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I just began studying for the SOA LTAM exam. There are 2 methods shown in the text for calculating $S_x(t)$, the survival function for a life age x to survive until at least time t:

Method 1:
$S_x(t) = \frac{S_o(x+t)}{S_o(x)}$

Method 2:
$$Sx(t) e^{\int_{0}^{t}{Ux+s} \space ds}$$, where $$Ux+s = \frac{\frac{d}{ds} (S_o(x+s))}{S_o(x+s)}$$

For the function $S_o(x) = (1-\frac{x}{120})^{\frac{1}{6}}$

applying method 1: $$\frac{S_o(x+t)}{S_o(x)}= \frac{(1-\frac{(x+t)}{120})^{\frac{1}{6}}}{(1-\frac{x}{120})^{\frac{1}{6}}} = (1-\frac{t}{(120-x)})^{\frac{1}{6}}$$

applying method 2, I get $$Sx(t) =(1-\frac{t}{(120-x)})^{\frac{120}{(720-x)}}$$

so they only give the same result when $x=0$, but I thought they should be the same result for all $x$?

Rhys Hughes
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user221943
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1 Answers1

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Apologies, I did not calculate $u_{x+s}$ correctly. The two approaches of course do produce the same result: $S_x(t) = (1-\frac {t}{(120-x)})^\frac 1 6 $.

Method 2 should have read: $S_x(t) = e^-{\int_{0}^{t}{u_{x+s}} \space ds}$

and the formula for $u_{x+s}$ should have read: $u_{x+s} = \frac {-\frac d {ds} S_x(x+s)} {S_ 0(x)}$

I had incorrectly calculated $u_{x+s} = \frac {-\frac d {ds} S_x(s)} {S_ x(s)} = -\frac d {ds} \space ln \space S_x(s)$

user221943
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