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Let's say $X$ has a log-normal distribution and there is one real positive number $c$. then is it right to say that $(X -c)$ also has some log-normal distribution? My feeling is that, it can't be, because $(X - c)$ may take negative value whereas a log-normal distribution is only defined on the positive domain. Can somebody disprove that?

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  • $\begingroup$ I think you are correct. I had to add 1 to my data to be able to use the Zipf distribution. $\endgroup$ – Damien Jul 12 '12 at 18:18
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The answer to your question is (essentially) no and your argument has the right idea. Below, we formalize it a bit. (For an explanation of the caveat above, see @whuber's comment below.)

If $X$ has a lognormal distribution this means that $\log(X)$ has a normal distribution. Another way of saying this is that $X = e^{Z}$ where $Z$ has a $N(\mu, \sigma^2)$ distribution for some $\mu \in \mathbb{R}, \sigma^2 >0$. Note that by construction, this implies that $X \geq 0$ with probability one.

Now, $X-c = e^Z - c$ cannot have a lognormal distribution because

$$ P(e^Z - c < 0 ) = P(e^Z < c) = P(Z < \log(c)) = \Phi \left( \frac{ \log(c) - \mu }{\sigma} \right) $$

which is strictly positive for any $c > 0$. Therefore, $e^Z - c$ has a positive probability of taking on negative values, which precludes $e^Z - c$ from being lognormally distributed.

In summary, the lognormal distribution is not closed under subtraction of a positive constant. It is, however, closed under multiplication by a (positive) constant, but that's an entirely different question.

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    $\begingroup$ +1 It may be worth noting that in some circles the term "lognormal distribution" may comprise the three-parameter version in which an additive location parameter is included. In this case, the answer--by explicit construction--is yes. $\endgroup$ – whuber Jul 12 '12 at 18:49

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