Such decisions are not normally made on the basis of testing, but on an understanding of the variables, the circumstances and the needs of the analysis.
For example, we would need to consider when it is more meaningful to us to use a geometric mean and when it is more meaningful to use an arithmetic mean.
If I am interested in a population mean, generally speaking I probably want to consider a sample arithmetic mean. However, if I have particular distributional assumptions, it's possible a geometric mean may (perhaps after some transformation, or in combination with other quantities) lead me to a better estimate of the population mean than the arithmetic mean (e.g. in a lognormal distribution, the geometric mean might be part of the calculation of a better estimate of the population mean than the arithmetic mean).
You seem to be suggesting that we can infer the best estimator from the data, as if we perform two stages of estimation, first of the distribution, and then from that choose a good estimator.
But in fact it's not at all clear that this is generally a good approach to estimation, except perhaps in certain classes of estimation problems, with particular structure (this is essentially a form of adaptive estimation). If you want to pursue that approach you'll likely need to narrow the scope of the problem and do some kind of study (either by looking at asymptotic properties or by use of simulation) of the properties of some proposed adaptive estimator.
You can compute the arithmetic mean of the log growth rate:
- Let $V_t$ be the value of your portfolio at time $t$
- Let $R_t = \frac{V_t}{V_{t-1}}$ be the growth rate of your portfolio from $t-1$ to $t$
The basic idea is to take logs and do your standard stuff. Taking logs transforms multiplication into a sum.
- Let $r_t = \log R_t$ be the log growth rate.
$$\bar{r} = \frac{1}{T} \sum_{t=1}^T r_t \quad \quad s_r = \sqrt{\frac{1}{T-1} \sum_{t=1}^T \left( r_t - \bar{r}\right)^2}$$
Then your standard error $\mathit{SE}_{\bar{r}}$ for your sample mean $\bar{r}$ is given by:
$$ \mathit{SE}_{\bar{r}} = \frac{s_r}{\sqrt{T}}$$
The 95 percent confidence interval for $\mu_r =
{\operatorname{E}[r_t]}$ would be approximately: $$\left( \bar{r} - 2 \mathit{SE}_{\bar{r}} , \bar{r} + 2 \mathit{SE}_{\bar{r}} \right)$$.
Exponentiate to get confidence interval for $e^{\mu_r}$
Since $e^x$ is a strictly increasing function, a 95 percent confidence interval for $e^{\mu_r}$ would be:
$$\left( e^{\bar{r} - 2 \mathit{SE}_{\bar{r}}} , e^{\bar{r} + 2 \mathit{SE}_{\bar{r}}} \right)$$
And we're done. Why are we done?
Observe $\bar{r} = \frac{1}{T} \sum_t r_t$ is the log of the geometric mean
Hence $e^{\bar{r}}$ is geometric mean of your sample. To show this, observe the geometric mean is given by:
$$ \mathit{GM} = \left(R_1R_2\ldots R_T\right)^\frac{1}{T}$$
Hence if we take the log of both sides:
\begin{align*} \log \mathit{GM} &= \frac{1}{T} \sum_{t=1}^T \log R_t \\
&= \bar{r}
\end{align*}
Some example to build intuition:
- Let's say you compute the mean log growth rate is $.02$. Then the geometric mean is $\exp(.02) \approx 1.0202$.
- Let's say you compute the mean log growth rate is $-.05$, then the geometric mean is $\exp(-.05) = .9512$
For $x \approx 1$, we have $\log(x) \approx x - 1$ and for $y \approx 0$, we have $\exp(y) \approx y + 1$. Further away though, those tricks breka down:
- Let's say you compute the mean log growth rate is $.69$, then the geometric mean mean is $\exp(.69) \approx 2$ (i.e. the value doubles every period).
If all your log growth rates $r_t$ are near zero (or equivalently $\frac{V_t}{V_{t-1}}$ is near 1, then you'll find that the geometric mean and the arithmetic mean will be quite close
Another answer that might be useful:
As this answer discusses, log differences are basically percent changes.
Comment: it's useful in finance to get comfortable thinking in logs. It's similar to thinking in terms of percent changes but mathematically cleaner.
Best Answer
It depends on what you want to use the average for.
A CPI of $x$ means it takes $x$ times as much money now to buy a basket of goods as it did during a reference period. (Sometimes values are multiplied by an arbitrary constant, such as $100$, so that a CPI of $x$ means it takes $x/100$ times as much money as before. The value of that constant does not matter: it merely establishes a unit of measurement of money. It will therefore be set to $1$ in the ensuing calculations.)
Suppose you want to measure how much people typically needed to spend during the year in toto. You would likely assume they would buy this same basket at regular intervals (rather than varying it or varying the rates at which they purchased it), so without loss of generality let's suppose they bought it once a month, spending $x_i$ during month $i$. Then they spent a total of $$S(x)=\sum_{i\in \{\text{January, February,}\ldots\text{, December}\}} x_i,$$ whence their mean monthly price $S(x)/12$ was the arithmetic mean of the monthly prices.
If you have data about rates of purchase you could instead compute a rate-weighted arithmetic mean.
Suppose you want to measure how much people with a fixed monthly income were able to purchase on average. This time assume they purchased as much as they could at regular intervals. During month $i$ they could purchase only $1/x_i$ as much as they could during the reference period. Now the total amount of goods purchased is $$H(x) = \sum_{i\in \{\text{January, February,}\ldots\text{, December}\}} \frac{1}{x_i}.$$ In this case the reciprocal of its mean--that is, the harmonic mean $12/H(x)$ of the monthly CPIs--is the right average to use.
Once again it would be possible to weight this harmonic mean by rates of purchase, if such information were available.
In any event, the harmonic mean cannot exceed the arithmetic mean (weighted or not, only provided both use the same weights). The two will be close when the ratio of the largest $x_i$ to the smallest $x_i$ is small--say, well under $2:1$. They will be identical when all the $x_i$ are equal.
The Wikipedia quotation does not apply because it refers to "normalized" values, or ratios, whose denominators could be varying or chosen arbitrarily. That is not the case with an index: the denominator is fixed during the (past) reference period. Its effect is only to establish the units in which money is expressed.