2018 CT6 April Q9c

[Laura]

Hi all,
I'm quite confused with this question.
How does part i and Ii differ from III and iv in terms of the risk model used? Or is the whole question using the collective risk model
Also I'm not sure how to go about obtaining the variance in iic)

Thanks in advance for the help!

 

[Andrew Martin]

Hello

The overarching structure of the models is the same in the sense that we have:

1. A Poisson distribution representing the number of infections for each individual patient (ie a collective risk model for each patient based on their Poisson parameter)
2. 100 patients in each study

The different between the two scenarios comes down to the Poisson parameter. In parts (i) and (ii), the Poisson parameter for each patient is either 0.1, 0.3 or 0.9 with the probabilities given in the table. These parameters can differ across patients, so we might have:

0.1, 0.3, 0.3, 0.1, 0.9, 0.1, 0.1, 0.3, .... etc for the 100 patients.

Or instead we might have:

0.3, 0.9, 0.9, 0.3, 0.3, 0.1, 0.9, ... etc for the 100 patients

The possible scenarios here are all the combinations of 0.1, 0.3 and 0.9 across the patients. When calculating the aggregate claims for all 100 patients, we take into account the variation in the Poisson parameter across patients based on the given distribution.

In part (iii), the Poisson parameter is the same for each patient, we just don't know which of the two it is. So we have two possible scenarios here, either all patients have parameter 0.2 or all patients have parameter 0.4. These are two 'extremes' compared to parts (i) and (ii) before, where we consider all the possible mixes of parameters across the patients. This contributes to why we end up with a larger variance for the aggregate claims in part (iii), even though the mean is the same.

Variance calculations

For part (ii)(c), we have from page 16 of the Tables:

Var(Si) = var(E[Si | lambda_i]) + E[var(Si | lambda_i)]
= var(lambda_i * m1) + E[lambda_i * m2]
= m1^2 * var(lambda_i) + m2 * E[lambda_i]
= 250^2 * 0.08 + (200 + 250^2) * 0.3 = 23810

The overall variance for the aggregate claims is then 100 * 23810 = 2,381,000

For part (iii), we have:

Var(S) = E[var(S | lambda)] + var(E[S | lambda])
= E[n * lambda * m2] + var(n * lambda * m1)
= 100 * 0.3 * (200 + 250^2) + 100^2 * 250^2 * 0.01 = 8,131,000

It looks like there is an error in the Examiners' report here, for m2 I think it uses 200 instead of 200 + 250^2.

The reason we can't take the same approach in part (iii) as we did in part (ii)(c), ie calculating the individual variances of Si and multiplying by 100, is that the Si's here are not independent, since they are all driven by the same unknown value of lambda. If lambda is known then they are independent (so independent conditional on lambda) and hence we can perform the calculations above.

Hope this helps!

Andy

 

[Laura]

Hello

The overarching structure of the models is the same in the sense that we have:

1. A Poisson distribution representing the number of infections for each individual patient (ie a collective risk model for each patient based on their Poisson parameter)
2. 100 patients in each study

The different between the two scenarios comes down to the Poisson parameter. In parts (i) and (ii), the Poisson parameter for each patient is either 0.1, 0.3 or 0.9 with the probabilities given in the table. These parameters can differ across patients, so we might have:

0.1, 0.3, 0.3, 0.1, 0.9, 0.1, 0.1, 0.3, .... etc for the 100 patients.

Or instead we might have:

0.3, 0.9, 0.9, 0.3, 0.3, 0.1, 0.9, ... etc for the 100 patients

The possible scenarios here are all the combinations of 0.1, 0.3 and 0.9 across the patients. When calculating the aggregate claims for all 100 patients, we take into account the variation in the Poisson parameter across patients based on the given distribution.

In part (iii), the Poisson parameter is the same for each patient, we just don't know which of the two it is. So we have two possible scenarios here, either all patients have parameter 0.2 or all patients have parameter 0.4. These are two 'extremes' compared to parts (i) and (ii) before, where we consider all the possible mixes of parameters across the patients. This contributes to why we end up with a larger variance for the aggregate claims in part (iii), even though the mean is the same.

Variance calculations

For part (ii)(c), we have from page 16 of the Tables:

Var(Si) = var(E[Si | lambda_i]) + E[var(Si | lambda_i)]
= var(lambda_i * m1) + E[lambda_i * m2]
= m1^2 * var(lambda_i) + m2 * E[lambda_i]
= 250^2 * 0.08 + (200 + 250^2) * 0.3 = 23810

The overall variance for the aggregate claims is then 100 * 23810 = 2,381,000

For part (iii), we have:

Var(S) = E[var(S | lambda)] + var(E[S | lambda])
= E[n * lambda * m2] + var(n * lambda * m1)
= 100 * 0.3 * (200 + 250^2) + 100^2 * 250^2 * 0.01 = 8,131,000

It looks like there is an error in the Examiners' report here, for m2 I think it uses 200 instead of 200 + 250^2.

The reason we can't take the same approach in part (iii) as we did in part (ii)(c), ie calculating the individual variances of Si and multiplying by 100, is that the Si's here are not independent, since they are all driven by the same unknown value of lambda. If lambda is known then they are independent (so independent conditional on lambda) and hence we can perform the calculations above.

Hope this helps!

Andy

Hi Andy,

Thanks and i understand the concept better now.
Just a further clarification: part b mentions to state the formula for the mean and variance of S_i conditional on lamba_i. Where can I find this formula?
Also for part iic, why can't we directly use the formula for variance derived for the compound poisson distribution? I.e Var(S) =lamba*m2?

Thanks!

 

[Andrew Martin]

For part b), these are the formula for the mean and variance of a compound distribution, which can be found on page 16 of the tables. It is also covered in Section 3.3 of Chapter 19.

For c), because lambda is a random variable, we have that Var(S | lambda_i) = lambda_i * m2. Only if lambda was a fixed number then we would have var(S) = lambda * m2.

We can find var(S) using the formula at the top of page 16 of the tables, ie:

Var(S) = Var(E[S | lambda_i]) + E[var(S | lambda_i)]

Andy