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

[Warren Weckesser]

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 at gmail.com 
> <mailto:josef.pktd at gmail.com>> wrote:
>
>     On Wed, Oct 14, 2009 at 4:22 AM, Mark Bakker <markbak at gmail.com
>     <mailto:markbak at 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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