2

How to formalize (in simple terms) when average of ratios would be greater than ratio of averages?
I found that sometimes the former is greater than the latter and sometimes the latter is greater than the former.
I want to understand in what cases one would be greater than the other.

What happens at asymptotic?
And what happens in day to day calculations? Which should I expect to be greater?

sds
  • 4,533
bilanush
  • 433
  • You are trying to compare $\frac{\frac ab +\frac cd}{2}$ and $\frac{\frac {a+c}{2}}{\frac {b+d}{2}}$ with a difference of $\frac{(b-d)(bc-ad)}{2bd(b+d)}$ and the sign of the difference depends on the sign of the numerator – Henry Jun 27 '22 at 15:14
  • Do you mean with two arbitrary datasets? Are there any assumptions on independence or lack of it? For datasets, $\overline x/\overline y$ vs $\overline{\left(x/y\right)}$ can have any relationship due to random variation. Though, there will be a set relationship between the expectations $E(X/Y)$ vs $E(X)/E(Y)$, but this can go either way depending on the underlying distributions. – jdods Jun 27 '22 at 17:52

3 Answers3

3

Let us start the conversation with just 4 numbers for simplicity.

Note that neither the Average of Ratios $$\frac{\frac{a}{b}+\frac{c}{d}}2 \tag{AR}$$ nor the Ratio of Averages $$\frac{\frac{a+c}2}{\frac{b+d}2}=\frac{a+c}{b+d} \tag{RA}$$ change when all 4 numbers are scaled together, to the relationship between AR and RA depends only on the projective point $a:b:c:d$.

Now, let us scale $(a,b)$ to $(\lambda a,\lambda b)$, then AR does not change and RA goes from $\frac{a}{b}$ for $\lambda=\infty$ to $\frac{c}{d}$ for $\lambda=0$.

This should give you intuition you are asking for: ratio of averages is the average of ratios weighted by their denominators: $$ \frac{a+c}{b+d} = \frac{b}{b+d}\times\frac{a}{b} + \frac{d}{b+d}\times\frac{c}{d}$$ while AR is their average with weights $\frac12$, and this generalizes directly to an arbitrary number of ratios.

Thus AR will be bigger when the bigger ratio has a bigger denominator.

Related:

sds
  • 4,533
0

Too long as a comment:

average of ratios

  1. Does the “average of $3{:}4$ and $5{:}6$” mean $4{:}5$ or $19{:}24$ ?

  2. What does the “average of $3{:}5{:}7$ and $1{:}2{:}3$” mean?

Given three fractions $\dfrac{x_1}{k_1},\dfrac{x_2}{k_2},\dfrac{x_3}{k_3},$

  • their arithmetic mean is $$\boxed{\frac13\left(\dfrac{x_1}{k_1}+\dfrac{x_2}{k_2}+\frac{x_3}{k_3}\right)}=\frac{x_1k_2k_3+k_1x_2k_3+k_1k_2x_3}{3\:k_1 k_2 k_3}.$$

This is what you mean by “average of ratios”.

ratio of averages

When they are assigned weights according to their denominators, then

  • their weighted arithmetic mean is $$\frac{k_1}{k_1+k_2+k_3}\left(\dfrac{x_1}{k_1}\right)+\frac{k_2}{k_1+k_2+k_3}\left(\dfrac{x_2}{k_2}\right)+\frac{k_3}{k_1+k_2+k_3}\left(\frac{x_3}{k_3}\right)\\=\boxed{\frac{x_1+x_2+x_3}{k_1+k_2+k_3}}\\=\frac{\frac13\left(x_1+x_2+x_3\right)}{\frac13\left(k_1+k_2+k_3\right)}.$$

This is what you mean by “ratio of averages”.

ryang
  • 38,879
  • 14
  • 81
  • 179
  • This an explanation but it doesn't adress the question and doesn't answer it – bilanush Jun 28 '22 at 18:12
  • My opening sentence means that the points I was making are collectively too much to be posted in the comment box (the usual way). P.S. I just edited my first line for typos. – ryang Jun 29 '22 at 03:55
0

You can use the Chebyshev inequality for sums to conclude:

$x_1 \le x_2 \le \cdots \le x_n$ and $\frac{y_1}{x_1}\le \frac{y_2}{x_2} \le \cdots \frac{y_n}{x_n}$ then

$$ \frac{\sum \lambda_i \frac{y_i}{x_i}}{\sum \lambda_i} \le \frac{\sum \lambda_i y_i}{\sum \lambda_i x_i}$$

If the ratios and the denominators are in reversed orders then the inequality is reversed.

Note that $\lambda_i$ are some positive weights, they could be $\frac{1}{n}$ if one takes equal weights. Now the standard Chebyshev inequality does not require the other numbers to be positive, but here we do a division, so we'd better have $\sum \lambda_i x_i >0$.

orangeskid
  • 53,909