Split up sum of products $\sum{a_i b_i}\approx(1/N)\sum{a_i}\sum{b_i}$ for uncorrelated summands? As the topic says, is $\sum{a_i b_i}\approx(1/N)\sum{a_i}\sum{b_i}$ possible when $a_i$ and $b_i$ uncorrelated? I have come across something like that very recently where this has been magically done without additional explanation and it makes intuitively sense. If I take a number example like 2*3 + 3*4 + 1*5 $\approx$ (6/3)*(12), but how can I justify this formally? Is there a name for this?
Thank you, M
 A: 
Here is a bare sketch of the solution. Please follow the commentaries to get an overview. In essence, $\sum{a_i b_i}\approx(1/N)\sum{a_i}\sum{b_i}$ are asymptotically equivalent for when $a_i$ and $b_i$ are uncorrelated. The reason for this is that if we multiply by 1/N both terms and take the probability limit, i.e.
$\sum{a_i b_i}/N\approx(1/N^2)\sum{a_i}\sum{b_i}$
$plim_{N->inf} \sum{a_i b_i}/N=plim_{N->inf}(1/N^2)\sum{a_i}\sum{b_i}$
by Markow Law of Large Numbers with unknown distribution it should be that
$E[a_i b_i]=E[a_i]E[b_i]$
which holds only for them being uncorrelated, i.e. Cov[a,b]=0.
The illustration provides an excel simulation of the above approximation formula. I checked the formula with a set of randomly created variables and how well the approximation works with increasing (sample size) and thus decreasing correlation. 
Note that for a small sample starting with 10 (0 on the x axis) we have a relatively high correlation of 0.4 and percentage error (defined as percentage difference between the two formulas) of 0.16. Then with an increasing sample size, the error becomes insignificant, while the unsurprisingly the correlation decreases exactly in tow with it. 
