Suppose $_{n}q_x$ is the $\textbf{probability}$ of dying between age $x$ and $x+n$.

And $_{n}M_x$ is the age-specific death $\textbf{rate}$ in the observed population in the interval $(x, x+n)$. The formula of $_{n}M_x$ is $$_{n}M_x=\frac{_{n}D_x}{_{n}P_x},$$

where $_{n}P_x$ is an estimate of the mid-year population between ages $x$ and $x+n$,

$_{n}D_x$ is the number of deaths in the same age interval during the year.

It seems to me $\frac{_{n}D_x}{_{n}P_x}$ is also a probability and $_{n}q_x=\frac{_{n}D_x}{_{n}P_x}=_{n}M_x$. But this is not the case. Rather, the relationship between $_{n}q_x$ and $_{n}M_x$ is $$_{n}q_x=\frac{2\times n\times _{n}M_x}{2+ n\times _{n}M_x}.$$

My question is: if $_{n}M_x=\frac{_{n}D_x}{_{n}P_x}$ and the definition of $_{n}q_x$ is the probability of dying between age $x$ and $x+n$, then why $_{n}M_x\ne _{n}q_x$?


1 Answer 1


The probability for somebody to die at age $x$ (that is how I interpret your $q_x$) is not only related to the probability of dying conditional on being of age $x$ but is also related to the probability of reaching the age of $x$. So that is why $_{n}M_x \ne {_{n}q_x}$.

Related questions:

Note especially the graph in that last question (repeated here below). On the left image is the age specific death rate, in the right image is the age distribution for the probability that somebody dies at a certain age.

death rates, survival rate, and probable age of death

The death rate is very high for people of high age, but that does not make the probability of dying at high age very probable.

Say, the death rate for people of hundred years old is 30% per year, does that make the probability for somebody to die at age hundred 30%?


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