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The force of mortality at age $x+t$, given survival to $x+t$ is given by

$\mu_{x+t}$ $=-\frac{d}{dt}[ln({}_{t}p_{x})]$

Given a life table I know how to calculate ${}_{t}p_{x}$, but then how can I calculate the force of mortality using the formula above?

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It's not 100% clear from your question what you need, but I'll mention some things that may help.

(Here $_tq_x$ is the probability a person alive at age $x$ will die by $x+t$, $q_x=\, _1q_x$ and $p_x=1-q_x$... unless I screwed up somewhere.)

$\mu_{x}$ is the force of mortality, i.e. the hazard rate.

$\mu_{x}=\lim_{t\to 0}\frac{_tq_x}{t} $

If you're working from a typical annual life table, one approximation may be found as follows:

  • $p_x=\exp(-\int_0^1\mu_{x+t}dt)$
  • $\int_0^1\mu_{x+t}dt\approx \mu_{x+\frac12}$ (e.g. via a first order Taylor approximation at the center of the interval)

Hence $p_x\approx\exp(-\mu_{x+\frac12})$. Therefore:

$\mu_{x}\approx\frac12 (\mu_{x-\frac12}+\mu_{x+\frac12})$ (say via Taylor again)

$\qquad\approx -\frac12 (\log(p_{x-1})+\log(p_{x}))$ (using the above result),

though other approximations are also used; here are a couple:

$\mu_{x+t}\approx\frac{q_x}{1-tq_x}$ (uniform deaths)

$\mu_{x+t}\approx\frac{q_x}{1-(1-t)q_x}$ (Balducci assumption)

This last assumption can produce some odd results though.

You might also assume the table is locally well approximated by a Gompertz assumption (say) and approximate $\mu_{x+t}$ for small $t$ by taking values from the table at say $x$ and $x+1$ to work out the parameters near age $x$.

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  • $\begingroup$ That is exactly what I was looking for, thanks. Do you have a reference for the approximation you wrote? $\endgroup$
    – Egodym
    Jul 1, 2015 at 3:40
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    $\begingroup$ I'll see if I can locate a suitable one for that, though I'll add some justification for it in any case. The other two I give can be found in a number of places; for example they're in "Modelling Longevity Dynamics for Pensions and Annuity Business", Pitacco et al $\endgroup$
    – Glen_b
    Jul 1, 2015 at 4:02
  • $\begingroup$ Great. I just saw a paper (belgianactuarialbulletin.be/articles/vol08/02-Luciano.pdf page 6) where they use $-ln(p_x) $ to approximate $\mu_x$. Is that a good approximation? $\endgroup$
    – Egodym
    Jul 1, 2015 at 4:08
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    $\begingroup$ If mortality is nearly constant near $x$, then perhaps, otherwise it will typically tend to be too high. See my derivation above for why. $\endgroup$
    – Glen_b
    Jul 1, 2015 at 4:13
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    $\begingroup$ There won't be a very good approximation there; the force of mortality will be very high (everybody dies); nevertheless if you use the formula, you'll get $\infty$, which would produce $p_x=0$, but practically speaking, the actual force of mortality will be finite (and $p_x$ wouldn't be exactly 0); eventually someone will live past the end of the table. $\endgroup$
    – Glen_b
    Jul 16, 2015 at 16:08

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