[SciPy-User] specify lognormal distribution with mu and sigma using scipy.stats

Armando Serrano Lombillo arserlom@gmail....
Wed Jul 21 12:15:28 CDT 2010


Ok, I had misunderstood that mu and sigma where the mean of the lognormally
distributed variable. So, this is what I should have written:

>>> mean = 10.0
>>> variance = 1.0
>>> mean_n = log(mean) - 0.5*log(1 + variance/mean**2)
>>> variance_n = log(variance/mean**2 + 1)
>>> d = lognorm(sqrt(variance_n), scale=exp(mean_n))
>>> d.stats()
(array(10.000000000000002), array(1.0000000000000013))

Thanks, Armando.

On Wed, Jul 21, 2010 at 5:48 PM, Warren Weckesser <
warren.weckesser@enthought.com> wrote:

> Armando Serrano Lombillo wrote:
> > Hello, I'm also having difficulties with lognorm.
> >
> > If mu is the mean and s**2 is the variance then...
> >
> > >>> from scipy.stats import lognorm
> > >>> from math import exp
> > >>> mu = 10
> > >>> s = 1
> > >>> d = lognorm(s, scale=exp(mu))
> > >>> d.stats('m')
> > array(36315.502674246643)
> >
> > shouldn't that be 10?
>
> In terms of mu and sigma, the mean of the lognormal distribution
> is exp(mu + 0.5*sigma**2).  In your example:
>
> In [16]: exp(10.5)
> Out[16]: 36315.502674246636
>
>
> Warren
>
>
>
>
> >
> > On Wed, Oct 14, 2009 at 3:20 PM, <josef.pktd@gmail.com
> > <mailto:josef.pktd@gmail.com>> wrote:
> >
> >     On Wed, Oct 14, 2009 at 4:22 AM, Mark Bakker <markbak@gmail.com
> >     <mailto:markbak@gmail.com>> wrote:
> >     > Hello list,
> >     > I am having trouble creating a lognormal distribution with known
> >     mean mu and
> >     > standard deviation sigma using scipy.stats
> >     > According to the docs, the programmed function is:
> >     > lognorm.pdf(x,s) = 1/(s*x*sqrt(2*pi)) * exp(-1/2*(log(x)/s)**2)
> >     > So s is the standard deviation. But how do I specify the mean? I
> >     found some
> >     > information that when you specify loc and scale, you replace x by
> >     > (x-loc)/scale
> >     > But in the lognormal distribution, you want to replace log(x) by
> >     log(x)-loc
> >     > where loc is mu. How do I do that? In addition, would it be a
> >     good idea to
> >     > create some convenience functions that allow you to simply
> >     create lognormal
> >     > (and maybe normal) distributions by specifying the more common
> >     mu and sigma?
> >     > That would surely make things more userfriendly.
> >     > Thanks,
> >     > Mark
> >
> >     I don't think loc of lognorm makes much sense in most application,
> >     since it is just shifting the support, lower boundary is zero+loc.
> The
> >     loc of the underlying normal distribution enters through the scale.
> >
> >     see also
> >
> http://en.wikipedia.org/wiki/Log-normal_distribution#Mean_and_standard_deviation
> >
> >
> >     >>> print stats.lognorm.extradoc
> >
> >
> >     Lognormal distribution
> >
> >     lognorm.pdf(x,s) = 1/(s*x*sqrt(2*pi)) * exp(-1/2*(log(x)/s)**2)
> >     for x > 0, s > 0.
> >
> >     If log x is normally distributed with mean mu and variance sigma**2,
> >     then x is log-normally distributed with shape paramter sigma and
> scale
> >     parameter exp(mu).
> >
> >
> >     roundtrip with mean mu of the underlying normal distribution
> >     (scale=1):
> >
> >     >>> mu=np.arange(5)
> >     >>> np.log(stats.lognorm.stats(1, loc=0,scale=np.exp(mu))[0])-0.5
> >     array([ 0.,  1.,  2.,  3.,  4.])
> >
> >     corresponding means of lognormal distribution
> >
> >     >>> stats.lognorm.stats(1, loc=0,scale=np.exp(mu))[0]
> >     array([  1.64872127,   4.48168907,  12.18249396,  33.11545196,
> >      90.0171313 ])
> >
> >
> >     shifting support:
> >
> >     >>> stats.lognorm.a
> >     0.0
> >     >>> stats.lognorm.ppf([0, 0.5, 1], 1, loc=3,scale=1)
> >     array([  3.,   4.,  Inf])
> >
> >
> >     The only case that I know for lognormal is in regression, so I'm not
> >     sure what you mean by the convenience functions.
> >     (the normal distribution is defined by loc=mean, scale=standard
> >     deviation)
> >
> >     assume the regression equation is
> >     y = x*beta*exp(u)    u distributed normal(0, sigma^2)
> >     this implies
> >     ln y = ln(x*beta) + u   which is just a standard linear regression
> >     equation which can be estimated by ols or mle
> >
> >     exp(u) in this case is lognormal distributed
> >
> >     Josef
> >     _______________________________________________
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> >
> >
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