Is it more correct to average numbers before taking log values, or averaging the log values..cite sources?

Question

I have 3 measurements, which are going to be "logged" (i.e) 3 values... 2,3,4 are to logged... the average of the 3 is log 3 = 0.477, or it correct to go...log 2 = .3, log 3 = .447 and log 4 =.602 and there average is 0.4496.... i know the difference is small, but to me it is important.

Answer 1

Keep in mind that what you're doing when you add the logs is multiplying the ratios. Then dividing by 3 takes the cube root of that. So averaging the values is the arithmetic average and averaging the logs is the geometric average. Which one is fits what you're doing best? You'll have to figure that out. I'd guess the geometric average based on the formula you've given.

Answer 2

Log Mean Average

Answer 3

I'm not sure what the answer is, but you need to look into the statistics of error propagation.



When you perform operations on a quantity that has a certain amount of uncertainty, you should carry the uncertainty with it through the operations.



Depending on which average you take, the overall uncertainty will be different.

Answer 4

Let y = log(4ln(1-a/b)...

Treat a, b and y as random variables.

You want to estimate the true mean (or average or expected values) of y.

(I) You can take samples of y (by taking samples of a and b and plugging them into the eqn.) and then average the samples of y.

(II) Alternately, you could take average values of a and b and plug them into the eqn. to get a value for y.



Considering the "contracting" nature of log functions (i.e., log(m-n) < m-n ) I can say intuitively that the std. deviation of y wil be smaller than the std. deviation of a/b. Therefore, without proof I suggest method (I) as described above.

Answer 5

It depends on your objective. They are different calculations and can not be comparable. So both are correct but it depends on what you want to do. Basically, you are asking which one is correct:



Log ((a+b+c)/3) or (log a + logb + log c)/3



They are not the same.