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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?

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    $\begingroup$ 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 ♦
    Sep 18, 2019 at 15:09
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    $\begingroup$ 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 $\sigma. \qquad$ $\endgroup$
    – Michael Hardy
    Sep 18, 2019 at 15:22

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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.)

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  • $\begingroup$ 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
    Sep 18, 2019 at 14:51
  • $\begingroup$ Nice answer! I was just a few seconds late in posting mine. 😃 $\endgroup$
    – Isabella Ghement
    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. $\endgroup$
    – Michael Hardy
    Sep 18, 2019 at 15:22
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    $\begingroup$ @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. $\endgroup$
    – Stephan Kolassa
    Sep 18, 2019 at 15:24
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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 = $\sqrt{\left[\exp\left(\sigma^2\right) - 1\right] \cdot \exp(2\mu + \sigma^2)}$

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    $\begingroup$ 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
    Jan 12, 2020 at 19:46

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