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I have a set of numbers (in my case: mean retention time (MRT) in the stomach (h)) of which I want to calculate the average gastric passage rate (/h). Gastric passage rate = 1/MRT.

My question is why 'the average of the calculated gastric passage rates of those numbers' is not the same as 'the calculated gastric passage rate of the averaged MRTs'. The next question is: what is the right way?

So for example:

$x = 5; 10; 4; 2.$ Average $= 5.25 h Rightarrow 1/5.25 = 0.19$/h

$1/x = 0.2; 0.1; 0.25; 0.5.$ Average $= 0.26$/h

So should I first take the average of the MRTs and then take the inverse for calculating the gastric passage rate (first way) or should I first take the inverse of all numbers for calculating the gastric passage rates and then take the average of that number (second way).

Thanks in advance!

Here's an everyday puzzle that may help.

If you travel from here to there at $30$ miles per hour and back at $60$ miles per hour, what is your average speed? Instinct says it should be the average, which would be $45$ miles per hour.

But speed is (total distance)/(total time). You don't have a distance given, but you can make one up. Suppose your destination was $60$ miles away. Then it took you $2$ hours to get there and $1$ to get back. You drove $120$ miles in $3$ hours so your average speed was $40$ miles per hour.

The moral of the story is that you can't naively average averages, and a rate is an average. So be careful when you have to compute an average rate.

In your case your MRT is like the reciprocal of the speed, whose units are hours/mile. In my example those are $2$ hours per $60$ miles for the slow trip and $1$ hour per $60$ miles mile for the fast return. You can average those to get the average number of hours per mile. The average is $1.5$ hours per $60$ miles. The reciprocal is $60$ miles per $1.5$ hours, or $40$ miles per hour.

So this is right:

take the average of the MRTs and then take the inverse for
calculating the gastric passage rate (first way)

You've encountered a phenomenon called the AM-HM inequality, $frac{1}{n}sum_{i=1}^n x_igefrac{n}{sum_{i=1}^nfrac{1}{x_i}}$ for $x_i>0$, with equality iff all $x_i$ are equal. In general, functions don't commute with taking expectations; this is the example with the function $1/x$.

What you do with your data depends on what form you assume the distribution of retention times has. For example, suppose you think it has an Exponential distribution, with rate parameter $lambda$. The fact that $1/lambda$ is the MRT is then $1/lambda=int_0^inftylambda xexp (-lambda x)mathrm{d}x$. I don't advise trying to estimate $lambda$ from averaged reciprocals, since $int_0^inftyfrac{lambda}{x}exp (-lambda x)mathrm{d}x$ is infinite.

Let's suppose for the sake of argument that we estimate $1/lambda$ as the dataset's mean retention time. This is an example of the method of moments. It can be shown maximum likelihood estimation would recommend the same estimator. Pages 3 and 4 here discuss what happens with Bayesian estimation, and it comes to much the same thing for a large sample size. So if I were you, I'd estimate $1/lambda$ as the mean GPR.

Given some positive values $x_1$, ..., $x_n$, their average is
$$frac{x_1+cdots+x_n}{n}=frac{1}{n}sum_{i=1}^nx_i,$$
and its inverse is
$$frac{1}{frac{1}{n}sum_{i=1}^nx_i}=frac{n}{sum_{i=1}^nx_i}.tag{1}$$
On the other hand, the average of the inverses is
$$frac{frac{1}{x_1}+cdots+frac{1}{x_n}}{n}=frac{1}{n}sum_{i=1}^nfrac{1}{x_i}tag{2}.$$
In general, these two expressions are not the same, as your calculations also show.

I wrote an incorrect and confusing bit here, after reading the question properly (and giving it some thought) I fully support @Ethan Bolkers answer.