Solution X2.8, CT6 assignment 2

Hi all

I have a query about the solution to X2.8. From the solutions:

When S = X + Y: (X = amount of an individual claim, Y is the expense associated with the claim)

- var(S) = 0.25n E[(X+Y)^2] = 0.25n [E(X^2 + 2E(X)E(Y) + E(Y^2)] due to X & Y being independent.



Question 1: Surely var(S) = var(X + Y) = var(X) + var(Y) + 2cov(X+Y) = var(X) + var(Y) due to X & Y being independent??!

or do I need to be using the var(S) = E(N)var(X) + var(N)E(X)^2 rule somehow? ie.

E(N) = var(N) = 0.25n

Therefore var(S) = 0.25n (var(X+Y) + E(X+Y)^2) = 0.25n E((X+Y)^2)?

Does this look right? It's just not intuitive to me whether it is or isn't correct.





Thanks for helping

 

hi there

Since N is Poisson this is a compound poisson distribution

Recall, from tables, that for compound poisson the moments are:
E(S) = lamda x m1
Var(S) = lamda x m2

where m2 is E(x^2) or in this case E( (X + Y)^2) )
So multiply out the (X+Y)^2 out and take the expecation through.

Hope this helps

 

thanks! those variances for compound distribution functions do tend to trip me up, thanks for your reply