Welcome to the hotseat. We've prepared a guide if you'd like to read more about how it works.

Independent and dependent rates

0 votes
27 views
asked Nov 4 in BUS 3024S - Contingencies by anonymous

BUS3024S Test 2 2007 (1).pdf (0,2 MB)

In the above question paper, I refer to question 5 (c). The memo says that we should see the biggest difference between the independent and dependent rates in the (dependent??) rates which are the lowest. 


\[q_{x}^d =\frac{(aq)_x^d}{1-0.5(aq)_x^{-d}}\]

where we have assumed uniform rates in the multiple decrement model and \[(aq)\] represents the dependent rate and \[(aq)_x^{-k}\] represents the total rate of decrement for all other decrements other than the decrement k.

Using the above relationship, how do we conclude the statement the memo has made?


1 Answer

0 votes
answered Nov 14 by Njabulo.Dube (1,850 points)

From Pieter: 

\( q_x^d = \frac{(aq)_x^d}{1-0.5(aq)_x^{-d}} \) looks incorrect.

\( (aq)_x^{-d} \) should be independent rates: \( q_x^{-d} \)

But that being said, remember independent rates always greater than or equal than dependent rates. So:

$$ \frac{(aq)_x^d}{q_x^d} <= 1 $$

and this expression equals: $$ \frac{(aq)_x^d}{q_x^d} = 1 - 0.5(q_x^{-d}) $$

where \( q_x^{-d} \) represents the sum of all independent decrement rates other than d, and ignoring the higher order terms (+(1/3)*(...)).

So, to minimise \( \frac{(aq)_x^d}{q_x^d} \), we need to maximise \( q_x^{-d} \) So the largest % difference will be seen for the decrement with the lowest independent rate, assuming UDD.

This is how I would've answered the question... Or alternatively, given that the question does not state when the decrements occur, (please check my assertion here. It feels that if one of the decrements should occur at the very end of the year after all the others, then it would have the biggest % difference between dependent and independent... 

commented Nov 15 by rohin_jain (460 points)

The formula I got which relates the dependent and independent rates comes from the ActEd notes. I am not sure where Pieter's version comes from? 

...