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The feedback I received from my initial post seemed to indicate that my question was ill-posed. Hence, I would like to clarify what I am doing and how I hope to achieve it.

I'm running some simulations on models with a material parameter $g(x)$ whose values, according to the literature I've read, are log-normally distributed in the spatial. The material parameter $g(x)$ corresponds to a physical property that is strictly positive and experimentally determined to be never larger or smaller than some threshold values $g_{\max},g_{\min}>0$. When I run my simulations, I would like to generate a distribution of values $g$ over the spatial domain such that

  1. All values of $g$ are strictly in the interval $[g_{\min},g_{\max}]$
  2. The distribution of values of $g$ is as close to a log-normal distribution as possible for a given mean $\mu$ and standard deviation $\sigma$.

I have at my disposal a pseudorandom number generator that creates a vector of normally distributed values with a given mean $\mu$ and standard deviation $\sigma$. I know that if the data set $X$ is normally distributed, then an exponential transformation of the data (in other words, $\exp(X)$) should produce a log normal distribution.

I hypothesize that I can create this log normal distribution by the following process:

  1. Make a normal distribution of values X with mean $\mu$ and standard deviation $\sigma$.
  2. Exponentiate the values to generate a new set $Y=\exp(X)$.
  3. Rescale and shift the values of $Y$ so that the minimum value of the new set is exactly $g_{\min}$ and the maximum is $g_{\max}$. I would do so by creating a new data set $Z=a + (b-a)Y$.

Would this give me the results I desire? That is, in the end, would the set $Z$ be log-normally distributed with all values lying in the interval $[g_{\min},g_{\max}]$?

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    $\begingroup$ Welcome to C.V. Paul. Let's start by 1. What do you mean 'make a normal distribution on the interval [0,1]'? Do you plan to generate generate $X\sim\mathcal{N}(0,1)$ and reject the 1/3 or so observations falling outside [-1,1] and square the rest? $\endgroup$ – user603 May 21 '13 at 22:15
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    $\begingroup$ It all depends on what exactly you mean by a "normal" and a "lognormal" distribution. The conventional mathematical definition of "normal" makes nonsense of this question, because the only normal distributions supported on finite intervals are the degenerate ones: one might go so far as to say there is no such thing as a "normal distribution on the interval [0,1]." Do you want to truncate your distributions, perhaps? If so, then why do you propose truncating twice in your procedure, when once will do? $\endgroup$ – whuber May 21 '13 at 22:16
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    $\begingroup$ en.wikipedia.org/wiki/Truncated_distribution $\endgroup$ – whuber May 21 '13 at 22:16
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    $\begingroup$ Rescaling it so all the values are between the limits means your $\mu$ and $\sigma$ no longer apply. Why specify them only to in effect destroy them? If the range is restricted, it is not lognormal. Why pound an infinite peg into a square hole? $\endgroup$ – Glen_b -Reinstate Monica May 22 '13 at 2:47
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    $\begingroup$ Check out some kind of beta distribution as a first approximation. Although you've revised your question, all the key comments on your first version still apply. $\endgroup$ – Nick Cox May 22 '13 at 7:38
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No. The data $Z$ clearly won't be distributed as log-Normal with parameters $\mu$ & $\sigma$; nor will they be distributed (conditional on $a$ & $b$) as any log-Normal, as you've introduced a location shift—which is not equivalent to changing a parameter value. Note also that you'll get a different distribution for each simulation of $Z$ & it will look very different at different sample sizes.

The discrepancy between the literature & experimental results should be the first thing to investigate. It could be that a truncated log-Normal is what you want, as @whuber suggested, & which you can easily get by discarding values of $Y$ outside $[g_\mathrm{min}, g_\mathrm{max}$]. Or perhaps another distribution entirely—@Nick's suggestion of a beta distribution is worth following up.

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    $\begingroup$ +1. But given that the literature suggests a lognormal is appropriate, betas will not be good substitutes because of their very different tail behavior. $\endgroup$ – whuber May 22 '13 at 15:07
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    $\begingroup$ @whuber +1. But given that experiments suggest a log-Normal is inappropriate, betas might be good substitutes because of their very different tail behaviour. $\endgroup$ – Scortchi - Reinstate Monica May 22 '13 at 15:38
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    $\begingroup$ +1 to everyone commenting. Poster has to decide: if the distribution is lognormal, it can't be confined to a finite interval; if the support is a finite interval, the distribution can't be lognormal. One assumption must be wrong. $\endgroup$ – Nick Cox May 22 '13 at 17:01

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