# Probability Vs Rate

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 $$_{n}M_x\ne _{n}q_x$$?

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 $$_{n}M_x \ne {_{n}q_x}$$.