Lognormal distribution parameters mu and sigma

Hi,

I want to generate a lognormal distribution for a set of data measurments using Simbiology model analyzer. I need to insert the mu and sigma for of this distribution as shown in the attachment. Is mu=log(mean) and sigma=log(Standard deviation)?

Thank you!

Answer 1

[John D'Errico]

In MATLAB, the tools that use a lognormal distribution have you provide mu and sigma in terms of the underlying normal distribution. For example...

help lognrnd

 LOGNRND Random arrays from the lognormal distribution.
    R = LOGNRND(MU,SIGMA) returns an array of random numbers generated from 
    the lognormal distribution with parameters MU and SIGMA.  MU and SIGMA 
    are the mean and standard deviation, respectively, of the associated 
    normal distribution.  The size of R is the common size of MU and SIGMA 
    if both are arrays.  If either parameter is a scalar, the size of R is 
    the size of the other parameter.
 
    R = LOGNRND(MU,SIGMA,M,N,...) or R = LOGNRND(MU,SIGMA,[M,N,...])
    returns an M-by-N-by-... array.
 
    The mean and variance of a lognormal random variable with parameters MU
    and SIGMA are
 
       M = exp(MU + SIGMA^2/2)
       V = exp(2*MU + SIGMA^2) * (exp(SIGMA^2) - 1)
 
    Therefore, to generate data from a lognormal distribution with mean M and
    Variance V, use
 
       MU = log(M^2 / sqrt(V+M^2))
       SIGMA = sqrt(log(V/M^2 + 1))
 
    See also LOGNCDF, LOGNFIT, LOGNINV, LOGNLIKE, LOGNPDF, LOGNSTAT, 
    RANDOM, RANDN.

    Documentation for lognrnd
       doc lognrnd

In that help, you can see how to choose mu and sigma, IF you know the mean and variance of the distribution you wish to see.

That is, if you know the mean and variance of the lognormal distribution you wish, choose mu and sigma as:

MU = log(M^2 / sqrt(V+M^2))

SIGMA = sqrt(log(V/M^2 + 1))

For example, choosing a lognormal with mean 2, and standard deviation 0.5 (so a variance of 0.25) we would see:

M = 2;
V = 0.25;
MU = log(M^2 / sqrt(V+M^2))

MU = 0.6628

SIGMA = sqrt(log(V/M^2 + 1))

SIGMA = 0.2462

X = lognrnd(MU,SIGMA,[1e8,1]);

mean(X)

ans = 2.0001

std(X)

ans = 0.5000

And that is exactly what we would expect from a random sample. I believe Simbiology would use the same version of a lognormal distribution.

[Day]

Thank you very much

Answer 2

[Bruno Luong]

Is mu=log(mean) and sigma=log(Standard deviation)?

No: https://en.wikipedia.org/wiki/Log-normal_distribution

If X is log-normal random variable

• mu is mean of log(X)
• sigma is standard deviation of log(X)
• log(X) is Gaussian random variable

[Day]

Thank you very much