2
$\begingroup$

I want to convert an instantaneous mortality rate that is reported per year (actual value = $0.58y^{-1}$) into a weekly mortality rate.

This answer gives the formula as $j=(1+i)^{1/12}-1$ where $j$ is monthly interest rate and $i$ is the annual rate.

But this answer gives the formula as $j=1-(1-i)^{1/365}$ where $j$ is daily mortality rate and $i$ is the annual rate again.

If I use both of these to calculate my weekly rate I get different answers.

   x = 0.58
    ((1 + x)^(1/52)) - 1
# Versus
    1 - (1 - x) ^ (1 / 52)

Where am I going wrong?

$\endgroup$
  • 2
    $\begingroup$ See answer by Vizag below. Main difference is that with interest, getting interest will increase your money, so there is (a bit) more existing money for the next interest payment to be based on. That means positive interest increases your money. With death/mortality, it's the other way around: As some people die during the first week, you now have less population from which some more people will die next week and later. That's why the $1-x$ formula is correct, not the $1+x$ formula. $\endgroup$ – Ingix Jun 12 at 11:56
  • $\begingroup$ So obvious when someone points it out. Thanks for your clarification. $\endgroup$ – adkane Jun 12 at 12:11
3
$\begingroup$

Let's break this down a little. Assume you had a normalized population of $1$ at the beginning of the year. The annual death rate is given to be $0.58$, i.e., $0.42$ population will be alive by the end of the year. To calculate the weekly mortality rate, let's go week by week. Assume the weekly death rate is $x$.

End of week $1$:

$1-x$ people are left.

End of week $2$:

$(1-x)\times(1-x)$ people are left.

And so on till End of week $52$,

$(1-x)^{52}$ people are left.

Now it must happen that:

$$(1-x)^{52} = 0.42 = 1-0.58$$

This will give you $x$.

$\endgroup$
  • 1
    $\begingroup$ Great, thanks for the clear explanation. $\endgroup$ – adkane Jun 12 at 12:10
  • $\begingroup$ No problem. You're welcome. $\endgroup$ – Vizag Jun 12 at 12:15
  • $\begingroup$ Please press the $\checkmark$ button below the answer to mark it as accepted. Thanks. $\endgroup$ – Vizag Jun 12 at 12:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.