When to use Mean(X/Y) versus Mean(X)/Mean(Y)?

Question

So, I understand that for two vectors of numbers X and Y the mean of X/Y is different from the mean of X divided by the mean of Y. But I'm not sure when it is appropriate to use each. Specifically, I need to calculate the average consumption per days active for a set of individuals, and I don't know which approach to use.

To illustrate, say there are 3 individuals: A, B, and C. Their consumption values are (1,3,7) respectively, and their days active are (3, 1, 6) respectively. If I individually calculate their consumption per days active I get 1/3, 3, and 7/6, which I can then take the mean of to get 1.5. But if I sum their consumption (11) and their active days (10) and then work out consumption per active days, I get 1.1. Which approach is 'correct'?

I hope I've expressed this ok, I wasn't really sure what the title should be which is why I've struggled to find answers because I'm not totally sure what to search. If anyone can explain the different properties of the two quantities and when using each might be appropriate I'd really appreciate it!

EDIT: I have found this question which seems to touch on a very similar issue, but I'm not totally satisfied with the answer. I think the second answer comes closest to what I'm looking for, but I'd just like to know when it's appropriate to use each (not specific to some simulation context like in the question) Usages of Mean(X/Y) vs. Mean(X) / Mean(Y)

Answer 1

If $x_i =$ number of items consumed on active days (for person $i$) and $y=$ number of active days (for person $i$), then...

• $\text{Mean}(x/y)$ is the average number of items consumed per person on a day in which that person is active. This gives equal weight to each person's data.
• $\frac{\text{Mean}(x)}{\text{Mean}(y)} = \frac{\text{Sum}(x)}{\text{Sum}(y)}$ is the total number of items consumed, divided by the total number of active person-days (1 person-day = 1 person active for one day), or the average number of items consumed per active person-day. This gives more weight to more active people, since they contribute more active person-days.

Which one you want depends on your goals, but I would imagine you're almost certainly more interested in people than in person-days.

Answer 2

I first consider $mean(x/y)$ versus $mean(x)/mean(y)$. As Eoin and dariober suggest, the more practically relevant quantity should be used, which is often the former, although the latter is not that outlandish. To illustrate both, let's modify the OP's example and imagine opening a bakery with bakers A, B and C.

1. A usecase for $mean(x/y)$. On the initial (test) day, A, B and C worked $X =(3, 1, 7)$ hrs and made $Y=(1, 3, 6)$ cakes, respectively; later on, each baker will work the same hours per month. The hourly productivity of worker $i$ is $z_i=y_i/x_i$, and the expected productivity $\sum z_i/n = 1.5$ cakes per hour/worker. If, say, A worked for 6 hrs and made 2 cakes on the test day, this expectation won't change.
2. A usecase for $mean(x)/mean(y)$. Suppose a single baker can work at a time, and the bakers do $X =(3, 1, 7)$ hrs per day. Then the bakery's expected hourly output, $\sum y_i/\sum x_i= 1.1$, is meaningful. Since $\sum y_i/\sum x_i= \sum w_i z_i$, where $w_1 = x_1/\sum x_i$ is the "timeshare" of worker 1, this is like the previous case, but weighted. Similar weighting could be based for example on customer expenditure.
3. Using $mean(x)/mean(y)$ when X and Y are not (fully) matched. Suppose we pay $X =(3, 1, 7)$ pounds per hour to $(A, B, C)$ and want to assess competing wages. This assessment is easy if the bakers moonlight for $Y=(1, 3, 6)$ pounds per hour at a competitor's. It is less so if only A and C do; then we have unequal samples or "missing" data. The sizes are equal but samples are still unmatched if $Y=(1, 3, 6)$ are randomly drawn IT wages. (Appropriate matching could be possible, but not always.)

The second ("robustness") aspect, implied by dariober and several commenters, is whether means are suitable for ratios in the first place. As suggested by @Nick Cox under the question, the geometric mean(s) might be better; see also. Alternatives include removal of outliers; the use of trimmed means; the use of $median(x)/median(y)$ or $median(x/y)$. Also, bootstrap estimation could be appropriate for ratios.

Answer 3

In my opinion mean(X/Y) is more meaningful because your experimental unit (not sure this is the correct term) is the individual, not the aggregate. Let's try to see this with a contrived example. Consider this data set:

individual cons  act ratio
         A    5    2   2.5
         B    5    2   2.5
         C    5    2   2.5
         D    5    2   2.5
         E    5    2   2.5
         F    5    2   2.5
         G    5    2   2.5
         H    5    2   2.5
         I    5    2   2.5
         J    5    2   2.5
         K 1000 2000   0.5

I would say that in this dataset the average ratio is just below 2.5 (~2.3 in fact) because you observe 10 individuals with ratio 2.5 and only one individual with ratio 0.5.

However, if you calculate mean(cons) / mean(act) you get an overall average ratio of ~0.52 because individual K dominates the dataset by having much higher values, but individual K is just one out 11 individuals. On the other hand, you may want to give more weight to K if you think its values are more reliable than the other individuals'.