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

Armando Serrano Lombillo arserlom@gmail....
Wed Jul 21 07:13:58 CDT 2010

```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?

On Wed, Oct 14, 2009 at 3:20 PM, <josef.pktd@gmail.com> wrote:

> On Wed, Oct 14, 2009 at 4:22 AM, Mark Bakker <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.
>
> http://en.wikipedia.org/wiki/Log-normal_distribution#Mean_and_standard_deviation
>
>
>
>
> 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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