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I have the following values $\overline{b}$, and $\overline{u}$

which is : $\overline{b} = \dfrac{b_1+b_2+b_3+b_4}{7}$, $\overline{u} = \dfrac{u_1+u_2+u_3+u_4}{7}$

is there any way I can get the following if I have only value of $\overline{b}, \overline{u}$?

$$\frac{b_1/u_1 + b_2/u_2 + b_3/u_3 + b_4/u_4}{7}$$

Basically I want to convert average of sums into sum of averages

Ali Caglayan
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Gaurav Shah
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  • I just want to add that the accepted answer is a very good method of attack. If you can't see why something is true, try to prove it's not true, which in this case means finding a counterexample - which means trying out numbers that might break. – djechlin Aug 18 '14 at 14:23

1 Answers1

14

It cannot be done. The average of the $\frac{b_i}{u_i}$ cannot be recovered from the average of the $b_i$ and the average of the $u_i$.

For suppose that $b_1=b_2=b_3=b_4=6$ and $u_1=u_2=u_3=u_4=3$. Then
$$\frac{b_1}{u_1}+\frac{b_2}{u_2}+\frac{b_3}{u_3}+\frac{b_4}{u_4}=8\tag{1}.$$

Suppose now that $b_1=b_2=b_3=b_4=6$, and $u_1=u_2=2$ and $u_3=u_4=4$. Then
$$\frac{b_1}{u_1}+\frac{b_2}{u_2}+\frac{b_3}{u_3}+\frac{b_4}{u_4}=9\tag{2}.$$

Note that in the two examples, the $b$'s were the same, and the $u$'s had the same sum, and therefore the same average.

But the sums of the $\frac{b_i}{u_i}$, and therefore their averages, are not the same.

André Nicolas
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