Lognormal and normal distributions

Question

I have a dataset, X, of real numbers, x, that I assume they follow a Lognormal distribution. Based on this, the distribution Y(y), y=LN(x), is Gaussian.

Compute the parameters of X and Y, the mean value, standard deviation value and ultimately the coefficient of variation value.

Questions:

Given
    X ~ Lognormal ⁡ (μX , σX)
    Y ~ Normal ⁡ (μY , σY)

> For Y: an estimate of the mean value is μY an estimate of the standard
> deviation value is σY an estimate of the CoV value is σY/μY
> 
> For X: an estimate of the mean value is exp(μX + σX^2/2) an estimate
> of the standard deviation value is SQRT[exp(σX^2) − 1]*exp(2μX +
> σX^2)] an estimate of the CoV value is SQRT[exp(σX^2) - 1]

1- Should μX = μY and σX = σY? Or are (μ , σ) calculated for each Dataset X and Y?

2- What are the expressions that relate the parameters, mean, standard deviation and CoV of X and Y?

Example:

import numpy as np
import pandas as pd

np.random.seed(0)
LNd = pd.DataFrame(np.random.lognormal(mean=0.0, sigma=1.0, size=1000000), columns=['Values'])

mX = np.mean(LNd.values)
sX = np.std(LNd.values)
print(mX, sX)

Nd = LNd['Values'].apply(lambda x: np.log(x))
Nd.columns = ['Values']

mY = np.mean(Nd.values)
sY = np.std(Nd.values)
print(mY, sY)

emX=np.exp(mY+(sY**2)/2)
esX=(np.exp(2*mY+2*(sY**2))-np.exp(2*mY+(sY**2)))**0.5
print(emX, esX)

emY=np.log(mX)-(sX**2)/2
esY=(np.log(1+(sX/mX)**2))**0.5
print(emY, esY)

Why is emy not equal to mY? In other references (wikipedia) there are other expressions but still emy is not equal to mY.

Answer 1

This is a good introductory question in Probability and Statistics, so I won't deprive you of the fun by answering it in detail.

First thing we need to understand is that Log normal distribution (X) ranges only on the positive real axis whereas the normal distribution (Y) ranges over the entire real axis. So, clearly,

1. $\mu_x \ne \mu_y$ You need to take the individual means of X and Y datasets. Read up about convex functions and figure out if $\mu_x >=< \mu_y$. Worth it!
2. Variance usually reduces by taking a log-transformation. So, we can expect that $\sigma_x > \sigma_y$. Now think, what happens to the Covariance because of this reduction in variance effect.