Individual standardised deviations test

Chapter 12 page 32 Rationale:

If there is heterogeneity within the age groups or deaths are not independent, the variance will be (respectively) smaller or greater than we would expect if our underlying model were correct.

Could someone give a more detailed explanation?

I would have thought that were there to be heterogeneity within the age groups
this would imply a greater variance in the standardised deviations.

Any help much appreciated.

Many thanks.

 

OK, having read that again, it's the variance observed compared to the variance of the underlying model being referred to (binomial, Poisson/Multi State). However can anyone explain how heterogeneity within an age group and deaths not being independent brings about lower/higher observed variance respectively?

Much appreciated.

 

OK, in answer to my own question let us consider the case of the binomial model.

Homogeneous assumption

Assume that all lives of a given age have the same constant chance of dying Q. Variance of the number of deaths D in population of N lives:

\[Var(D) = NPQ \text{ where } P = 1-Q\]

Heterogeneity observed

N lives of the same age are made up of r homogeneous groups. ith group has ni lives that have a chance qi of dying within that year.

\[N = \sum n_i \text{, } Q = (1/N)*\sum n_iq_i \text{, } D = \sum d_i\]
\[E(D) = \sum E(d_i) = \sum n_iq_i = NQ\]
\begin{eqnarray}
Var(D) & = &
\sum Var(d_i)
\\& = &
\sum n_iq_i(1-q_i)
\\& = &
NQ - \sum n_i q_i^2
\\& = &
....
\\& = &
NPQ -\sum n_i (q_i-Q)^2
\\&<&
NPQ
\end{eqnarray}

This shows that heterogeneity in mortality lowers the variance of the number deaths.

P.S. How do I left align the eqnarray environment above?

 

For completeness, the lack of independence --> higher observed variance is shown in Chapter 13 where duplicate policies are considered.

 

Hi
How to get second last step? After "..." Step

 

Something like the following:

\begin{eqnarray}
\\NQ - \sum n_i q_i^2 & = &
NQ - \sum n_i (q_i-Q)^2 + Q^2\sum n_i - 2Q\sum n_i q_i
\\& = &
NQ - \sum n_i (q_i-Q)^2 - NQ^2
\\& = &
NPQ -\sum n_i (q_i-Q)^2
\\&<&
NPQ
\end{eqnarray}

 

(Y) thanks!