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CT5 April 2018 Q13

A

AMD

Made first post
I've been trying to answer the multiple decrement table for Q13 but haven't been able to understand how in the examiners report they have gone from
q_x^d to (aq)x^d and q_x^s to (aq)_x^s. I thought it was just using the formulas for mu and sigma but I can't recreate the same answer.
How come the (aq) for death is the same as qd and what calculation has been done to get from qx^s to to (aq)^s?
Thanks
 
I think it’s because surrenders are only allowed at the end of the year, you don’t use the formula. That would be if they could withdraw any time during the year

(aq)x^d stays the sameas qx, I think because there aren’t any other decrements that operate evenly over the year.

(aq)x^s is the probability they survived over that year multiplied by the probability they surrendered at the end of it ( qx^s value)

So year 1

(1-0.0100221)*0.1 = 0.098998

Year 2

(1-0.0113344)*0.05 = 0.049433

Year 3

They can’t surrender at the end of the year because the contract is finished then, so zero
 
Good answer, Noreen. This is the way we proceed when death is the only decrement operating over the year (meaning the dependent and independent probabilities of death coincide).
 
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