## Question 21, Spring 2007 MLC May 29, 2007

Posted by Peter in Exam 3/MLC.

This question was received with a bit of controversy because of its wording.

21. You are given the following information about a new model for buildings with limiting age ω.

1. The expected number of buildings surviving at age x will be $l_x = (\omega - x)^\alpha$, x < ω.
2. The new model predicts a 33.3% higher complete life expectancy (over the previous DeMoivre model with the same ω) for buildings aged 30.
3. The complete life expectancy for buildings aged 60 under the new model is 20 years.

Calculate the complete life expectancy under the previous DeMoivre model for buildings aged 70.

The problem with the way this question reads lies in the phrase “previous DeMoivre model” mentioned in item 2, and at the end of the question. Usually, one does not relegate essential information to an offhandedly casual and parenthetical remark. A properly posed question should read “…previous model, which is DeMoivre….” This eliminates the ambiguity of whether the word “previous” belongs to “DeMoivre” or to “model,” the latter being the intended meaning. This issue is further exacerbated by the fact that the new model is a modified/generalized DeMoivre, and that both models share the same ω. Together, this leads to a confusingly written question, because the author did not take care to make it absolutely clear (preferably in a separately listed item) that the old model was DeMoivre (or equivalently, α = 1). That said, the solution is as follows:

Solution: We first compute the complete life expectancy of a building aged (x) under the new model, noting that the old model has α = 1:

${\setlength\arraycolsep{2pt} \begin{array}{rcl}\displaystyle\overset{\circ}{e}_x(\alpha) &=& \displaystyle\int_0^{\omega - x} \!\!_t p_x \, dt = \int_0^{\omega - x} \frac{l_{x+t}}{l_x} \, dt = \int_0^{\omega - x} \!\left(1 - \frac{t}{\omega - x}\right)^{\!\alpha} dt \\ &=& \displaystyle\left[\frac{\omega-x}{\alpha+1} \left(1 - \frac{t}{\omega - x}\right)^{\alpha+1}\right]_{t=0}^{\omega-x} = \frac{\omega-x}{\alpha+1}. \end{array}}$

Then item (2) gives the condition $\overset{\circ}{e}_{30}(\alpha) = \frac{4}{3} \overset{\circ}{e}_{30}(1)$, from which we obtain

$\displaystyle \frac{\omega-30}{\alpha+1} = \frac{4}{3} \cdot \frac{\omega - 30}{2},$

and hence α = 1/2. Item (3) then gives the condition $\displaystyle \overset{\circ}{e}_{60}(1/2) = \frac{\omega-60}{\frac{1}{2}+1} = 20,$ so ω = 90. Therefore,

$\overset{\circ}{e}_{70}(1) = \frac{90-70}{2} = 10.$