# Calculating Expected Number of Deaths from Life Tables

Normally, if analysing cohort data, I would expand on time using the Lexis macro (if using SAS) and then calculate the expected number of deaths as the life table central mortality rate (mx) matched on age, sex, year multiplied by the exposed time to risk (y) for each cell (i.e. mx*y=expected deaths).

However, I have some code downloaded from Paul Dickman that calculates the expected number of deaths as:

-log(1-mx^length)*(y/length) where length is the interval length (usually 1).

I don't really follow this calculation, and was wondering if someone could talk me through it and explain why one method is better than the other. In comparison of the two methods, the results are very similar but his results in lower risk compared to the population than my method does once summed together.

Thanks!

Keep in mind this isn't a derivation from first principles in any way nor is it a formal proof. Rather, I'm just trying to use intuition to make sense of the equation.

Let's assume the length in the equation is equal to 1 (i.e. a 1 year interval). Now this gives (as you've described it):

$$-\text{log}(1-m_{x})E_{x}$$

Now, we can make the following approximations:

$$\mu_{x+1/2}\approx-\text{log}(p_{x})\approx m_{x}$$

From here:

\begin{align} 1-m_{x}&=1+\text{log}(p_{x})\\ -\text{log}(1-m_{x})&=-\text{log}(1+\text{log}(p_{x})) \end{align}

Now, we know that, for small $y$:

$$\text{log}(1+y)\approx y$$

So:

\begin{align} -\text{log}(1-m_{x})&=-\text{log}(1+\text{log}(p_{x}))\\ &=-\text{log}(p_{x})\\ &=\int_{0}^{1}\mu_{x+t}\,dt \,\,(\approx m_{x}) \end{align}

Therefore, this is the cumulative hazard across age interval $[x,x+1]$ i.e. the cumulative risk of death. So, if we multiply this by the total at-risk in this interval, it should give the expected number of deaths:

$$d_{x}=E_{x}\int_{0}^{1}\mu_{x+t}\,dt$$

or

$$d_{x}\approx E_{x}m_{x}$$

• So, the way I would calculate it is an approximation then, and using the full formula would give a more accurate estimate? Thanks for this! Really helpful.
– DC2
Commented Jan 29, 2017 at 16:28
• @DC2 There would be a good reason for the formula he uses. Maybe that's it. Commented Feb 2, 2017 at 9:02