Hello all!
I'm having some difficulty addressing a database problem with data of Sizes (cm).
I want to calculate the geometric mean of the values of the size of a number of species in a group of regions. My data set is divided in columns (regions) and lines (species).
When I attempt to calculate the geometric mean in Excel (it's too big a database to easily work it in R), I get a problem with overflow in an intermediate step (the formula being "Potence(Produce(RANGE);1/RangeSize").
There is just too much cells to be multiplied and, as such, I cannot get the geometric mean.
I was wondering, is it correct to calculate the geometric mean in each region and afterwards perform another mean measure (arithmetic mean?) to calculate the geometric mean of the group of regions?
I've been advised to use a Weighted arithmetic mean, since it uses the Sample size as a ponderation.
I hope I've been clear, and would gladly accept any help or advice.
Thank you so much!
Usually, weighted geometric mean of a finite number of positive numbers x_1,...,x_n is dominated by the corresponding weighted arithmetic mean, thanks to Jensen's inequality (see the book of Walter Rudin, "Real and complex analysis"). In Jensen's inequality equality occurs if and only if x_1=…=x_n. A simple proof of Jensen's inequality (and of the assertion related to the case when equality occurs) can be done by means of strict concavity of the function "ln" on (0, infinity).
Hello,
it does not seems right to make the arithmetic means of a finite number of geometric ones.
Instead, generalize the following process :
Say u1=product(a1,a2,...aN)^(1/N) and u2=product(b1,b2,...,bN)^(1/N) are geometric means of two groups of size N.
Your goal is to compute v=product(a1,a2,...,aN,b1,...,bN)^(1/(2N)), the global geometric mean.
Thus, you can divide the pb of size 2N into two pb of size N, but you have to do
v=(u1xu2)^(1/2) to get the right geometric mean in the end.
A function in Visual Basic should be appropriate to compute your big geometric mean.
Gabriel
Thank you all for your answers!
Gabriel Thomas, is this also correctly applied when N differs between u1 and u2?
Would the final potence of ^(1/2) change in the final step?
Piotr Kot , would that imply applying a log function to the whole data set prior to calculating the geom means?
Thank you all in advance!
Francisco.
Hello,
for the logarithms apply the log to all the raw data and use the summation laws (log (ab) =log a + log b ; log(a^n)=n log a ) to recover the log of the means. In the very end take the exponential.
For the case u1=(a1a2...aN1)^(1/N1) and u2=(b1b2....bN2)^(1/N2) you have then v=(a1...aN1b1...bN2)^(1/(N1+N2))=u1^(1/(N1+N2)).u2^(1/(N1+N2)) which should help. You have to generalize to manage your array of NXM data.
Gabriel
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