I think it will, though.
Calum's right. Get the MGF of the compound binomial (using p16 formula). Log it to obtain the CGF of the compound Binomial Differentiate it 3 times and set t=0 to get the skewness of the compound binomial. More likely to be asked coefficient of skewness rather than skewness. This will be messy. See Subject 106, September 2004, Question 9(i) where they did it for an individual risk model (which is a special case of the compound binomial).
for the exam, I am just wondering if memorizing the skew(x) is best instead of messy derivations.... what would you advise?
In an ideal world, you would have the time to sit down and make sure you're able to derive from scratch the raw and central moments, likelihoods, and generating functions of all the common distributions. Sooner or later, being able to do this will earn you some marks in an exam. In the real world...
Thank you. I really never know how much is expected to be derived and when it is sufficient to just apply ready formulas.... Is there something like exam techniques, where I could learn about it?
They won't expect you to know the skew of compound distributions and so marks will be available for its derivation. A list of CT6 proofs that could be asked can be found on this thread here.
