I have two series of trading profit results. I use the geometric mean to calculate the average in percent (CAGR). I would like to divide it by the standard deviation by combining the two series, but I´m having trouble calculating the combined standard deviation.

I found this thread, but it looks like it only works with arithmetic mean, not geometric:

Is it possible to find the combined standard deviation?

Is there a way to calculate the combined standard deviation here?


I got the geometric mean and corresponding geometric standard deviation for two separate data series. I now want to calculate the geometric mean and geometric standard deviation of the combined data series, using only the two mean/stdev values (not the original data points). Example:

Geometric mean 1: 0.070

Geometric stdev 1: 0.051

Number of datapoints: 7


Geometric mean 2: 0.053

Geometric stdev 2: 0.049

Number of datapoints: 16

I know how to calculate the combined geometric mean, but how can I calculate the combined stdev from the above values?

  • 1
    $\begingroup$ I can see that you might want the geometric mean of two series, but why do you think you want the geometric SD of two series? You have just two numbers for each case; if you are concerned that they may not be close, then quote their ratio, or both numbers. Otherwise put, there is not much to summarize here. $\endgroup$
    – Nick Cox
    Oct 21, 2014 at 10:44
  • $\begingroup$ "Geometric series" suggests standard elementary algebra; in statistics a "geometric series" is not defined or characterised by your wanting to take a geometric mean. $\endgroup$
    – Nick Cox
    Oct 21, 2014 at 10:45
  • $\begingroup$ Maybe I was a bit unclear. I'm trying to calculate the risk adjusted CAGR by dividing CAGR with the standard deviation for a combined series of data. So, I got two CAGR values and two standard deviations, one for each series. I know how to calculate the combined CAGR, but I've not been able to figure out how to get the combined stdev. $\endgroup$
    – Macce
    Oct 21, 2014 at 12:18
  • $\begingroup$ CAGR is jargon unknown to me and more generally I am still unclear on what you seek here. (I don't think this is mainstream "mathematical statistics" so you may need different tags to attract attention: some people only follow certain tags.) Perhaps you should show a sample of real or at least realistic data and what you have done and what you want to do. $\endgroup$
    – Nick Cox
    Oct 21, 2014 at 12:27
  • $\begingroup$ I was interpreting "series" literally, as if wanted e.g. the mean of two series for a series of years; it appears that you use it to mean any set of data values. As I understand it, geometric mean and SD are just like mean and SD generally, just that you work on the logarithms and exponentiate at the last step. Combining SDs is the same, but note that you need covariance as well as variances. $\endgroup$
    – Nick Cox
    Oct 21, 2014 at 14:10

1 Answer 1


Geometric mean of group A and B is defined as follows: $$\eqalign{ & {G_x} = \root n \of {{a_1}{a_2}...{a_n}} \cr & {G_y} = \root m \of {{b_1}{b_2}...{b_m}} \cr} $$ $${z_i} = the{\text{ }}elements{\text{ }}of{\text{ }}combined{\text{ }}groups$$ So $$\eqalign{ & {G_z} = \root {n + m} \of {{G_x}^n \times {G_y}^m} \cr & \ln {G_z} = \frac{1}{{n + m}}\left( {n\ln {G_x} + m\ln {G_y}} \right) \cr} $$

Standard deviation of combined series is defined as follows; $${\sigma _z} = \exp \left( {\sqrt {\frac{{\sum\limits_p {{{\left( {\ln {z_i} - \ln {G_z}} \right)}^2}} }}{p}} } \right)$$ Where $$p = n + m$$ $${z_i} = {\text{are }}{a_i}{\text{ or }}{{\text{b}}_j}$$ As a result, the ${\sigma _z}$ can be written as: $${\sigma _z} = \exp \left( {\sqrt {\frac{{\sum\limits_n {{{\left( {\ln {a_i} - \ln {G_z}} \right)}^2} + \sum\limits_m {{{\left( {\ln {b_j} - \ln {G_z}} \right)}^2}} } }}{p}} } \right)$$

$${\sigma _z} = \exp \left( {\sqrt {\frac{{\sum\limits_n {\ln {a_i}^2 - 2\ln {G_z}\sum\limits_n {\ln {a_i} + \sum\limits_n {\ln {G_z}^2 + \sum\limits_m {\ln {b_i}^2 - 2\ln {G_z}\sum\limits_m {\ln {b_i} + \sum\limits_m {\ln {G_z}^2} } } } } } }}{p}} } \right)$$ Where $$\eqalign{ & \ln {\sigma _x} = \sqrt {\frac{{\sum\limits_n {{{\left( {\ln {a_i} - \ln {G_x}} \right)}^2}} }}{n}} \cr & \ln {\sigma _x} = \sqrt {\frac{{\sum\limits_n {\ln {a_i}^2 - 2\ln {G_x}\sum\limits_n {\ln {a_i} + \sum\limits_n {\ln {G_x}^2} } } }}{n}} \cr} $$ With $n\ln {G_x} = \sum\limits_n {\ln {a_i}} $ So $$\eqalign{ & \ln {\sigma _x} = \sqrt {\frac{{\sum\limits_n {\ln {a_i}^2 - 2n\ln {G_x}^2 + n\ln {G_x}^2} }}{n}} \cr & \sum\limits_n {\ln {a_i}^2} = n\left( {{{\ln }^2}{\sigma _x} + {{\ln }^2}{G_x}} \right) \cr} $$ Similarly $$\sum\limits_m {\ln {b_j}^2} = m\left( {{{\ln }^2}{\sigma _y} + {{\ln }^2}{G_y}} \right)$$ Therefore $${\sigma _z} = \exp \left( {\sqrt {\frac{{\sum\limits_n {\ln {a_i}^2 - 2\ln {G_z}\sum\limits_n {\ln {a_i} + \sum\limits_n {\ln {G_z}^2 + \sum\limits_m {\ln {b_i}^2 - 2\ln {G_z}\sum\limits_m {\ln {b_i} + \sum\limits_m {\ln {G_z}^2} } } } } } }}{p}} } \right)$$ $${\sigma _z} = \exp \left( {\sqrt {\frac{{n\left( {{{\ln }^2}{\sigma _x} + {{\ln }^2}{G_x}} \right) - 2n\ln {G_z}\ln {G_x} + n\ln {G_z}^2 + m\left( {{{\ln }^2}{\sigma _y} + {{\ln }^2}{G_y}} \right) - 2m\ln {G_z}\ln {G_y} + m\ln {G_z}^2}}{p}} } \right)$$ $$\ln {\sigma _z} = \sqrt {\frac{{n\left( {{{\ln }^2}{\sigma _x} + {{\ln }^2}{G_x}} \right) - 2n\ln {G_z}\ln {G_x} + n\ln {G_z}^2 + m\left( {{{\ln }^2}{\sigma _y} + {{\ln }^2}{G_y}} \right) - 2m\ln {G_z}\ln {G_y} + m\ln {G_z}^2}}{p}} $$


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