I'm doing some reading and this is the definition I got from DeGroot's book: enter image description here

Does that mean the parameters are the same? For example, assume X is lognormally distributed and Y is normally distributed where Y = log(X). Is this saying that X and Y have the same mean and SDs even though they are different shaped distributions? If not, what distribution is μ and σ referring to?

In other words, if someone says X is lognormally distributed with mean μ and SD σ, do I need to do any conversion so that the mean and SD are in normal terms?

  • 4
    $\begingroup$ Do not confuse parameters of a distribution family with moments. Although $\mu,\sigma$ parameterize the Lognormal distributions, they are not their means or standard deviations. $\endgroup$
    – whuber
    Commented Sep 18, 2019 at 15:09
  • 1
    $\begingroup$ They have the same parameters, but they don't have the same mean or the same standard deviation. The two parameters, $\mu$ and $\sigma,$ that are the mean and standard deviation of $\log X,$ are not the mean and standard deviation of $X.$ But the mean and standard deviation of $X$ are functions of $\mu$ and $\sigma. \qquad$ $\endgroup$ Commented Sep 18, 2019 at 15:22

2 Answers 2


assume X is lognormally distributed and Y is normally distributed where Y = log(X)

This is where you are confused. You don't make assumptions on two distributions, one of which just happens to be the log of the other.

Instead, you start with a distribution $X$. Then you consider $\log X$. If $\log X\sim N(\mu,\sigma^2)$, then we say that the original distribution $X$ is lognormal with parameters $\mu$ and $\sigma^2$.

(And then the mean of $X$ is $\exp\left(\mu+\frac{\sigma^2}{2}\right)$, for instance, so the parameters are certainly not the same. This is also why it is better to speak of the "parameters" of a lognormal, rather than of the "mean and SD" - because it's very easy to get confused whether these refer to the actual mean or the log-mean, same for SD.)

  • $\begingroup$ Ok thanks for clarifying. So generally when people provide the parameters, such as μ and σ, that refers to the distribution of Y or log(X). To get the mean of the lognormal distribution requires a conversion. $\endgroup$
    – confused
    Commented Sep 18, 2019 at 14:51
  • $\begingroup$ Nice answer! I was just a few seconds late in posting mine. 😃 $\endgroup$ Commented Sep 18, 2019 at 14:59
  • $\begingroup$ But the parameters are the same. Then mean and the standard deviation and many other things are the same, but those two parameters are the same. $\endgroup$ Commented Sep 18, 2019 at 15:22
  • 1
    $\begingroup$ @MichaelHardy: yes, the parameters are the same, by definition. I just wince a bit every time someone calls $\mu$ the "mean parameter of the lognormal", because it's only the log-mean, and it's so easy to confuse them. $\endgroup$ Commented Sep 18, 2019 at 15:24

Wikipedia has a nice article on log-normal distributions: https://en.m.wikipedia.org/wiki/Log-normal_distribution. The article reveals that the log-normally distributed X and the normally distributed log(X) have different means and standard deviations.

If X follows a log-normal distribution with parameters $\mu$ and $\sigma$, then $\mu$ and $\sigma$ represent the mean and standard deviation of the distribution of log(X), which is normal. In other words, the mean and standard deviation of the normally distributed log(X) are:

Mean of $\log(X)=\mu$

SD of $\log(X) = \sigma$

The mean and standard deviation of the log-normally distributed X are as follows:

Mean of X = $\exp(\mu + \sigma^2/2)$

SD of X = $\sqrt{\left[\exp\left(\sigma^2\right) - 1\right] \cdot \exp(2\mu + \sigma^2)}$

  • 1
    $\begingroup$ Isabella Ghement's answer is good.. Just wanted to point out that in that answer, SD of X has a typo. exp[𝜎^2−1] should be (exp[𝜎^2]−1). Was doing Monte Carlo sampling to verify the mean and SD of a log-normal distribution and my SD did not match with the above expression. Verified the wiki page linked and noted the typo. PS: Actually wanted to add a comment to @Isabella Ghement answer but don't have necessary credentials to do that. Adding a new answer instead. $\endgroup$
    – Ajay A
    Commented Jan 12, 2020 at 19:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.