As $Y$ is log-normal we've $Y\sim \mathbb{LN}\big(\exp(\mu+\sigma^2/2),\exp(2\mu+\sigma)(\exp(\mu^2)-1)\big)$.

Now I define $e = Y - \mathbb{E}(Y) = Y - \exp(\mu+\sigma^2/2)$.

As $e$ is the centered version of $Y$ I would expect that $e$ has a first moment of zero. I'm not sure about the distribution because $e$ can obtain negative values so log-normal is out of the question. Looking at some empirical results it seems that normal is not a proper fit either (top left is the histogram of $\hat{u}=\log(Y)-\widehat{\mu}$, top right is $\exp(\hat{u})$, bottom left corresponds to $\hat{e} = Y - \exp(\hat{\mu}+s_{\hat{u}}^2/2)$. The red line gives the log-normal $\mathbb{LN}(\exp(\frac{0.5^2}{2}),\exp(0.5^2)(\exp(0.5^2)-1))$ and the blue line the normal $\mathbb{N}(\bar{\hat{e}}=0.36,s_{\hat{e}}^2=2.67)$. We see that neither does fit the observed distribution properly: enter image description here

n <- 100000
mu <- 1
sigma <- 0.5
e <- rnorm(n,mu,sigma)
y <- exp(e)
data <- data.table(y=y,e=e)
mod <- lm(log(y)~1)
s <- data[,sqrt(mean(u^2))]

u <- data[,u]
mean.u <- mean(u)
sd.u <- sd(u)
h1 <- hist(u,freq=FALSE,ylim=c(0,1),main="u = log(y)-y_hat")
lines(xfit, yfit, col="red", lwd=2)

u.exp <- data[,exp(u)]
h1 <- hist(u.exp,freq=FALSE,ylim=c(0,1),main="exp(u)")
lines(xfit, yfit, col="red", lwd=2)

e <- data[,e]
h2 <- hist(e,freq=FALSE,ylim=c(0,0.5),main="e = y-exp(mu-s^2/2)")
mean.e <- exp(mean.u+sd.u^2/2)
var.e <- exp(2*mean.u+sd.u^2)*(exp(sd.u^2)-1)
sd.e <- sqrt(sd.e)
mean.e <- mean(e)
sd.e <- sd(e)
lines(xfit, yfit, col="red", lwd=2)
lines(xfit, yfit2, col="blue", lwd=2)
  • 3
    $\begingroup$ You must be using an unusual definition of "$\mathbb{LN}$", because for most people a two-parameter lognormal variable cannot be negative, whereas obviously $X-\mathbb{E}(X)$ can be negative and therefore is not a Lognormal variable, at least not as commonly understood. Please tell us what your notation means. $\endgroup$
    – whuber
    Commented Jan 23, 2016 at 20:59
  • $\begingroup$ Indeed it is a little misleading. I was thinking that $e$ (centered version of $Y$) is log-normal even though the distribution is not defined for negative values. I was guessing that a shift of the distribution by its first moment to the left will not change its shape and therefore will just be a "centered" version of the log-normal (which is just not true). I edited the initial post to make things more clearer $\endgroup$
    – Druss2k
    Commented Jan 25, 2016 at 1:33
  • 2
    $\begingroup$ I have reopened your post but you still need to make some changes. For example, you seem to be parameterizing the lognormal such that its parameters are the mean and variance of the lognormal. This is not the convention for the lognormal (the $\mu$ and $\sigma^2$ parameters are the mean and variance of the normal distribution of the log of a lognormal random variate). If you're following the usual parameterization, for your question you'd say $LN(\beta,\sigma^2)$ -- or if you're using an unusual parameterization you should say so. $\endgroup$
    – Glen_b
    Commented Jan 25, 2016 at 1:46
  • $\begingroup$ Thanks. I'll change it to match the usual conventions with $\mu$ and $\sigma^2$. $\endgroup$
    – Druss2k
    Commented Jan 25, 2016 at 1:52
  • 3
    $\begingroup$ What is wrong with a mere change of variable? Since the density of the log-normal distribution is known in closed form, so is the density of a location shift of that distribution. $\endgroup$
    – Xi'an
    Commented Jan 25, 2016 at 5:34

2 Answers 2


The answer is simply $$ f_E(e) = \frac{1}{ [e+\exp(\mu+\sigma^2/2)]\sigma \sqrt{2 \pi}}\exp\left[-\frac {(\mbox{ln}[e+\exp(\mu+\sigma^2/2)] - \mu)^{2}} {2\sigma^{2}}\right] $$ This can be proven using the "change of variables" property $$ f_E(e) = \left| \frac{d}{de} (g^{-1}(e)) \right| \cdot f_Y(g^{-1}(e)) $$

Where $f_Y$ is the log normal density function $$ f_Y(y) = \frac{1}{ y\sigma \sqrt{2 \pi}}\exp\left[-\frac {(\mbox{ln}y - \mu)^{2}} {2\sigma^{2}}\right] $$ and $g$ is the monotonic function defined by $$ g(y) = y-\exp(\mu+\sigma^2/2) $$ Note $g^{-1}(e) = e+\exp(\mu+\sigma^2/2)$ and so $ \frac{d}{de} (g^{-1}(e))=1$. Thus $f_E(e) = f_Y(g^{-1}(e))$.


One can derive that $e = Y - \mathbb{E}(Y)$ is log-normal. To see this, define $\mathbb{E}(Y):=\mu$, $Var(Y) := \sigma^2$ and simply write out the probability density functions (pdfs):

\begin{align} f_Y(ln(x)) &= \frac{1}{\sigma \sqrt{2\pi}}e^ \frac{- \left( {x - \mu } \right)^2}{2 \sigma^2} \\ f_e(x) &= f_Y(x - \mu) \end{align}

Now $f_e(ln(x)) = f_Y(ln(x - \mu)) = \frac{1}{\sigma \sqrt{2\pi}}e^ \frac{- \left( {(x - \mu) - \mu } \right)^2}{2 \sigma^2} = \frac{1}{\sigma \sqrt{2\pi}}e^ \frac{- \left( {x } \right)^2}{2 \sigma^2}$~$N(0,\sigma^2)$.

So the logarithm of $e = Y-\mu$ is standard normal, i.e. $ln(e)$~$N(0,\sigma^2)$.

  • 2
    $\begingroup$ This is either incorrect or needs more explicit explanation, for the reasons given in my comment to the question. $\endgroup$
    – whuber
    Commented Jan 25, 2016 at 14:37
  • 1
    $\begingroup$ It would be a shifted lognormal distribution, not a standard log-normal, at least if you accept Wikipedia as a source. Under related distributions: "...$X+c$ is said to have a shifted log-normal distribution..." $\endgroup$ Commented Jan 25, 2016 at 22:26
  • 1
    $\begingroup$ Yes, I will have to revise this. My problem is that I don't consider the range of x. (I.e., x - $\mu$ cannot be negative the way I derive things.) I will look into a fix... Thanks for the remarks! $\endgroup$
    – Jeremias K
    Commented Jan 26, 2016 at 10:16

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