Chapter 6/7 Question

[Rob]

I've been looking at practice question 6 in revision booklet 3:

claim amounts: X ~ Exp (1/5000)
# claims: N ~ Poi(5)
XOL reinsurance with retention limit 6000

Let Z be the amount paid by the reinsurer on individual claim i.e.
Z = 0 , X < 6000
Z = X - 6000 , X >= 6000

Now if Sr is aggregate claim paid by reinsurer, then
E[Sr] = E[N]E[Z] --- happy with this

but the answer in the booklet then states:
var[Sr] = E[N]E[Z^2]

Where does this expression for var[Sr] come from?

I was using var[Sr] = E[N]var[X-M]+var[N](E[X-M])^2 from pg 16 tables.

Any help would be much appreciated.

Thanks.

 

[RyuVI]

'I was using var[Sr] = E[N]var[X]+var[N](E[X])^2 from pg 16 tables.'

It's the same thing. When # of Claims has a Poission distribution then E[N] = Var[N] and the above equation becomes E[N]{Var[X]+(E[X])^2}

but Var[X] + (E[X])^2 = E[X^2] hence we get: var = E[N]E[X^2]

We can use this shortcut and others whenever we have a compound poission distribution ie when N~Poi. Look again at page 16 and you will see three useful equations when dealing with compound poisson.

(Note how Variance = lambda x m2 ie var = E[N]E[X^2])

 

[Rob]

Thanks.

Spotting that E[N] = Var [N] (for a Poisson) would have helped me!