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I have the following issue. With a Monte Carlo simulation I have generated a data set, x values and their frequencies. When I plot the histogram I noticed that the shape of the distribution is skewed. The distribution of ln(x) is almost a perfect Gaussian distribution. As far as I have understood this right a quantity is ln-normal distributed when ln(x) is normal distributed. So what then I calculated $\mu$ and $\sigma$ of the distribution of ln(x). As I mentioned I assumed that ln(x) is normal distributed so that its PDF is $y=A\times e^{-0.5\times(ln(x)-\mu)^2/\sigma^2}$.

Now, how do I get the standard deviation of x (being ln-normal distributed) and its expected value?

kjetil b halvorsen
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Zorg
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    You could consult this Wikipedia article - it has all the necessary formulas (derivations are a matter of a few simple integrations). – Roger V. Nov 14 '19 at 10:28
  • i alreday had a look at the this, but the problem is that in these they use the PDF of X and it would be $1/(\sigma\sqrt{2\pi}\times x)\times e^{-0.5\times(ln(x)-\mu)^2/\sigma^2}$ – Zorg Nov 14 '19 at 14:53
  • I am not sure that I understand your problem: the expectation of $\ln(x)$ is $\mu$, while its variance is $\sigma^2$. Whereas the mean of $x$ is $\exp(\mu + \sigma^2/2)$ and variance is $[\exp(\sigma^2) - 1]\exp(2\mu + \sigma^2)$. – Roger V. Nov 14 '19 at 15:30
  • I have added the derivation details to my answer, in case this is what posed difficulty. – Roger V. Nov 14 '19 at 16:06
  • https://stats.stackexchange.com/search?q=lognormal+standard+deviation – whuber Nov 14 '19 at 21:47
  • @Vadim with mean you mean the median? – Zorg Nov 18 '19 at 13:42
  • By mean I mean the mean - the mathematical expectation of the variable. In our case $\mathbb{E} [\ln(x)] = \mu$, whereas $\mathbb{E} [x] = \exp(\mu + \sigma^2/2)$. – Roger V. Nov 18 '19 at 13:54

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