April 2006 Question 8 (ii) & (iii)

[Paapi]

For part(ii ) you are asked to calculate the posterior distribution of theta.
I thought you would derive the posterior the normal way wich is prior multiplies by the likelihood interms of x (as is always the case).
But here they have taken X = total number of claims in year i = n1Y1/c, and then shown the likelihood interms of this new X = n1Y1 /c

How do you know that you need to use this result of X?

In part (iii) The first line to this solution is -

E(Y|y1,y2) = c(1+ r)^2 * E(X3|y1,y2)

Whats the logic behind this result?? This question seems to be the hardest in the paper.

 

[John Lee]

For part(ii ) you are asked to calculate the posterior distribution of theta.
I thought you would derive the posterior the normal way wich is prior multiplies by the likelihood interms of x (as is always the case).
But here they have taken X = total number of claims in year i = n1Y1/c, and then shown the likelihood interms of this new X = n1Y1 /c

How do you know that you need to use this result of X?

In part (iii) The first line to this solution is -

E(Y|y1,y2) = c(1+ r)^2 * E(X3|y1,y2)

Whats the logic behind this result?? This question seems to be the hardest in the paper.

This was a messy question. I do recommend our ASET or revision notes (from which I copy bits of this).

How would you know? It's not clear - but it does say given y1 and y2 where the Yi's are the total claim amounts but we have the number of claims has a Poisson.

For year one:

• the average total claim amount per policy was y1
• the overall total claim amount for the whole group was y1n1
• the overall total number of claims was y1n1/c (as each claim is c)

So X1 = y1n1/c ~ Poi(n1θ)

Similarly since claims are (1+r) more expensive in year 2 we will have:

X2 = y2n2/c(1+r) ~ Poi(n2θ)

And for year 3 we will have:

X3 = y3n3/c(1+r)² ~ Poi(n3θ)