# Homework Help: Bi-geometrical mean using logs -- don't get the same result

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1. Jan 27, 2017

### ducmod

1. The problem statement, all variables and given/known data
Hello!
I am trying to compute the bi-geometrical mean on data that contains negatives.
But before that I wanted the test the formula that accounts only for positive values using the sum of their logarithms. By doing so I don't get the result I compute by using the "usual" geo mean formula:
for values between 0 and 1: ((1+a1) * (1+a2) * ....(1+ aN)) ^(1/n) - 1
for other positive values: (last value / initial value) ^ (1/number of periods) - 1
2. Relevant equations
Here is the example:
1,062 1,252 1,587 1,934 2,519
Geometric mean = (2519 / 1062) ^ (1/5) - 1 = 18.86%
3. The attempt at a solution
Using the formula log g = sum of logs of each value (see picture attached)
log g = (LOG(1062)+LOG(1252)+LOG(1587)+LOG(1934)+LOG(2519))/number of periods
I get geo mean = 0.505, which is nowhere close to 18.86%
What am I doing wrong?

Thank you!

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2. Jan 27, 2017

### Staff: Mentor

You have three completely different formulas computing different things here. I don't see why you expect two of them to give the same result.

(last value / initial value) ^ (1/(number of periods)) - 1 for sorted values gives you some information how much the numbers grow per step. Where "number of periods" should probably be the number of data points minus one.

Averaging the logs (the result is not 0.505) gives you some information about the average value, but not about the differences between the values. Calculate its exponential to get back to the original scale.

((1+a1) * (1+a2) * ....(1+ aN)) ^(1/n) - 1 is yet another parameter, telling you something about the average. It is similar to the logarithmic average but not identical.

3. Jan 27, 2017

### ducmod

As to the first formula: (last value / initial value) ^ (1/(number of periods)) - 1 it is a compound average growth rate, which should give the same result as the geometric average, and I wouldn't say that it is for a sorted set of data, because "sorted" implies that values grow sequentially (second is bigger than the first one, etc), which is not the case usually and which is why this formula is used to compute growth rate; the last formula is a geometric average computed for values between 0 and 1 (this is why 1 is added, to smooth those values).

4. Jan 27, 2017

### Staff: Mentor

An average growth rate and an average of the values are different things.

Your growth rate doesn't change if you multiply all values by 1000, but the average of the values does change.

5. Jan 27, 2017

### ducmod

Sorry but all this doesn't help me to understand and solve the whole issue. Your answers are very vague, and they leave me very confused.

6. Jan 27, 2017

### Ray Vickson

It ought to yield identical results to the log form, except maybe for roundoff errors, because
$$\log \left[ \prod_{i=1}^n A_i \right]^{1/n} = \frac{1}{n}\sum_{i=1}^n \log A_i .$$

7. Jan 27, 2017

### Ray Vickson

You really are confusing yourself by mixing up several different concepts. The two series s1 = {1, 1.2, 1.44} and s2 = {1000,1200, 1440} both have the same (geometric) growth rate r = 0.2 (20%), but their numerical values are orders of magnitude different. Furthermore, the calculation $(1440/1000)^{1/3} -1 \doteq 1.1292 -1 = 0.1292$ does not give a value anywhere near the true growth rate 0.20.

I can think of no significance whatsoever for the quantity $(a_n/a_1)^{1/n}-1$ that you have called the "geometric mean". In mathematics the concept of geometric mean is well-defined (and has been for hundreds of years); it is not what you are claiming it to be.

8. Jan 27, 2017

### Staff: Mentor

Don't forget the 1 that gets added and subtracted.
Example: a0=1, a1=2. The formula with +1 gives $\sqrt{6}-1 \approx 1.45$, the log formula gives 1.16. Let a0 go to zero and the log average diverges while the other one stays finite.

I don't understand what is unclear.
You have three different formulas calculating different things.