I am running a regression that includes consumption and disposable income.

It was suggested that I take mean of the logarithm of my data rather than logarithm of the mean. What is the difference?

• Presumably in one case you takes logs and then the mean (which is the geometric mean when anti-logged) and in the other you reverse the ordering of operations. They are not the same, as indeed you hint. – mdewey Dec 7 '16 at 15:38

There is a potential confusion in terminology here, as this question, for example, seems to take "log-mean" to be the mean of the logs.

Putting aside that confusion, here's a simple example. Say you have 3 measurements with values of 1, 10, and 100.

Their mean value is $\frac{111}{3}$=37. The base 10 logarithm of 37 is 1.57, which is the log of their mean value in the original scale.

The base 10 logarithms of the original data are 0, 1, and 2; the mean of the logarithms is 1, corresponding to a value of 10 in the original scale.

If a log transformation of the data is appropriate then you should typically do the transformation on the original data first, whatever you call that process.

• This brings to mind Jensen's inequality which applies to convex or concave functions. The logarithm falls into this class.. So the mean of the log does not equal the log of the mean. If I recall correctly equality holds only under some degenerate condition (x is a constant). – Michael R. Chernick Dec 7 '16 at 21:47

Consumption and disposable income are often though of as exponential growth processes, something like $x_t=x_0e^{rt}$, where $r$ - rate of growth. If you take a log, you get $\ln x_t=\ln x_0+rt$, and the function becomes linear. So we linearize the model by taking a log. That's the motivation.

In reality there's always a stochastic noise in the data. To see how it's handled let's first write the above equation in a difference form: $\Delta \ln x_t=\ln x_t -\ln x_{t-1}=r$. One way to make this process stochastic is to add noise to the rate of change as follows: $$\Delta \ln x_t=r+\varepsilon_t\\\varepsilon_t\sim\mathcal N(0,\sigma^2)$$

Suppose that we agree with this process. We want to estimate the parameters of the process. We have the following for the rate of growth: $$E[\Delta \ln x_t]=r$$ which leads to an obvious estimator $$\hat r=\frac 1 n\sum_{t=1}^n\Delta \ln x_t$$

So, for the random walk with a drift process like above the mean of the logarithm makes a sense in the estimator of the growth rate.