Average of ratios compared to ratio of averages

Question

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?

Answer 1

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:

• Simple question regarding ratio and average
• Ratio of Averages vs Average of Ratios

Answer 2

Too long as a comment:

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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”.

Answer 3

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 $\sum \lambda_i x_i >0$.