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Chapter 9 Q9.5

G

Gemma

Made first post
Hi,

In the Chapter 9 Practice Questions, the solution for question 5, includes calculations for P_82(0.75) and P_82(2.75), to estimate the number of lives at the start of these intervals. Can someone please explain how the weights of 0.25 and 0.75 are derived the answer?

Thanks,
Gemma
 
Hello

The weights come from the linearity assumption. We have the values at time 0 and time 1 (P82(0) and P82(1)). We want to calculate the value at time 0.75. Imagine a plot of P82(t) with t on the x-axis and P82(t) on the y-axis.

Let's consider P82(0) as the 'start' so then if we go linearly to P82(1), then at any point along the way, the proportion that we travel in the x direction (ie along t) must be the same we travel in the y direction (along P82(t)). So we have:

P82(0.75) = P82(0) + (0.75 - 0) / (1 - 0) * (P82(1) - P82(0))

where (0.75 - 0) / (1 - 0) is the proportion we travel in the x-direction, which we apply to the y-direction. Multiplying this by (P82(1) - P82(0)) gives the distance we go in the y-direction.

This simplifies to:
P82(0) + (0.75 - 0) / (1 - 0) * (P82(1) - P82(0)) = P82(0) + 0.75 * P82(1) - 0.75 * P82(0)
= 0.25 P82(0) + 0.75 * P82(1)

Hope this helps!

Andy
 
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