Understanding the shifted log-normal distribution

I have difficulties understanding why a third parameter (the shift) is necessary to describe the log-normal distribution.

Let's say we have a normal random variable X, if I shift this variable by an additive constant b, X+b is still normally distributed and can still be described by the two parameters $$\mu$$ and $$\sigma$$.

Why is it not the same for log-normal random variables? By definition, a random variable X has a shifted log-normal distribution with shift $$\theta$$ if log(X + $$\theta$$) ~ N($$\mu$$,$$\sigma$$). However, if X + $$\theta$$ ~logN($$\mu$$,$$\sigma$$), then also X has a log-normal distribution X ~logN($$\mu'$$,$$\sigma'$$).

Accordingly, the two parameters should suffice to describe any shifted log-normal fully, or not?

• Did you prove or have link about "However, if X + θ ~logN(μ,σ), then also X has a log-normal distribution X ~logN(μ′,σ′)." ? – user158565 Dec 2 '18 at 22:01
• I thought this follows from Glen_b's answer here - is my understanding hence incorrect that shifting a lognormal distribution by an additive constant results in a log-normally distributed variable X+b? stats.stackexchange.com/questions/156277/… – user24544 Dec 2 '18 at 22:14
• The question and answers are related to moments, did not mention the distributions. – user158565 Dec 2 '18 at 22:18
• @Glen_b What do you think? I do not think that $X+\theta$ and $X$ follow lognormal simultaneously given $\theta \ne 0$. – user158565 Dec 2 '18 at 22:27
• Hence log(X) ~ N($\mu$,$\sigma^2$) implies that X+c is a shifted log-normal, i.e. log(X+c) ~ N($\mu+c$,$\sigma^2$), but the reverse is not correct? – user24544 Dec 2 '18 at 22:27

1 Answer

By definition, a random variable X has a shifted log-normal distribution with shift $$\theta$$ if log(X + $$\theta$$) ~ N($$\mu$$,$$\sigma$$).

In the more usual notation, that would correspond to a lognormal with shift $$-\theta$$.

However, if X + $$\theta$$ ~logN($$\mu$$,$$\sigma$$), then also X has a log-normal distribution X ~logN($$\mu'$$,$$\sigma'$$).

This is not the case, as we'll see.

To keep things clear, let us distinguish between the two parameter lognormal (with parameters $$\mu$$ and $$\sigma^2$$) and a shifted (i.e. three parameter) lognormal ($$\delta,\mu,\sigma^2$$); if $$\delta=0$$ we get the two-parameter lognormal as a special case. The density for the three parameter lognormal is:

$$\frac {_1}{^{(x-\delta)\sigma {\sqrt {2\pi }}}}\, e^{-{\frac{1}{2\sigma ^{2}} {\left(\ln (x-\delta)-\mu \right)^{2}}}},\quad x>\delta, \:\delta,\mu\in \mathbb{R},\,\sigma>0$$

[Here a positive $$\delta$$ shifts up by $$\delta$$, corresponding to the negative of your $$\theta$$; a positive $$\theta$$ corresponds to a shift down by $$\theta$$. I'll stick with the more common convention.]

It is the case that if you already have a shift (location-parameter) in the model, then adding a shift parameter would do nothing. For example, $$\mu$$ plays this role in the normal distribution, so there would be no point in adding a shift parameter to a normal distribution; it would simply be combined with the $$\mu$$ term.

However, in the lognormal, $$\mu$$ is not a shift parameter. It is a scale parameter; it stretches and compresses rather than shifts. Meanwhile $$\sigma$$ is a shape parameter, controlling how skewed/heavy tailed the lognormal distribution is.

One quick way to see that the shift parameter does something different to the two parameters already there (assuming you don't wish to follow through the algebraic manipulations on the density), is to use the fact that the log of any two parameter lognormal variate is itself distributed as a normal (the three parameter lognormal doesn't share this property in general, as we'll see).

With the normal, if we apply a shift we move the density up or down along the x-axis, which simply alters its $$\mu$$ parameter and leaves us with another normal. With the two parameter lognormal, altering the $$\mu$$ parameter leaves us with another two parameter lognormal but does not simply shift the values. It does have the property that if we then take logs, we get back to a normal. Shifting the normal and then exponentiating to a two parameter lognormal is different from shifting the two parameter lognormal.

[The issue boils down to the fact that addition and exponentiation are not commutative, so shifting the lognormal doesn't work like shifting the normal.]

We can immediately see that if we supply a negative shift ($$\delta<0$$ in a three parameter lognormal) that we can't take logs to get back to a normal -- some of the density applies to negative values of $$x$$. We might briefly entertain the notion that positive arguments might somehow work but we can readily determine that it cannot be the case via simulation, or more directly, even just by considering the lower limit:

The log of a three-parameter lognormal variate with $$\delta>0$$ will have a smallest possible value of $$\ln \delta$$; therefore it cannot be normal, since all normal variates range over the whole real line.

Alternatively, in a simulation, the steps are:

1. generate data from a normal distribution with some $$\mu,\sigma$$

2. exponentiate, to a corresponding two-paramater lognormal with the same parameters

3. shift the distribution up by a substantial amount (say, twice the mean of the lognormal), so that it has a clear impact on the location

4. take logs and note that the result is clearly not normal.

For a large sample from a standard normal and a shift parameter of $$\delta=2e^\frac12\approx 3.3$$, we obtain:

These histograms are what we get at step 1 and 4 respectively.

We can clearly see we don't get a normal back out (it's skewed, for starters), so the shift parameter is not doing the same thing as changing $$\mu$$ would*. The lowest value in the $$\log(x)$$ sample displayed on the right, is $$1.19498$$, just above the lower limit of $$\frac12+\log(2)\approx 1.19315$$

* we can actually see that back in the formula for the density, where $$\delta$$ is inside the $$\log(x..)$$ part but $$\mu$$ is outside it, so they clearly don't just add together.

Appendix:

#R code for the above (fewer simulations than I did, but enough)
#
z <- rnorm(1000000,0,1) # 1 million normals
delta <- 2*exp(1/2)
y <- exp(z)             # 2-parameter lognormal
x <- y + delta          # shifted (i.e. 3-parameter) lognormal
par(mfrow=c(1,2))
hist(z,n=200,col="skyblue",bord="skyblue",freq=FALSE)
hist(log(x),n=200,col="lightgreen",bord="lightgreen",xlim=c(1,5),freq=FALSE)

• Thank you for writing such a beautiful answer. I understand that shifting in the normal "space" on the x-axis is not a shift in the log-normal space, where changes in $\mu$ scale the distribution. I also see in my own data, that only my shifted data leads to a normal distribution of log(data) and else the log of the data leads to a skew-normal. The point which is still confusing me is this (e.g. from wikipedia): "If X ~ logn($\mu$,$\sigma$) then X+c is said to be a shifted lognormal" -- how is this not opposed to your arguments at the end & the simulation, if by def. then log(X+c) is normal? – user24544 Dec 3 '18 at 13:43
• I don't see any disagreement between what you quoted and anything in my answer. Can you help me understand where you feel the two disagree? ... if log(X+c) is normal then X+c is two parameter lognormal but X is not (it's shifted lognormal) – Glen_b -Reinstate Monica Dec 3 '18 at 23:58
• Thanks! It might sound trivial, but I think the problem was that I did not understand the definition correctly: I read it as "Y is a shifted log-normal if for ∀ s: (Y+s) ~ N", when in fact the reading is that "Y is a shifted log-normal iff ∃s s.t. (Y+s) ~ N" - correct? If I understand correctly, what you meant with addition and exponentiation not being commutative can easily be seen by: X = exp(sigma + mu*Z) + s is not a normal? Has the fact however, that log(X) ==> log(X)+c is a shifted log-normal be proven by showing that the distribution of log(X) + c can be written as the 3-param logN? – user24544 Dec 4 '18 at 21:01
• You seem to have picked up a new confusion about what a lognormal is. Neither of your statements in the second sentence are correct. The part where you mention commutativity is also odd; if $Z$ is normal, then $µ+σZ$ is normal, so then (clearly) $\exp(µ+σZ)$ is not normal (whether or not you shift it). Neither $\log(X)$ nor $\log(X)+c$ will be lognormal (of any kind). .... Instead: of $Z$ is normal, then $\exp(Z)$ is lognormal (ordinary two parameter lognormal) and $\exp(Z)+c$ is shifted lognormal (three parameter lognormal). – Glen_b -Reinstate Monica Dec 4 '18 at 23:07
• 1. Is this correct: "Y is a shifted log-normal if for ∀ s: (Y+s) ~ N", when in fact the reading is that "Y is a shifted log-normal iff ∃s s.t. (Y+s) ~ N" - correct? 2. Here I was not referring to the commutativity, but solely on the statement from wikipedia: Has the fact however, that log(X) ==> log(X)+c is a shifted log-normal be proven by showing that the distribution of log(X) + c can be written as the 3-param logN? How do we know this implication is correct? – user24544 Dec 5 '18 at 10:30