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I have a problem that I can't seem to figure out on my own due to limited knowledge of mathematics, and I was hoping I could seek help from the community. I would like to use LogNormalDistribution to generate random variables bound by min and max. I fail to do so, and I believe it is due to the way I derive Mean and StandardDeviation. Here is a simplified version of what I want to achieve but using a UniformDistribution instead.

size = 10^4; 
bound = {10^-7, 10^-4}; 

NumberLinePlot @ Interval @ bound

enter image description here

Histogram @ RandomVariate[UniformDistribution @ bound, size]

enter image description here

Based on the above, I can generate random numbers using an UniformDistribution. I have tried to derive a PDF for the above. However, it fails with an error message. Note, I use Win10 and Mathematica 11 for this work.

Show[
 Histogram[
   RandomVariate[UniformDistribution @ bound, size], 
   30, 
  "PDF"
 ],
 Plot[
  PDF[UniformDistribution @ bound, x], 
  {x, bound}
 ]
]

Skeleton is not a Graphics primitive or directive.

What I would like to be able to do is generate random numbers within a pre-defined boundary using a LogNormalDistribution. I have tried to achieve it by exploring Mean and StandardDeviation functions. However, I fail to achieve the desired outcome. In terms of the Mean, I would like it to remain sloped to the left towards the min. However, it can be anywhere within the first 15% of the CDF.

I do appreciate your help, and thank you for your willingness to assist me with the above.

EDIT 1 - PDF Fix for the above example, based on @kglr contribution

Show[
  Histogram[
    RandomVariate[UniformDistribution@bound, size], 
    30, 
    "PDF"
  ],
  Plot[
    PDF[UniformDistribution@bound, x], 
    {x, ## & @@ bound}
  ]
]
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    $\begingroup$ try Show[Histogram[RandomVariate[UniformDistribution@bound, size], 30, "PDF"], Plot[PDF[UniformDistribution@bound, x], {x, ## & @@ bound}]]? $\endgroup$
    – kglr
    Commented Dec 30, 2019 at 17:21
  • $\begingroup$ Hello, @kglr, and thank you - this actually does work. I am not sure why; I will need to read-up about it. $\endgroup$
    – e.doroskevic
    Commented Dec 30, 2019 at 17:23
  • 2
    $\begingroup$ e.doroskevis, correct syntax is Plot[function[x], {x, a, b}] not Plot[ function[x], {x, {a,b}}] (this is what you have in your code). {x, ##&@@round} (or, equivalently, {x, Sequence @@ round}) gives {x, 10^-7, 10^-4} which is the form Plot wants in its second argument. $\endgroup$
    – kglr
    Commented Dec 30, 2019 at 23:59
  • $\begingroup$ Thank you, @kglr. I see what I did wrong. $\endgroup$
    – e.doroskevic
    Commented Dec 31, 2019 at 0:55

1 Answer 1

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@Chris gave you most if not all that you needed. Here is a follow-up to that answer:

dist = TruncatedDistribution[{xmin, xmax}, LogNormalDistribution[μ, σ]]
(* Mean  *)
mean = Simplify[Mean[dist], Assumptions -> {0 < xmin < xmax, σ > 0}]

Mean

(* 15th percentile  *)
x15 = Simplify[InverseCDF[dist, 15/100], Assumptions -> {0 < xmin < xmax, σ > 0, μ ∈ Reals}]

15th percentile

You ask for determining the combination of parameters (xmin, xmax, μ, and σ) that result in mean < x15. I don't think there are any combinations of the parameters that will result in that inequality being true.

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