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.
- "the confidence interval gives you an interval such that the probability that the mean is in this interval is 95%"
This is not quite true. Indeed, in this framework, the unknown mean is fixed (either it lies inside the interval or it lies outside, there is no probability attached to it). Instead, you should say that the probability that the interval (which is random) covers the true mean equals 95%. Or that there is a 5% chance that the interval does not cover the true mean.
- "Ok, but the confidence interval with using the approximate technique above gives (45.7467, 48.617). Yet for λ=47.18 I get P(X≥49)=0.41469. That I don't get." (cf. comment below question)
$(45.7467, 48.617)$ is meant to provide information concerning the mean of the Poisson distribution, i.e. $\lambda$. Not about an observation from the Poisson distribution.
Perhaps you are interested in a prediction interval...
- my answer to your comment below
$\Pr(X > 330)$ is the probability that the first Poisson variable (with mean 326) returns a count larger than the mean of the second Poisson variable. It does not answer your question (which I can rephrase: are the two means equivalent?).
To convince yourself, you can compute Pr(Pois(326) > 326), i.e. just as if the second Poisson has the same mean as the first ("equivalence").
What you might want to do instead is to construct a (90%) confidence interval for the mean difference and check that it lies within some equivalence margins. This is called an equivalence test (for means) and this approach is known as TOST.
Best Answer
The standard procedure (Hahn & Meeker, section 7.2.2) exploits the basic relationship between Poisson and Chi-squared variates; namely, when $F_{\lambda}$ is the Poisson PDF of parameter $\lambda$ and $G_{\nu}$ is the Chi-squared PDF of parameter $\nu,$ then for any $k\in\{0,1,2,\ldots\},$
$$1-F_\lambda(k) = G_{2k+2}(2\lambda).\tag{1}$$
An upper confidence limit of size $1-\alpha$ for $\lambda$ based on observing a Poisson variable $K_\lambda$ is, by definition, a function $u$ for which
$$1-\alpha = \inf_{\lambda\in\mathbb{R}^+}\Pr(\lambda \le u(K_\lambda)).$$
If we choose a suitable inverse of $u$ and write $k=K_\lambda$ for the observed value, we may exploit $(1)$ to re-express this criterion as
$$\eqalign{ 1-\alpha &= \inf_{\lambda\in\mathbb{R}^+}\Pr(u^{-1}(\lambda) \le K_\lambda) \\ &= \inf_{\lambda\in\mathbb{R}^+}1-F_\lambda(k)) \\ &= \inf_{\lambda\in\mathbb{R}^+}G_{2k+2}(2\lambda), }$$
with unique solution
Similar reasoning arrives at a lower $1-\alpha$ confidence limit
One of the many possible two-sided confidence interval procedures splits the risk between the upper and lower endpoints by using $[\lambda_{-}(\alpha/2), \lambda_{+}(\alpha/2)].$
When $k=0,$ the function $G_{0},$ or the distribution of a "chi-squared variate with zero degrees of freedom," has to be understood as the distribution of the constant zero, whence "$G^{-1}_0(\alpha)$" is always zero no matter what $\alpha\gt 0$ may be. In this case $G_{2k+2} = G_2$ is the Exponential distribution with scale factor $2,$ entailing
$$\lambda_{+}(\alpha/2) = G^{-1}_2(1-\alpha/2) = -2\log(\alpha/2).$$
For instance, with $\alpha=5\%$ this UCL is $7.38,$ whereas the one-sided upper confidence limit for the same $\alpha$ is only $3.00.$ If you are tempted to use the latter because it produces a shorter confidence interval, consider these simulation results for a large range of $\lambda$ (from $0.1$ to $1,000,$ after which a Normal approximation will work well):
"Coverage" is the proportion of samples for which the confidence interval, nominally set at $1-\alpha = 95\%,$ includes $\lambda.$ Each red point in this plot summarizes 400,000 independently simulated samples. The gray graph is the calculated coverage based on Poisson probabilities only.
The discreteness of the Poisson distributions causes the actual coverage to oscillate, but a trend is clear: coverage really is close to the nominal value for large $\lambda,$ but can be substantially greater for small $\lambda.$
Some of the conclusions we may draw are
The foregoing analysis produces confidence intervals with the correct coverage.
The coverage tends to be higher than intended (greater than $1-\alpha$) when $\lambda$ is smaller than $10$ or so, approaching $100\%$ in the limit as $\lambda\to 0.$
In retrospect this behavior is obvious: because the confidence limits depend only on $k,$ the limits for $k=0$ have to be fairly large to allow for the possibility that $\lambda$ is fairly large. Consequently, when $\lambda$ actually is small, the coverage must be greater than the nominal coverage.
If you know (or assume) $\lambda$ is small at the outset, you could modify this procedure accordingly to produce confidence intervals that tend to be shorter.
Reference
G. J. Hahn and W. Q. Meeker (1991), Statistical Intervals. A Guide for Practitioners. J. Wiley & Sons.
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