2007 #47 b) LAS(4M Xs 1M)

[piemedia]

the answer said censored losses divided by 6, but isn't it by 10? So, it should be 1,670,000. right?

[booyah81]

There are only six claims in the excess layer.

[Vorian Atreides]

You could divide by 10, but then you'd need to adjust for the probability of being in the layer--that is, divide this answer by 6/10.

The final result is the same as if you just divided by only those claims that penetrate the layer.

[chopsuey]

I am confused here. So are we only to include the prob of being in the layer when we are adding this to LAS(1.0M) to obtain LAS(5.0M)? This still trips me up.

[dukeblue2002]

This is tripping me up, as well. The logic of this solution seems inconsistent with the logic used in the solution to 2010/5 #31.

Is the difference between the two problems that in 2007 #47 we are only calculating the LAS in the excess layer, while in 2010 #31 we are using the LAS in the calculation of an ILF? If that's the reason we need to multiply the LAS by Prob(loss in layer), I'm fine with that explanation...not sure this is correct though.

Also, the solution to 2010 jumps from the LAS ($136,199) to LAS*P(loss in layer) = $86,540 without showing this step. In fact, the formula it uses to get to $136,199 is wrong since the denominator should be (637 + 561 + 407) instead of (1,390 + 1,136).