abbracciopj

2022-06-20

Interpret Logarithmic Values like Mean and Std

let's say i have a series of data containing prices on the log scale. How would i interpret a arithmetic mean of 0.55 and a std of 0.69 (both metrics are computed with the log prices).

Is there an intuitive explanation in terms of percentage change or anything similar?

let's say i have a series of data containing prices on the log scale. How would i interpret a arithmetic mean of 0.55 and a std of 0.69 (both metrics are computed with the log prices).

Is there an intuitive explanation in terms of percentage change or anything similar?

timmeraared

Beginner2022-06-21Added 22 answers

Let ${x}_{i}$ your observations. Then the (observed) arithmetic mean ${\hat{\mu}}_{\mathrm{log}}$ of the logarithm of those is

$\frac{\sum _{i=1}^{n}\mathrm{log}({x}_{i})}{n}=\frac{\mathrm{log}\left(\prod _{i=1}^{n}{x}_{i}\right)}{n}=\mathrm{log}\left(\prod _{i=1}^{n}{x}_{i}^{1/n}\right)$

which is the logarithm of the (observed) geometric mean of the original values (i.e. $\mathrm{exp}({\hat{\mu}}_{\mathrm{log}})$ is the geometric mean of the ${x}_{i}$).

The (observed) standard deviation of these values is

${\hat{\sigma}}_{\mathrm{log}}=\sqrt{\frac{\sum _{i=1}^{n}(\mathrm{log}({x}_{i})-{\hat{\mu}}_{\mathrm{log}}{)}^{2}}{n-1}}$

but the ${}^{2}$ makes it a bit more difficult to transform the way we transformed ${\hat{\mu}}_{\mathrm{log}}$ above. However, $\mathrm{exp}({\hat{\sigma}}_{\mathrm{log}})$ does have an interpretation as a measure of deviation from the (geometric) mean, not in terms of difference but in terms of ratio.

When you have a lot of observations where exactly half of them are −1 and exactly half of them are 1, then the arithmetic mean is 0, and the standard deviation is a little larger than 1 (it tends to 1 as the number of observations grows), which is the observed deviation from the arithmetic mean in all cases.

In exactly the same way, for a long list of observations where exactly half are $\frac{1}{2}$ and exactly half are 2, the geometric mean $\mathrm{exp}({\hat{\mu}}_{\mathrm{log}})$ is 1, and the "geometric standard deviation", $\mathrm{exp}({\hat{\sigma}}_{\mathrm{log}})$ is a little larger than 2 (and tends to 2 as the number of observations grows), since all the observations deviate from the geometric mean by a factor of 2.

In your case, this means that you have observed a geometric mean of $\mathrm{exp}(0.55)\approx 1.73$, and a standard multiplicative deviation from that mean of $\mathrm{exp}(0.69)\approx 1.99$. So what would usually be an interval of "mean plus or minus a standard deviation" now becomes

$[1.73/1.99,1.73\cdot 1.99]=[0.87,3.46]$

$\frac{\sum _{i=1}^{n}\mathrm{log}({x}_{i})}{n}=\frac{\mathrm{log}\left(\prod _{i=1}^{n}{x}_{i}\right)}{n}=\mathrm{log}\left(\prod _{i=1}^{n}{x}_{i}^{1/n}\right)$

which is the logarithm of the (observed) geometric mean of the original values (i.e. $\mathrm{exp}({\hat{\mu}}_{\mathrm{log}})$ is the geometric mean of the ${x}_{i}$).

The (observed) standard deviation of these values is

${\hat{\sigma}}_{\mathrm{log}}=\sqrt{\frac{\sum _{i=1}^{n}(\mathrm{log}({x}_{i})-{\hat{\mu}}_{\mathrm{log}}{)}^{2}}{n-1}}$

but the ${}^{2}$ makes it a bit more difficult to transform the way we transformed ${\hat{\mu}}_{\mathrm{log}}$ above. However, $\mathrm{exp}({\hat{\sigma}}_{\mathrm{log}})$ does have an interpretation as a measure of deviation from the (geometric) mean, not in terms of difference but in terms of ratio.

When you have a lot of observations where exactly half of them are −1 and exactly half of them are 1, then the arithmetic mean is 0, and the standard deviation is a little larger than 1 (it tends to 1 as the number of observations grows), which is the observed deviation from the arithmetic mean in all cases.

In exactly the same way, for a long list of observations where exactly half are $\frac{1}{2}$ and exactly half are 2, the geometric mean $\mathrm{exp}({\hat{\mu}}_{\mathrm{log}})$ is 1, and the "geometric standard deviation", $\mathrm{exp}({\hat{\sigma}}_{\mathrm{log}})$ is a little larger than 2 (and tends to 2 as the number of observations grows), since all the observations deviate from the geometric mean by a factor of 2.

In your case, this means that you have observed a geometric mean of $\mathrm{exp}(0.55)\approx 1.73$, and a standard multiplicative deviation from that mean of $\mathrm{exp}(0.69)\approx 1.99$. So what would usually be an interval of "mean plus or minus a standard deviation" now becomes

$[1.73/1.99,1.73\cdot 1.99]=[0.87,3.46]$