Q&A Bank Part 2 - 28:

The aggregate claims process for a risk is a compound Poisson process. The expected number of claims each year is 0.5 and individual claim amounts have the following distribution:
P[claim amount =1] = 0.5
P[claim amount =2] =0.25
P[claim amount =3] =0.25
Let U(t) denote the insurer’s surplus at time t . The insurer’s surplus at time 0 is 0.5 and the insurer charges a premium of 1 each year to insure this risk.
Calculate the probability[U(t) < 0 for t = 1 or 2].

In the solution, A table is given to find the sum of the probabilities
P(S(1)<1.5) , P(S(2)<2.5).

My doubt is the whether the table is designed to apply the Convolutions concept by treating S(1) & S(2) as independent random variables? Thats why the probabilities are multiplied ? Please clarify.

Alternative to this can i follow the below approach to find the indivdual probabilities :

P(S(1) < 1.5) = P(S(1) = 0) + P(S(1) = 1)

We know P(S(1) = 0) = P(N=0) //N ~ Poi(0.5)
And P(S(1)=1) = P(N=1) * P(X=1) // X is claim amount, P(X=1)= 0.5.

P(S(2) < 2.5 ) = P(S2=0) + P(S2=1) + P(S2=2)

P(S2=2) = P(N=1)*P(X=2)+P(N=2)*P(X=1)*P(X=1) //N ~ Poi(0.5)

By following this approach, i am getting the final ans 0.388. In the material it is given as 0.366.
Is this procedure can be followed? Please tell me.

Thanks in advance,

Nageshwar.

 

Please help me in understanding this.

 

In the solution they have set the situation up like a tree diagram structure though it has been presented in table format. They have made sure to consider only the limbs of the tree that satisfy both non ruin at t = 1 and at t = 2. All other limbs lead to ruin at t = 1 or t =2.

As we normally do with tree diagrams, all probabilities on one path leading to an outcome are multiplied together. They need not be independent though they maybe. In our case they are not independent.

 

Your method is questionable. For S2 even if it is less than 2.5, if a claim of 2 happens in the first year would that not mean ruin? It seems so to me. I would go with the Q&A method.

 

Hello...do we add the P[ U(1) < 0] with P [ U(2) < 0] together?
Thank you.

 

Nice insight zivanaik!!