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The mean has good mathematical properties. The mean of a sum is the sum of the means. For example, if $y$ is total income, $u$ is "earned income" (wages and salaries), $v$ is "unearned income" (interest, dividends, rents), and $w$ is "other income" (social security benefits and pensions, etc.). Clearly, a person's total income is the sum of the incomes he or she receives from each source $y_i = u_i + v_i + w_i$. Then $$ \overline{y} = \overline{u} + \overline{v} + \overline{w}. $$ So it doesn't matter if we take the means from each income source and then add them together to find the mean total income, or add each individual's incomes from all sources to get his/her total income and then take the mean of that. We get the same value either way.

I've been trying to prove this, but it doesn't make sense to me.

e.g. $$ \frac{3 + 4 + 2}{3} = 3 $$ $$ \frac{6 + 14}{2} = 10 $$ $$ 3 + 10 \neq \frac{9 + 20}{2} $$

$ 3 + 10 $ is the sum of the means

$ \frac{9 + 20}{2} $ is the mean of the sums which are $3+4+2=9$ and $6+14=20$

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  • $\begingroup$ Can you explain how your counterexample is intended, by statingwhat the numbers are instead of just listing them? $\endgroup$ – 6005 Jul 25 '16 at 18:41
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    $\begingroup$ It only works if the populations are the same. $\endgroup$ – Doug M Jul 25 '16 at 18:42
  • $\begingroup$ No $(9+20)/2$ is not the mean of the sums. The sums would be $3+6=9$, $4+14$ and $2+?$. Notice the sums are not even defined here because one set of numbers has three points, the other has only two. $\endgroup$ – smcc Jul 25 '16 at 18:47
  • $\begingroup$ @smcc Why is it necessary to pair them? $\endgroup$ – John Jul 25 '16 at 18:50
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In your example, you have $u_1, u_2, u_3$, $v_1, v_2$, and you have correctly showed that $$ \text{mean}(u_1,u_2,u_3) + \text{mean}(v_1,v_2) $$ is not necessarily equal to $$ \text{mean}(u_1 + u_2 + u_3, v_1 + v_2), $$ so in that sense you are exactly correct.

However, this is not what the statement was intended to express. What is intended is that if you have two (or more) lists with the same number of elements, and you take the mean of each list and sum them, that will be the same as summing the corresponding elements and then taking the mean. So if we have lists $u_1, u_2, u_3$ an $v_1, v_2, v_3$, it is saying that $$ \text{mean}(u_1 + u_2 + u_3, v_1 + v_2 + v_3) = \text{mean}(u_1,v_1) + \text{mean}(u_2,v_2) + \text{mean}(u_3,v_3). $$ Notice how in the phrase "sum of the means", the individual means must take elements of the same index -- we take the mean of $u_1, v_1$ and the mean of $u_2, v_2$ for example, rather than mean of $u_1, v_1, v_2$ or $u_1, u_2$ or anything else.

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    $\begingroup$ I'm having difficulty understanding how "sum of the means" implies summing the means of corresponding elements of the same index. The different number of sources of income that make up each variable (e.g. $u$ is "earned income" (wages and salaries) and $v$ is "unearned income" (interests, dividends, rents)) make this even more difficult to derive from the author's writing. $\endgroup$ – John Jul 25 '16 at 18:57
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    $\begingroup$ @John I agree the book is unclear -- and the statement "the sum of the means is the mean of the sums" is flatly ambiguous. The book is giving a helpful (or not-so-helpful, in your case) mnemonic phrase to describe this property of means, but what exactly it means must be derived from a more careful reading. $\endgroup$ – 6005 Jul 25 '16 at 19:02
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    $\begingroup$ The sources of income, though, should make this a bit more clear. It's saying that if you want to find the average income, you can just find the average earned income, average unearned income, and average other income, and add those three results. The indices $u_i, v_i, w_i$ are a big hint as well that there is the same number of each. You should think of $u_i, v_i, w_i$ as being the incomes for month $i$. $\endgroup$ – 6005 Jul 25 '16 at 19:04
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    $\begingroup$ Actually, the way how the book phrases it, it is not just unclear, but mathematically wrong, as you pointed out in your answer(+1). But also from an educational/pedagogic perspective, I would argue that such a book should be pulled "from the shelves"...A non math person would definitely draw false conclusions here... $\endgroup$ – imranfat Jul 25 '16 at 19:57
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Suppose you have observations on income from work $w_i$ and income from benefits $b_i$ for a population of $n$ people

The text is simply saying that

$$\frac{1}{n}\sum_{i=1}^nw_i+\frac{1}{n}\sum_{i=1}^nb_i=\frac{1}{n}\sum_{i=1}^n(w_i+b_i)$$

where the left hand side is the sum of the mean work income and mean benefit income, while the right side is the mean of the sum of work income and benefit income.

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The statement is not precise enough. You didn't say at what mathematical level it is intended to be. The most general formulation is, if you have a number of random variables defined on the same measure space and they all have finite averages (expected values), then their sum also has finite average and the "average of the sum equals the sum of the averages" holds. In your counter-example the two "random variables" are defined over different sets, let alone different measure spaces, so the claim is not intended to cover that case.

I agree that mathematical texts should be more careful in their formulations - otherwise they are bound to cause confusion. Good question!

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