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According to Wikipedia (page about lognormal distribution), if $X \thicksim N(\mu, \sigma^2)$ then $Y=e^X \thicksim \mathrm{logN}(\mu, \sigma^2)$.
But the support of $\mathrm{logN}$ is just $(0,+\infty)$
So assuming $N(0,\cdot)$ I would have a fair chance that an instance of $X$ be negative. What would happen in this case? $e^\text{sth. neg}$ is a very real number, but according to the formula for the density of the lognormal distribution, it's probability is not defined, because the logarithm of $x$ is taken.
I assume the error is in my head and not on Wikipedia, so what am I missing here?

Update:
I was very confused. I'm not sure how, but I guess I thought $e^{sth. neg.}$ was a negative number itself because it's on the left side of the 0.
Or I somehow inserted $x \in X$ into $\frac1{x\sqrt{2\pi\sigma^2}}e^{-\frac{(\ln x-\mu)^2}{2\sigma^2}}$ instead of the (positive!) $e^x=y\in Y$.
I guess I wasn't expecting that symmetric values like $-x$ and $x$ would enter the $\mathrm{logN}(x,\mu,\sigma^2)$ at different points (namely $e^{-x}$ and e^x)
So thank you all very much for putting up with that and apologies for wasting everyone's time!

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    $\begingroup$ What the heck is $e^{sth. neg}$ ? $\endgroup$ Jul 24, 2015 at 18:05
  • $\begingroup$ btw there's no problem with $\log(e^X)$... $\endgroup$ Jul 24, 2015 at 18:06
  • $\begingroup$ @DavidC.Ullrich $e^x$ where $x \in (-\infty,0)$ $\endgroup$
    – user2740
    Jul 24, 2015 at 20:11

3 Answers 3

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If $x$ is a negative number then $e^x$ is a positive number less than $1$.

The logarithm of a positive number less than $1$ is a negative number.

$e^\text{something negative}$ is a positive number less than $1$. It has a logarithm. Its logarithm is negative.

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  • $\begingroup$ "If $x$ is a real number then $e^x$ is a positive number less than $1$"? $\endgroup$ Jul 24, 2015 at 20:34
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The terminology may be a bit confusing, but when we say $Y$ is lognormally distributed, that means the logarithm of $Y$ is normally distributed. In other words if $X \sim \operatorname{Normal}(\mu,\sigma^2)$, then $\log Y = X$, and $Y = e^X \sim \operatorname{LogNormal}(\mu,\sigma^2)$. Then it becomes clear that if $X \in (-\infty, \infty)$, then $Y = e^X \in (0, \infty)$.

There were times when I thought, "shouldn't $Y$ be called 'Exponential Normal' rather than 'Log Normal?'" After all, what you are doing to produce $Y$ from $X$ is to take the exponential (anti-log) of $X$. But lognormal makes its own sort of sense too, I suppose: $Y$ is "log"-normal because its logarithm is normal. Hey, I didn't make up the rules.

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When we write $e^X \sim logN$ we don't mean that we need to take the logarithm of the normal r.v. Rather, we mean that $X \sim \mathcal{N}(\mu, \sigma^2)$ and if we let $Y = e^X$, then $Y$ is log-normal. Note that since $X \in \mathbb{R}$, we must have $Y = e^X \in (0, +\infty)$.

Now, note that since $Y$ is log-normal, we have $\ln Y \sim \mathcal{N}$, which is well-defined: since $Y \in (0, +\infty)$ as we argued above, $\ln Y \in \mathbb{R}$...

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