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I'd like to know how to generate two LogNormalDistribution with the same standart deviation and, if possible, get the associated CDF.

To contextualize, I'm trying to reproduce this plot:

enter image description here

Reference of the paper: https://www.sciencedirect.com/science/article/pii/S0031920106001610

Thanks in advance

Note: This plot was obtanied by numerically solving the equations $M(t) = M_{eq} -(M_{eq}-M_0) e^{-t/\tau}$, with $M_0$ = 0, $M_{eq}$ = 1 and I supposed that (although I don't know how to do it) the $\tau$ distributions was used as an input. The objective of the plot is to show that independently of the $\tau$ distribution, the resultant plot of $M(t) / M_{eq}$ is almost the same

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  • $\begingroup$ Does MixtureDistribution do what you need? $\endgroup$ – mikado May 3 at 16:35
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LogNormalDistribution is a built-in distribution so you can use PDF and CDF.

Plot[
 {PDF[LogNormalDistribution[1, 0.25], x],
  8 PDF[LogNormalDistribution[3, 0.25], x], 
  CDF[LogNormalDistribution[1, 0.25], x] + 
  CDF[LogNormalDistribution[3, 0.25], x]},
 {x, 0, 200},
 PlotRange -> {-0.01, 2},
 ScalingFunctions -> {"Log", None}
]

PDFs and CDF of LogNormalDistribution.

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  • $\begingroup$ Thanks, MassDefect. I appreciate your help. Now that I know how to do this plot, the next step is to try to fit experimental data with a CDF similar to this of your plot and, as an output, get the associated PDF(s). $\endgroup$ – José Augusto Devienne May 3 at 20:50
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I understand that the point is how much the variability of the $\tau$ parameter will affect the shape of the $M(t)=M_0\cdot E^{-t/\tau}$ function.

$\tau$ obay a certain probability distribution. We nead to obtain $\tau$ values for the probabilities $p$ that set the boundaries of ever wider confidence intervals. This is done using InverseCDF[]. I have selected the fallowing probabilities: {0.125,0.25,0.375,0.5,0.625,0.75,0.875}. For respectve $\tau$ values, we define the function $M(t)$.

Now, instead of a single $M(t)$ function, we plot a family of functions that differ in the $\tau$ parameter, so that the 'central'* function (the one with $p=0.5$) is marked in black and its less and less likely variants are gray.

Such a plots ware crated for the $\tau$ obaying the LogNormalDistribution[μ,σ]** - where different values of $\mu$ and $\sigma$ were tried.

pp = Table[i, {i, 0.125, 1 - 0.125, 0.125}]
\[Mu]\[Mu] = Table[i, {i, 1, 5, 1}];
\[Sigma]\[Sigma] = {0.5, 1, 2};
dist1 := LogNormalDistribution[\[Mu], \[Sigma]]
\[Tau][p_] := InverseCDF[dist1, p]
M[t_, p_] := E^(-t/\[Tau][p])
MM := Table[M[t, p], {p, pp}]
rowh = Table[StringJoin["\[Mu]=", ToString[i]], {i, \[Mu]\[Mu]}];
colh = Table[
   StringJoin["\[Sigma]=", ToString[i]], {i, \[Sigma]\[Sigma]}];
TableForm[
 Table[Plot[MM, {t, 0, 25}, 
   PlotStyle -> {Gray, Gray, Gray, Black, Gray, Gray, Gray}, 
   Filling -> {1 -> {7}, 2 -> {6}, 3 -> {5}}, 
   FillingStyle -> {Gray, Opacity[0.5]}, 
   PlotRange -> {0, 
     1}], {\[Mu], \[Mu]\[Mu]}, {\[Sigma], \[Sigma]\[Sigma]}], 
 TableHeadings -> {rowh, colh}]

enter image description here

Similarly for NormalDistribution[μ,σ], but here some of $\tau$'s fell below $0$ causing exponential explosion of $M$.

enter image description here

*) Please note that this one corresponds to the median of the $\tau$ distribution and not to mode (the most probable value).

**) Please pay attention to the meaning of parameters in log-normal distribution - they are non-trivial!

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  • $\begingroup$ Thanks, Druid. The problem is exactly this that you presented: how to show that, independetly of the distribution of tau, which may be related with a probabilistic distribution, the M(t) plot should be almost the same. But the resultant M(t) curve should be just one. Other fellow users have been helping me, and I'm still confusing on how the author plotted the equation for the given M0 and Meq parameters and had taken as an output the tau-distribution OR if they putted as an input the tau-distribution and noticed that the final M(t) curve is the same. $\endgroup$ – José Augusto Devienne May 3 at 21:01
  • $\begingroup$ But since you're dealing with the distribution of argument, you have to accept the distribution of results, not the single result. Each graph illustrates the distribution of the $M(t)$ function-forms resulting from the $\tau$ distribution. The individual plots correspond to different $\tau$ distributions - differing by two parameters - therefore they form a two-dimensional array. $\endgroup$ – Druid May 4 at 11:11

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