**Comment** on lognormal and normal distributions. The variance of data tends to decrease when you take logs of the values. Perhaps the most common case is that, if $X_1$ is lognormal, then $X_2 = \ln(X_1)$ is normal, which has smaller variance than $X_.1$ Also, if it exists, $X_3 = \ln(X_2)$ may have a still smaller variance. Below we begin with a random sample (using R) of $n = 10^4$ observations from a lognormal distribution with parameters $\mu = 50, \sigma = 2.$ (It is customary to use lognormal parameters that match parameters of the related normal distribution. You can look at [Wikipedia](https://en.wikipedia.org/wiki/Log-normal_distribution) on 'lognormal distributions' for details.) We show means and standard deviations for distributions of $X_1, X_2,$ and $X_3.$ set.seed(720); n = 10^4 x2 = rnorm(n, 50, 2); x1 = exp(x2); x3 = log(x2) mean(x1); mean(x2); mean(x3) [1] 3.686093e+22 [1] 50.01289 # aprx E(X2) = 50 [1] 3.911481 sd(x1); sd(x2); sd(x3) [1] 2.308712e+23 [1] 1.997261 # aprx SD(X2) = 2 [1] 0.04004551 Then we show histograms of the three samples. At the left, notice that it is difficult to make an informative histogram of $X_1$ because it is so severely skewed to the right. In the center panel, we overlay the density function of $\mathsf{Norm}(\mu =50, \sigma=2);$ which is symmetrical. At the right, notice that taking (natural) logs once again results in a slightly left-skewed distribution. [![enter image description here][1]][1] _Notes:_ (1) The the support of a lognormal distribution is $(0, \infty).$ A normal distribution may take negative values. If the lognormal distribution is truncated to $(1, \infty)$ so that the normal distribution is truncated to $(0,\infty),$ then the natural log of that "normal" distribution exists. The distribution $\mathsf{Norm}(50, 2)$ has almost no probability below $0,$ so the truncation would have little practical effect in this example. (2) R code for the figure above: par(mfrow=c(1,3)) hist(x1, prob=T, br=50, col="skyblue2") hist(x2, prob=T, col="skyblue2") curve(dnorm(x,50,2), add=T, col="red") hist(x3, prob=T, col="skyblue2") par(mfrow=c(1,1)) (3) However, it is _not always true_ that taking logs gives a smaller variance. If $X_2 \sim \mathsf{Unif}(0,1),\, X_1 = e^{X_2},$ and $X_3 = \ln(X_2),$ then R code similar to the code for the lognormal example gives the following results: set.seed(720); n = 10^5 x2 = runif(n); x1 = exp(x2); x3 = log(x2) var(x1); var(x2); var(x3) [1] 0.2411124 [1] 0.08316279 # aprx V(X2) = 1/12 [1] 1.01091 [![enter image description here][3]][3] [1]: https://i.sstatic.net/ICW5Y.png [2]: https://i.sstatic.net/h4Wm4.png [3]: https://i.sstatic.net/Trosu.png