Ideally, I would like to know the fraction of the mean of $x$ over the sum of the mean of $x$ and the mean of $y$:

$$\text{Fraction of interest} = \frac{\bar{x}}{\bar{x}+\bar{y}}$$

where $$\bar{x} = \frac{\sum_{i=1}^I x_i }{I}$$.

However, I have the fraction of the geometric mean of $x$ over the sum of the geometric mean of $x$ and the geometric mean of $y$:

$$\text{Fraction I have} = \frac{\prod_{i=1}^I{x_i}}{\prod_{i=1}^I{x_i}+\prod_{i=1}^I{y_i}}$$.

Is there a good interpretation for the fraction I have? I lose intuition when it comes to geometric means. Alternatively, is there a way to connect the two fractions?

Note (thanks to comments notice that): $x_i, y_i, i=1,\ldots,I$ always $>0$.

To further add to this, $x_i$ and $y_i$ are observations from an experiment. It is assumed that $log(x_i)$ and $log(y_i)$ follow a normal distribution. I'm interested in the difference between $x_i$ and $y_i$, across replicates (i.e., I have $R$ experiments where in each I measure $x_i$'s and $y_i$'s) hence it is convenient for me to model $log(x_i/y_i)$ as a normally distributed random variable (with mean and variance as the sample mean and sample variance of $log(x_i/y_i)$ over $I$). I then use a regression model (without going into too much details) where the response in every experiment $r$ of $R$ is the mean $log(x_i/y_i)$ over $I$ (roughly, $\sum_{i=1}^I log(x_i/y_i)/I = \alpha + \beta X$).

Having estimated $\alpha$ (i.e., the mean response) I would like to express $\frac{\bar{x}}{\bar{x}+\bar{y}}$ as a function of $\hat\alpha$.

So what I have is: $\hat\alpha = \sum_{i=1}^I log(x_i/y_i)/I = log(\frac{\prod_{i=1}^I{x_i}}{\prod_{i=1}^I{y_i}})$

Therefore, $e^{I\alpha} = \frac{\prod_{i=1}^I{x_i}}{\prod_{i=1}^I{y_i}}$

Adding 1 to both sides and skipping a few steps gets me that:

$\frac{e^{I\alpha}}{1 + e^{I\alpha}} = \frac{\prod_{i=1}^I{x_i}}{\prod_{i=1}^I{x_i}+\prod_{i=1}^I{y_i}}$

So basically I'm trying to either get an intuition of how $\frac{\prod_{i=1}^I{x_i}}{\prod_{i=1}^I{x_i}+\prod_{i=1}^I{y_i}}$ relates to $\frac{\bar{x}}{\bar{x}+\bar{y}}$ or some function to connect the two.

  • $\begingroup$ can we assume that $\prod_i^I y_i>\prod_i^I x_i$? $\endgroup$
    – user603
    Mar 23, 2014 at 22:04
  • $\begingroup$ Generally not. Are you thinking about assuming that $\prod_{i=1}^I{x_i}$ is negligible relative to $\prod_{i=1}^I{y_i}$ ? $\endgroup$ Mar 23, 2014 at 22:09
  • $\begingroup$ no not necearly negligeable, just smaller. And by the way, it'll be interesting to know if $y_i>0$ and $x_i>0$ always $\endgroup$
    – user603
    Mar 23, 2014 at 22:12

1 Answer 1


There's not really enough information to exploit the likely shortcuts that tend to crop up in particular applications, so if you have more details, please give them. For example, depending on what this relates to, information about the distribution of the terms or dependence between the x's and y's may be relevant.

One trick that is sometimes useful is to divide through by the numerator on one or both of those formulas. So that leaves you (in either case) with something of the form



$$R=\frac{\bar y}{\bar x}$$

or equivalently (since the count appears to be the same for both)

$$R=\frac{\sum y_i}{\sum x_i}$$

in the first case, and


where $\stackrel{\sim}{x}$ is the geometric mean, or equivalently

$$R=\frac{\prod y_i}{\prod x_i}\,$$ .

If any of those forms are convenient for your problem they may make life easier.

Now the relationship between arithmetic and geometric means tends to come up in a number of applications. I'll assume we're dealing with non-negative variables. The geometric mean is smaller (and generally, also less variable) than the arithmetic mean, unless all the values are equal.

It may often be the case that the ratio, $R$ will be less variable as well.

If you have a particular situation in mind you might consider using simulation to investigate the relationship between the two more closely. In particular, if one of the two forms is convenient to work with, often the other will be inconvenient. So consider, for example, $x_i$'s and $y_j$'s all independent and lognormally distributed. Then the geometric means are reasonably nice, but arithmetic means not so nice to manipulate algebraically. Simulation is good for that kind of circumstance.

With the geometric $R$, you might find it convenient to work on the log-scale.

Finally, one might expand one or the other term (perhaps working with $R$ for the arithmetic version or log-$R$ for the geometric) in a Taylor-series in terms of the other one, in order to investigate their relationship.


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