# The sum of independent lognormal random variables appears lognormal?

I'm trying to understand why the sum of two (or more) lognormal random variables approaches a lognormal distribution as you increase the number of observations. I've looked online and not found any results concerning this.

Clearly if $X$ and $Y$ are independent lognormal variables, then by properties of exponents and gaussian random variables, $X \times Y$ is also lognormal. However, there is no reason to suggest that $X+Y$ is also lognormal.

HOWEVER

If you generate two independent lognormal random variables $X$ and $Y$, and let $Z=X+Y$, and repeat this process many many times, the distribution of $Z$ appears lognormal. It even appears to get closer to a lognormal distribution as you increase the number of observations.

For example: After generating 1 million pairs, the distribution of the natural log of Z is given in the histogram below. This very clearly resembles a normal distribution, suggesting $Z$ is indeed lognormal. Does anyone have any insight or references to texts that may be of use in understanding this?

• Are you assuming equal variances for $X$ and $Y$? If you simulate xx <- rlnorm(1e6,0,3); yy <- rlnorm(1e6,0,1), then the log of the sum does not look very normal any more. – Stephan Kolassa Oct 5 '16 at 7:53
• I did assume equal variances - I'll try another with unequal variance and see what I end up with. – Patty Oct 5 '16 at 8:01
• With variances of 2 and 3, I got something that still looked a bit normal, albiet with what looks like a tiny tiny skew. – Patty Oct 5 '16 at 8:05
• Looking through previous questions may be helpful. Here and here are potentially useful papers. Good look! – Stephan Kolassa Oct 5 '16 at 9:51

## 4 Answers

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 here).

I have also in the past sometimes pointed people to Mitchell's paper

Mitchell, R.L. (1968),
"Permanence of the log-normal distribution."
J. Optical Society of America. 58: 1267-1272.

But that's now covered in the references of Dufresne.

But while it holds in a fairly wide set of not-too-skew cases, it doesn't hold in general, not even for i.i.d. lognormals, not even as $$n$$ gets quite large.

Here's a histogram of 1000 simulated values, each the log of the sum of fifty-thousand i.i.d lognormals: As you see ... the log is quite skew, so the sum is not very close to lognormal.

Indeed, this example would also count as a useful example for people thinking (because of the central limit theorem) that some $$n$$ in the hundreds or thousands will give very close to normal averages; this one is so skew that its log is considerably right skew, but the central limit theorem nevertheless applies here; an $$n$$ of many millions* would be necessary before it begins to look anywhere near symmetric.

* I have not tried to figure out how many but, because of the way that skewness of sums (equivalently, averages) behaves, a few million will clearly be insufficient

Since more details were requested in comments, you can get a similar-looking result to the example with the following code, which produces 1000 replicates of the sum of 50,000 lognormal random variables with scale parameter $$\mu=0$$ and shape parameter $$\sigma=4$$:

res <- replicate(1000,sum(rlnorm(50000,0,4)))
hist(log(res),n=100)


(I have since tried $$n=10^6$$. Its log is still heavily right skew)

• Can you please add the parameters (or code snippet) used to make the histogram in the figure? – altroware Sep 21 '18 at 11:34
• That was two years ago, I don't recall what the lognormal parameters were. But let us apply simple logic. You wouldn't need to worry about the $\mu$ parameter, since it only affects the values on the x-axis scale, not the shape (something convenient like $\mu=0$ would be used). So that leaves the $\sigma$ parameter as the only one with any impact on the shape. Assuming $\mu=0$ and working back roughly from the scale in the histogram above we get that $\sigma$ must be in the ballpark of $4$ or so (NB beware how skew this is). And just trying $4$ gives a pretty similar appearance to the above. – Glen_b Sep 22 '18 at 0:03
• So: res <- replicate(1000,sum(rlnorm(50000,0,4))); hist(log(res),n=100) ... if you try it a few times you'll see the scale jumps around a little but the general picture is about right. Note that the population moment-skewness of the component lognormals is $26.5$ billion -- the population mean will exceed almost every generated value in most of your samples. – Glen_b Sep 22 '18 at 0:04

It's probably too late, but I've found the following paper on the sums of lognormal distributions, which covers the topic. It's not lognormal, but something quite different and difficult to work with.

Lognormal law is widely present on physical phenomena, sums of this kind of variable distributions are needed for instance to study any scaling behavior of a system. I know this article (very long and very strong, the beginning can be undertood if you are not specilist!), "Broad distribution effects in sums of lognormal random variables" published in 2003, (the European Physical Journal B-Condensed Matter and Complex Systems 32, 513) and is available https://arxiv.org/pdf/physics/0211065.pdf .

The adviced paper by Dufresne of 2009 and this one from 2004 together with this useful paper cover the history on the approximations of the sum of log-normal distribution and gives sum mathematical result.

The problem is that all the approximations cited there are found by supposing from the depart that you are in a case in which the sum of log-normal distributions is still log-normal. Then you can compute the $\mu$ and the $\sigma$ of the global sum in some approximated way. But this doesn't give you the conditions that you have to fulfill if you want that the sum is still log-normal.

Maybe [this paper] (http://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=6029348) give you in a particular case a kind of central limit theorem for the sum of log-normals but there is still a lack of generality. Anyway the example given by Glen_b it's not really appropriate, because it's a case where you can easily apply the classic central limit theorem, and of course in that case the sum of log-normal is Gaussian.

But is true as said in the paper cited just above that even in the limit $n\to \infty$ you can have a log-normal sum (for example if variables are correlated or sufficiently not i.i.d.)

• You say that in my example "you can easily apply the classic central limit theorem" but if you understand what the histogram is showing, clearly you can't use the CLT to argue that a normal approximation applies at n=50000 for this case; the sum is so right skew that its log is still heavily right skew. The point of the example was that it's even too skew to approximate by a lognormal (or that histogram would look very close to symmetric). A less skew approximation (such as the normal) would be *worse*/ – Glen_b Aug 19 '18 at 23:32
• I agree, but probably in you example either numerical convergence of the sample is not reached (1000 trials are too few) or statistical convergence is not reached, (50 000 addends are too few), but for in the limit to infinity the distribution should be Gaussian, since we are in CLT conditions, isn't it? – Mimì Aug 21 '18 at 10:30
• The 1000 samples is more than sufficient to discern the shape of the distribution of the sum -- the number of samples we take doesn't alter the shape, just how "clearly" we see it. That clear skewness isn''t going to go away if we take a larger sample, it's just going to get smoother looking. Yes, 50,000 is too few for the sum to look normal -- it's so right skew that the log still looks very skew. It may well require many millions before it looks reasonably normal. Yes, the CLT definitely applies; it's iid and the variance is finite, so standardized means must eventually approach normality. – Glen_b Aug 21 '18 at 11:13