From a randonly generated normal distribution of values v:

v = RandonVariate[NormalDisrtibution[56.6*10^-9,2.5*10-9]] 

I created another distribution of values and appended all in a list called vol

Clear[vol]; vol = {};

Do[AppendTo[vol, v[[i]]^3], {i, Length[v]}]

From the list vol I generated another distribution, which I called f($\tau$), being $\tau$ a function of v such as

$\tau = \tau_0 Exp \left[\frac{M_s B_c}{2kT} v \left(1-\frac{B}{B_c}\right)^2 \right]$


\[Tau]0 = 10^-10; \[Mu]0 = 4*Pi*10^-7; Bk = 
 10*10^-3; Ms = 157*10^3; kb = 1.38*10^-23; T=300; B=100*10^-6;

When I generate an histogram evaluating $\tau$ for each vol the result is

Histogram[tau0, "Log"]

enter image description here

The apparent mean value is ~ 10^6 based on the figure, but when I use the function Mean the result is different

In[707]:= Mean[tau0]

Out[707]= 3.72233*10^11

Does anyone know why? Any help would be great!

Thanks in advance!

  • 2
    $\begingroup$ Since you are plotting on a log scale, your visual interpretation will be off here. $\endgroup$ May 14, 2020 at 16:55

1 Answer 1


The transformed distribution of v has a non-vanishing third moment (i.e skew) so the median is not the same as the mean - meaning the new distribution is not symmetric about the 50% mark. This could explain why there's a lot of extra probability mass in the upper half of your logarithmic histogram if you plug the numbers in.

taudist = 
  TransformedDistribution[t0 Exp[(Ms Bc)/(2 k T) v (1 - B/Bc)], 
   v \[Distributed] NormalDistribution[\[Mu], \[Sigma]]];

Moment[taudist, 3] // InputForm

E^((3*(B - Bc)*Ms*(-4*k*T*\[Mu] + 3*(B - Bc)*Ms*\[Sigma]^2))/(8*k^2*T^2))*t0^3

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