The first part of this question seeks to find the probability of at least one claim occurring in a year where the claim frequency are Poisson distributed. The value of claims are a constant of 2.5m The solution attempts to equate the risk premium = 2.5 * (1 - Probability (0 claims)) Why would the above formula be correct? Shouldn't it be risk premium = 2.5 * (P(0 claims) + P(1 claim)) ; as at most 1 claim can arise
'At least one claim' means 'either 1 claim, or 2 claims, or 3 claims, or 4 claims ... and so on.' In short, it means 'any possibility EXCEPT no claims'. So we do indeed need 2.5*[1-P(N=0)].
Hi Katherine I understand the question itself asks the probability of 'At least one claim' and the formula [ 1 - P(0)] will solve this, however my doubt is how we were able to come up with the equation risk premium = 2.5 * (1 - Probability (0 claims)) It states in the question, that the premium is for covering only 1 risk. So doesn't that imply that the frequency of claims can either take 0 OR 1 ? Wouldn't the above equation be suitable for a premium that covers at least 1 risk? So my main doubt is related to how we arrived at the above equation.
The questions doesn't say the reinsurance covers only one risk. In fact, whether it does or not is immaterial, because there is no reason why a single risk cannot give rise to more than one claim. Indeed, we're told that the number of claims is Poisson.
"A reinsurance company writing aggregate excess of loss business is attempting to calculate the premium rates for a layer of £2.5m excess of £1.5m for coverage of the first claim occurring within the policy year" Doesn't the above imply that only the first claim is being covered by the reinsurance?
Yes, it sounds like the reinsurance only pays out for a single hit to the layer. You'll see the question then goes on to consider the premium required to cover two hits to the layer. Perhaps you are confused by the fact that this is agg XL? Remember, one event may give rise to many direct claims, but the reinsurer would still experience this as a single claim against the layer.
I'm still mulling over your question George. We're told that the number of claims to the layer are Poisson (and remember the Poisson distribution takes values 0,1,2,3,...). I recognise your confusion, this doesn't seem to agree with the statement that the reinsurer can take at most one hit to the layer. Therefore, your only alternative is to interpret the question to mean 'the number of underlying events with the potential to hit the layer is Poisson'. Think of it this way: Imagine that your direct claims follow a Poisson distribution, and your XL reinsurance has no reinstatements. Does that help?
My interpretation was if a claim occurs, it will be a full loss to the layer, so the risk premium is £2.5m multiplied by probability of any number of claims occurring…the result is the same as they only pay out on the first claim, i.e. loss is capped at £2.5m. So it’s an irrelevance how many claims occur, as cover is only provided for the first claim (and we know if it hits the layer it burns through it fully). ….Where this falls down is for the second part…if you are now covering the first 2 losses, and each loss is a full loss to the layer, then shouldn’t the risk premium be 2*2.5*P(2 or more claims)! Or alternatively, it doesn’t matter how many losses after the first loss occur…as reinsurer will always pay out £2.5m (assuming no reinstatements) So I think I am also a bit confused with the mechanics of this question, it’s the distinction between “first claim” and “one loss” and “two losses” etc! Or is it that if a claim occurs…there’s a chance that it doesn’t hit the layer?
Thanks Katherine for taking time to try clarifying this. I would say my main confusion arises from the use of "first claim" in the question. If this wasn't part of the questions, then I can simply assume that the frequency follows a poisson distribution. And also, as brought up by @triosfall , why is the second loss distributed differently.
Yes that's right. Just imagine claims are Poisson but the cover only pays out on the first one (or two) claims. Your suggestion ignores the possibility of having exactly one claim. The solution calculates the additional premium required to pay out on the second claim (and of course, the reinsurance only pays out once for that second claim). So you don't need to multiply by 2. Hopefully that answers your question too George? The second loss is not distributed differently, the point is that we're calculating the premium for the two scenarios separately. Not quite. The Poisson distribution relates to events that are big enough to hit the layer.
