PDA

View Full Version : SOA Sample Questions #8

Doc Holiday
10-27-2007, 09:52 PM
I'm not getting the EPV for this problem.

How does E[X^2 given theta] = 200(theta^2)
I get 100(theta^2)

Maybe I'm just burnt out from all these practice problems ..... and sad for missing all the halloween parties going on .... ah, the life of an actuarial student .....

Doc Holiday
10-27-2007, 09:53 PM
oh yeah, x is Exponentially distributed with mean 10(theta)

Islander
10-28-2007, 09:47 PM
E[X^2] for exponential is 2*(theta)^2.

So E[X^2 given theta] = 2*(10theta)^2 = 200(theta^2).

Hang in there!

Doc Holiday
10-29-2007, 12:02 AM
oh right, I keep on thinking of the Variance ...

Jonnyboy
04-08-2012, 03:52 AM
I wonder if someone can guide my intuition on this question, as I still don't understand where the solution comes from. So we assume that the claim size, X, is exponentially distributed with mean 10Θ, and that the claim count, N, is a Poisson distribution with mean Θ. We want to calculate the conditional variance of the aggregate loss, S, conditioned on Θ; that is, we want to find Var(S|Θ).

If we were simply looking for Var(S|N) we would get N*Var(X), but since we are conditioning on Θ, N can actually vary, which means we are not dealing with a gamma distribution.

The solution says that Var(S|Θ) = E(N|Θ)*E(X²|Θ). Where does this come from?

Essentially, what this implies is that if we have an exponential severity distribution and a Poisson frequency distribution, then the variance of the resulting aggregate loss distribution is given by the first moment of the frequency distribution multiplied by the second moment of the severity distribution.

Were we supposed to know this? What key concept from the syllabus am I missing here?

We were supposed to be familiar with the Poisson/Gamma pair, but in that situation the Poisson Parameter followes a Gamma distribution. In this case, a Gamma parameter follows a Poisson distribution...

______

Oh damn, I just reviewed my notes and a "key special case" where N ~ Poisson(Θ) we have Var(S)= Θ*E[X²]. Man I need to remember these things... T-52 days and counting...

joni308
04-08-2012, 12:37 PM
I wonder if someone can guide my intuition on this question, as I still don't understand where the solution comes from. So we assume that the claim size, X, is exponentially distributed with mean 10Θ, and that the claim count, N, is a Poisson distribution with mean Θ. We want to calculate the conditional variance of the aggregate loss, S, conditioned on Θ; that is, we want to find Var(S|Θ).

If we were simply looking for Var(S|N) we would get N*Var(X), but since we are conditioning on Θ, N can actually vary, which means we are not dealing with a gamma distribution.

The solution says that Var(S|Θ) = E(N|Θ)*E(X²|Θ). Where does this come from?

Essentially, what this implies is that if we have an exponential severity distribution and a Poisson frequency distribution, then the variance of the resulting aggregate loss distribution is given by the first moment of the frequency distribution multiplied by the second moment of the severity distribution.

Were we supposed to know this? What key concept from the syllabus am I missing here?

We were supposed to be familiar with the Poisson/Gamma pair, but in that situation the Poisson Parameter followes a Gamma distribution. In this case, a Gamma parameter follows a Poisson distribution...

______

Oh damn, I just reviewed my notes and a "key special case" where N ~ Poisson(Θ) we have Var(S)= Θ*E[X²]. Man I need to remember these things... T-52 days and counting...

You had to review your notes for that? Aren't you the one who could derive everything on your own. Forgive me if I'm mistaken but I think that this formula is perhaps the only one that everyone can derive on their own.

But I guess it happens to the best. I once had a professor who was a genius, more than 100 publications, and did not know how to integrate x. Now every time I see your name I'll remember him.

Jonnyboy
04-08-2012, 01:23 PM
You had to review your notes for that? Aren't you the one who could derive everything on your own. Forgive me if I'm mistaken but I think that this formula is perhaps the only one that everyone can derive on their own.

But I guess it happens to the best. I once had a professor who was a genius, more than 100 publications, and did not know how to integrate x. Now every time I see your name I'll remember him.

No, you're correct. I'm not sure what happened, the wording of the problem, fatigue, but I just flat out missed the boat. My mind was thinking about Buhlmann credibility and sort of compartmentalized, ignoring other things I knew (like the overall variance of an aggregate distribution S with a Poisson Frequency distribution).