Question about 'lognrnd' function
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Generally, we say 'log-normal distribution with standard deviation XdB'. When I generate lognormally distributed random number, R, with dtandard deviation 8dB using 'R=lognrnd(mu, sigma)', is sigma 8 or ln8 (linear scale of 8dB)?
4 Comments
Daniel Shub
on 4 Aug 2012
I agree we don't have the tools to do a proper merge. I still think it is useful flag duplicates since it can help people who might spend time duplicating an answer, helps people who are looking for an answer, and lets the OP know that we do not appreciate being tricked.
Answers (3)
Oleg Komarov
on 3 Aug 2012
Edited: Oleg Komarov
on 3 Aug 2012
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...
So, sigma is the standard deviation of
log(R)
1 Comment
Fernanda Suarez Jaimes
on 12 Mar 2020
Do you know how to run a regression of a time series with lognrnd?
Daniel Shub
on 3 Aug 2012
Why not just test it:
R = lognrnd(0, 8, 1e6, 1);
std(R)
ans = 1.8915e+13
std(log(R))
ans = 7.9976
0 Comments
the cyclist
on 6 Aug 2012
I'm hesitant to say that it is in the dB scale, because dB is generally the base 10 logarithm, which is not the case here. However, the input parameters are the mean and standard deviation of the (natural) log of variable.
So, for example, if
>> r = lognrnd(3,7,1000000,1);
then
>> mean(log(r))
will be about 3, and
>> std(log(r))
will be about 7.
You can see details by typing
>> doc lognrnd
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