How to generate lognormal distribution with specific mean and std in python?

Question

I need to generate a lognormal distribution with mean=1 and std=1. That is:w~logN(1,1). I need the variable w has mu=1 and sigma=1. However, when I use scipy.stats.lognorm, I have trouble on manipulating the parameters s,loc,sigma. The code is as follows:

import numpy as np
from scipy.stats import lognorm

lo = np.log(1/(2**0.5))
sig = (np.log(2.0))**0.5
print(lognorm.stats(s=sig,loc=lo,scale=1.0,moments='mv'))

The result is:

(array(1.06763997), array(2.))

This is clearly not I want. I want the mean=1 and sigma=1.

Could anyone please tell me how to manipulate with s,loc, and scale to get desired results?

Can anyone give me some hints about this? – lalala8797 Jul 13 '21 at 11:35 — lalala8797, Jul 13 '21 at 11:35
Anyone here for helping this? – lalala8797 Jul 13 '21 at 13:01 — lalala8797, Jul 13 '21 at 13:01

charelf · Accepted Answer · 2022-03-12T17:45:50.917

Edit: maybe look at this answer instead: https://stackoverflow.com/a/8748722/9439097

Its probably too late now, but I have an answer to your problem. I have no idea how the lognormal really works and how you could mathematiclaly derive values to arrive at your desired result. But you can programatically do what you want using standardisation.

Example:

I assume you have something like this:

dist = scipy.stats.lognorm.rvs(0.2, 0, 1, size=100000)

plt.hist(dist, bins=100)
print(np.mean(dist))
print(np.std(dist))

which outputs:

mean: 1.0200
std:  0.2055

Now I have no idea what parameters you would need to feed into lognorm to get mean 1 and std 1 like you desired. I would be interested in that. However you can standardise this distribution.

Standardisation means that the final distribution has mean 0 and std 1.

dist = scipy.stats.lognorm.rvs(0.2, 0, 1, size=100000)

# standardisation to get mean = 0, std = 1
dist = (dist - np.mean(dist)) / np.std(dist)

plt.hist(dist, bins=100)
print(f"mean: {np.mean(dist):.4f}")
print(f"std:  {np.std(dist):.4f}")

mean: 0.0000
std:  1.0000

And now you can reverse this process to get any mean you want. Say you want mean = 123, std = 456:

dist = scipy.stats.lognorm.rvs(0.2, 0, 1, size=100000)

# standardisation to get mean = 0, std = 1
dist = (dist - np.mean(dist)) / np.std(dist)

# get desired mean + std
dist = (dist * 456) + 123

plt.hist(dist, bins=100)
print(f"mean: {np.mean(dist):.4f}")
print(f"std:  {np.std(dist):.4f}")

outputs

mean: 123.0000
std:  456.0000

The shape itself is the same as initially.

Careful, this is not a lognormal distribution anymore and likely inappropriate to use if you are after the properties that the lognorm is typically used for (e.g. positive support). If standardization is an option, one would likely rather just define a normal distribution in the first place. — mcsoini, Sep 03 '22 at 06:14

mcsoini · Answer 2 · 2022-09-03T06:15:33.383

I got confused by the parameterization of the scipy lognorm distribution too and ended up reverse engineering its built-in calculation of the mean and variance, solving for the input parameters. Here you go:

import numpy as np
from scipy.stats import lognorm

mu = 1     # target mean
sigma = 1  # target std

a = 1 + (sigma / mu) ** 2
s = np.sqrt(np.log(a))
scale = mu / np.sqrt(a)
print(f"s={s:.4f} scale={scale:.4f}")

mu_result, sigma_result = lognorm.stats(s=s, scale=scale, moments='mv')
print(f"mu={mu_result:.4f}, sigma={sigma_result:.4f}")

Result:

# scipy lognorm parameters we have to use ...
s=0.8326 scale=0.7071
# ... to obtain the target distribution properties.
mu=1.0000, sigma=1.0000

PDF:

import matplotlib.pyplot as plt
x = np.linspace(0, 3, 301)
plt.plot(x, lognorm.pdf(x, s=s, scale=scale))

How to generate lognormal distribution with specific mean and std in python?

2 Answers2