Chapter 20 Acted Practice Q 20.6

[Bill SD]

Hi,
I have two questions on Q20.6:

For part Q20.6 (ii) why is the summation of probability, q, necessary for all i's (from 1 to 500)? If probabilities are constant for all risks, i, then should be able to simplify to 500 * q? (similar to pg 15 in the Acted course notes?)

For part Q20.6 (iv), could someone briefly explain where the formula for the MGF of compound Poisson distribution comes from? The solution shows: exp[q(Mx(t)-1)] but this is different to the formula in the Tables pg 16. Apologies if this topic was covered in CS1 and its just my memory!

Many thanks in advance for your help

 

[Andrew Martin]

Hello

For part (ii), we haven't yet said that the probabilities are the same for each risk. This only happens from part (iii) on. This is similar to pages 15/16 of the notes. We first explain the general case, where the probabilities are different for each risk, and then talk about the special case where they are all the same.

For part (iv), page 16 of the tables gives the general equation for the MGF of a compound distribution. In part (iv) we have applied that for a compound Poisson distribution (so we know the distribution of N).

From page 7 of the tables, the MGF of a Poisson distribution is:

\(e^{\mu(e^t - 1)} \)

So, plugging this into the equation on page 16:

\( M_S (t) = M_N [logM_X (t)]
= e^{\mu(e^{logM_X (t)} - 1)}
= e^{\mu(M_X (t) - 1)} \)

If the rate is q, then we have:

\( = e^{q(M_X (t) - 1)} \)

I hope this helps

Andy