Question about 'lognrnd' function

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Sooyeol Im
Sooyeol Im 2012-8-3
评论： Fernanda Suarez Jaimes 2020-3-12
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)?
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Daniel Shub
Daniel Shub 2012-8-4
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.
Oleg Komarov
Oleg Komarov 2012-8-6
@Daniel and the cyclist: thanks.

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回答（3 个）

Oleg Komarov
Oleg Komarov 2012-8-3
编辑：Oleg Komarov 2012-8-3
Reading carefully the documentation of lognrnd():
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)
Daniel Shub
Daniel Shub 2012-8-3
Why not just test it:
R = lognrnd(0, 8, 1e6, 1);
std(R)
ans = 1.8915e+13
std(log(R))
ans = 7.9976
the cyclist
the cyclist 2012-8-6
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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