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
This is not necessarily about arithmetic or geometric mean. This is also about simple or continuous return. Consider this:
$100(1+0.1)(1-0.1)=100(1-0.01)=99$
$100 e^{0.1}e^{-0.1}=100$
In the first case I assumed that your 10% return is a simple return. In the second case I assumed it's continuous return. Both are used a lot in finance in different situations.
Going back to the arithmetic and geometric returns, the rule of thumb is that if you're using the return for a single period forecast, then you use arithmetic return. If you're using it for multi period return, i.e. with compounding, use geometric return. This is not an absolute rule like NEwton's law of mechanics, of course