50

Although it may appear that the mean of the log-transformed variables is preferable (since this is how log-normal is typically parameterised), from a practical point of view, the log of the mean is typically much more useful. This is particularly true when your model is not exactly correct, and to quote George Box: "All models are wrong, some are useful" ...


45

The (right) tail of a distribution describes its behavior at large values. The correct object to study is not its density--which in many practical cases does not exist--but rather its distribution function $F$. More specifically, because $F$ must rise asymptotically to $1$ for large arguments $x$ (by the Law of Total Probability), we are interested in how ...


31

The gamma and the lognormal are both right skew, constant-coefficient-of-variation distributions on $(0,\infty)$, and they're often the basis of "competing" models for particular kinds of phenomena. There are various ways to define the heaviness of a tail, but in this case I think all the usual ones show that the lognormal is heavier. (What the first person ...


30

As for qualitative differences, the lognormal and gamma are, as you say, quite similar. Indeed, in practice they're often used to model the same phenomena (some people will use a gamma where others use a lognormal). They are both, for example, constant-coefficient-of-variation models (the CV for the lognormal is $\sqrt{e^{\sigma^2} -1}$, for the gamma it's ...


28

Some points to start with: i) these distributional conventions are at best approximations. They can be convenient models, but we shouldn't confuse that with the actual distribution of stock prices or returns. ii) stock prices are typically increasing (but in any case, have changing mean; the mean isn't stable). So when we're talking about the distribution ...


26

Recall that $e^x\geq 1+x$ $E\left[e^{Y}\right]=e^{ E(Y)} E\left[e^{Y- E(Y)}\right]\geq e^{E(Y)} E\left[1+{Y- E(Y)}\right] = e^{E(Y)}$ So $e^{E(Y)}\leq E\left[e^{Y}\right] $ Now letting $Y=\ln X$, we have: $e^{E(\ln X)}\leq E\left[e^{\ln X}\right]=E(X)$ now take logs of both sides $E[\ln (X)]\leq\ln[E(X)]$ Alternatively: $\ln X = \ln X - \ln \mu+\ln\...


23

If we consider "approximation" in a fairly general sense we can get somewhere. We have to assume not that we have an actual normal distribution but something that's approximately normal except the density cannot be nonzero in a neighborhood of 0. So let's say that $a$ is "approximately normal" (and concentrated near the mean*) in a sense that we can ...


22

Let me answer in reverse order: 2. Yes. If their MGFs exist, they'll be the same*. see here and here for example Indeed it follows from the result you give in the post this comes from; if the MGF uniquely** determines the distribution, and two distributions have MGFs and they have the same distribution, they must have the same MGF (otherwise you'd have a ...


22

There is something puzzling in those results since the first method provides an unbiased estimator of $\mathbb{E}[X^2]$, namely$$\frac{1}{N}\sum_{i=1}^N X_i^2$$has $\mathbb{E}[X^2]$ as its mean. Hence the blue dots should be around the expected value (orange curve); the second method provides a biased estimator of $\mathbb{E}[X^2]$, namely$$\mathbb{E}[\exp(...


21

The underlying model is $$E[\log Y] = \beta_0 + \beta_1 x_1 + \cdots + \beta_k x_k$$ or, in terms of error terms $\varepsilon_i,$ $$\log Y_i = \beta_0 + \beta_1 x_{1i} + \cdots + \beta_k x_{ki} + \varepsilon_i.\tag{*}$$ When we assume the conditional distribution of $\log Y$ is Normal, then the Ordinary Least Squares (OLS) estimate of $\log Y$ also is ...


20

This approximate lognormality of sums of lognormals is a well-known rule of thumb; it's mentioned in numerous papers -- and in a number of posts on site. A lognormal approximation for a sum of lognormals by matching the first two moments is sometimes called a Fenton-Wilkinson approximation. You may find this document by Dufresne useful (available here, or ...


18

Here are my two cents from an advanced data analysis course I took while studying biostatistics (although I don't have any references other than my professor's notes): It boils down to whether or not you need to address linearity and heteroscedasticity (unequal variances) in your data, or just linearity. She notes that transforming the data affects both ...


17

Consider flipping your questions around. Begin with uncorrelated data - I generated this data randomly, so these variables are independent; my y is normal and my x is log(1+X1) where X1 is a mixture of several geometric distributions chosen to give a roughly similar appearance to your plot: The y variable is symmetric and the x-variable is mildly skew, but ...


17

It's been a long time since I took a physics class, so let me know if any of this is incorrect. General description of moments with physical analogs Take a random variable, $X$. The $n$-th moment of $X$ around $c$ is: $$m_n(c)=E[(X-c)^n]$$ This corresponds exactly to the physical sense of a moment. Imagine $X$ as a collection of points along the real line ...


15

This is a difficult problem. I thought first about using (some approximation of) the moment generating function of the lognormal distribution. That doesn't work, as I will explain. But first some notation: Let $\phi$ be the standard normal density and $\Phi$ the corresponding cumulative distribution function. We will only analyze the case lognormal ...


14

The answer to your question is (essentially) no and your argument has the right idea. Below, we formalize it a bit. (For an explanation of the caveat above, see @whuber's comment below.) If $X$ has a lognormal distribution this means that $\log(X)$ has a normal distribution. Another way of saying this is that $X = e^{Z}$ where $Z$ has a $N(\mu, \sigma^2)$ ...


14

By the Delta method, the variance of a function of an RV is approximately equal to the variance of the RV times the squared derivative evaluated at the mean. Hence $$\mathrm{var}(\log(X)) \approx \frac{1}{ \left[E(X)\right] ^2 } \mathrm{var}(X)$$ and there you have it. Your derivation was right of course.


13

I thought I'd throw up some figs showing that both user29918 and Xi'an's plots are consistent. Fig 1 plots what user29918 did, and Fig 2 (based on same data), does what Xi'an did for his plot. Same result, different presentation. What's happening is that as T increases, the variances becomes huge and the estimator $\frac{1}{n} \sum_i x_i^2$ becomes like ...


13

You can generate a sample by first generating a normally distributed value, then take the exponent of that, then use that as the parameter in a Poisson distribution and take a sample from that distribution. The resulting samples of this three-step process will be Poisson-Lognormally distributed.


12

I think it will be helpful to separate the question into two parts: What is the functional form of your empirical distribution? and What does that functional form imply about the generating process in your network? The first question is a statistics question. If you've applied the methods of Clauset et al. for fitting the power-law distribution and those ...


12

These data have a short tail compared to a lognormal distribution, not unlike a Gamma distribution: set.seed(17) par(mfcol=c(1,1)) x <- rgamma(500, 1.9) qqnorm(log(x), pch=20, cex=.8, asp=1) abline(mean(log(x)) + .1,1.2*sd(log(x)), col="Gray", lwd=2) Nevertheless, because the data are strongly right-skewed, we can expect the largest values to play an ...


12

It seems that you "know" or otherwise assume that you have two quantiles; say you have that 42 and 666 are the 10% and 90% points for a lognormal. The key is that almost everything is easier to do and understand on the logged (normal) scale; exponentiate as little and as late as possible. I take as examples quantiles that are symmetrically placed on the ...


12

Let m and s be the mean and sd of $X$ on the original scale. The appropriate mean and sd on the log scale can be found after a little algebra to be $E(\log(X)) = \log(m) - \frac{1}{2} \log [ (s/m)^2 +1]$ $sd(\log(X)) = \sqrt{\log [(s/m)^2 +1]}$


12

The two estimators you are comparing are the method of moments estimator (1.) and the MLE (2.), see here. Both are consistent (so for large $N$, they are in a certain sense likely to be close to the true value $\exp[\mu+1/2\sigma^2]$). For the MM estimator, this is a direct consequence of the Law of large numbers, which says that $\bar X\to_pE(X_i)$. For ...


11

The Bayesian approach to your problem would be to consider the posterior probability over models $M \in \{ \text{Normal}, \text{Log-normal} \}$ given a set of data points $X = \{ x_1, ..., x_N \}$, $$P(M \mid X) \propto P(X \mid M) P(M).$$ The difficult part is getting the marginal likelihood, $$P(X \mid M) = \int P(X \mid \theta, M) P(\theta \mid M) \, d\...


11

Yes, the gamma distribution is the maximum entropy distribution for which the mean $E(X)$ and mean-log $E(\log X)$ are fixed. As with all exponential family distributions, it is the unique maximum entropy distribution for a fixed expected sufficient statistic. To answer your question about physical processes that generate these distributions: The ...


11

If the logs of the data are really drawn from normally distributed populations with constant variance (but possibly different means), then the original data must have come from lognormal distributions with possibly different scales (due to differences in $\mu$, where adding something to the means on the log-scale has a multiplying effect on the original ...


10

We can use an entirely analogous technique to the one typically used to calculate the moments of a lognormal. In particular, note that if $\newcommand{\E}{\mathbb E}X \sim \mathrm{Bin}(n,p)$ and $Y = e^X$, then $Y^k = e^{k X}$. But, $\E e^{kX} = m_X(k)$ where $m_X(t)$ is the moment-generating function of $X$ evaluated at $t$. Hence, $$ \E e^{k X} = m_X(k)...


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