Solved – Weighted average of percentage increase

mean

I believe I have quite a simple problem, but want clarification on whether it is the best method to use.

Say I have have insurance line, and the net income (£) from this business for 2011, 2012, 2013 is 1,000,000; 4,500,000; and 5,000,000 respectively.

Hence, the percentage increase from 11-12 is 350%, and from 12-13 is 11.1%.

Now, the average of these is 180.5%, which is a representation of the average percentage increase across all years, but it has been suggested to me to instead represent the data by a “weighted” average.

I am curious… would the weights be the difference in income from each year?

i.e. Weighted Avg. = $((3.5)*350\% + (0.5)*11.1\%)/4 = 307\%$

or would the weights possibly the values themselves? Or maybe there is another option? Maybe a standard average is sufficient?

Thanks very much for your inputs.

Best Answer

The weights could be almost anything, depending on why someone wanted them added. However, the arithmetic mean of the percentages is not the best method for averaging growth rates.

You have 1, 4.5, 5 in successive years and, as you noted, the growth in the first two years is 350% and in the second two years, only 11.1%; again as you noted, the arithmetic mean of these (expressed as proportions) is $\frac{3.5 + .111}{2} = 1.805$. Yet that is not the average growth: $1*1.805*1.805 = 3.205$ but your final value is 5.

You could express these as ratios of 4.5 and 1.11, then the average is 2.8. But $2.8*2.8 = 7.84$, again, not 5.

What you want is the geometric mean of the ratios: $(4.5*1.11)^{0.5} = 2.23$ and, indeed, $2.23*2.23 = 5$ (except for rounding error).

Perhaps the weights were meant to reproduce this?

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Solved – Confusion regarding weighted average

I was writing my own weighted average algorithm yesterday and it applies to what you’re looking to do. Your problem is what you're calling returns_rate and the logic behind it. For these examples we have a new field called weighted_score or you could call it weighted_returns if you wanted. The point is that its weighted.

id | name | count | returns | weight | weighted_score
-----------------------------------------------------
1     A      150      10       0.2         2
2     B      250      50       0.3         15
3     C      350      60       0.4         24
4     D      550      100      0.4         40
5     E      150      80       0.2         16
6     F      50       10       0.1         1
-----------------------------------------------------
sums                           1.6         98
weighted_average                           61.3

Logic:

weighted_score = returns*weight
weighted_average = sum(weighted_scores)/sum(weights)

I see that you have some logic to determine the weight however your upper limit is 40%. There really should be 0% through 100% in there. You can also calculate the weight on the fly for any given data set by finding the highest “count” and dividing the “count” in each row by that highest number and this gives you the appropriate 0% - 100% weights.
Example:

id | name | count | returns | weight | weighted_score
-----------------------------------------------------
1     A     150      10       0.2727         2.7
2     B     250      50       0.4545         22.7
3     C     350      60       0.6364         38.2
4     D     550      100      1.0000         100
5     E     150      80       0.2727         21.8
6     F     50       10       0.0909         0.9
-----------------------------------------------------
sums                          2.7273         186.3636
weighted_average                             68.3

Logic:

weight = count/550

Another simpler option that doesn’t require so much processing trying to figure out the appropriate weight is to just create a static variable to use as your ceiling. Any count lower than the ceiling will be weighted with lower importance and anything above the ceiling is weighted at 100%.

id | name | count | returns | weight | weighted_score
-----------------------------------------------------
1     A     150      10       0.4983         2.7
2     B     250      50       0.8306         22.7
3     C     350      60       1.0000         38.2
4     D     550      100      1.0000         100
5     E     150      80       0.4983         21.8
6     F     50       10       0.1661         0.9
-----------------------------------------------------
sums                          3.9934         248.0399
weighted_average                             62.1

Logic:

if(count < 301) weight = count/301
if(count > 301) weight = count/count

Solved – Weighted average vs “unweighted” average in probability

What you are calling the "weighted average" is the only proper way to calculate the percentage of satisfied customers in both shops. Taking average of percentages (what you call "unweighted" average) will give you useless results if your samples differ in size.

Imagine extreme case: you have two shops A and B, in shop A there was one customer and he was not satisfied and in shop B there were 100 customers and 90 of them were satisfied -- would you conclude that 45% of customers of both shops were satisfied? Obviously not!

The smaller sample is much less reliable, so it should not be included in the final estimate with the same weight as the larger one. Speaking more formally, estimates from the smaller sample has a greater error.

Two shops: A and B, in A there are two customers, one happy, in B, there were 100 and 75 were happy. So they have 50% and 75% happy customers, so the average is 62.5%. Using your example from the comment:

Standard error for the estimate from the first shop is $\frac{0.5(1-0.5)}{\sqrt 2} = 0.18$, while for the second one $\frac{0.75(1-0.75)}{\sqrt{100}} = 0.02$. So the possible deviation from the true proportion of satisfied customers in shop A is much larger. If the samples differ that much in their reliability, you shouldn't give them equal trust and weight them equally when taking their average.

You can easily convince yourself why using pooled mean is a much better idea than using the raw one by conducting a simple simulation study. I simulate two shops, with $n=8$ and $n=100$ customers, both having the same proportion of satisfied customers. If you compare the results obtained using raw mean ("unweighted") and the pooled ("weighted") estimates, you'll see that when using the raw mean the variability of the difference between the estimate and the true value is much greater. Saying this in simple English: using raw mean you are at greater risk of obtaining the wrong estimate than in the case of pooled mean.

For much more advanced methods of pooling different probabilities, you can check Combining probabilities/information from different sources thread or this one Combining two estimates .

Best Answer

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Solved – Confusion regarding weighted average

Solved – Weighted average vs “unweighted” average in probability

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