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I have been working with lognormal distributions as a proposal distribution for some MCMC routines which require the proposal distribution to have some pre-defined sample mean and variance, $m$ and $v$, respectively. Considering that I'm using scipy.stat.lognorm to generate a sample from the proposal distribution, I have to define $\mu$ and $\sigma^2$, parameters of the underlying normal distribution.

I've determined $\mu$ and $\sigma^2$ using the relation: \begin{equation} \mu=\ln\left(\frac{m}{\sqrt{1+\frac{v}{m^2}}}\right), \qquad \sigma^2=\ln\left(1+\frac{v}{m^2}\right) \end{equation}

This have worked fine when $m \geq 1$, however when $m < 1$ the variance of the generated sample becomes smaller than it was supposed to be ($v$) and I am not sure why.

As an example, running the code bellow

def get_underlying_normal_params(m, v):
    m2 = m * m
    nrm_mean = np.log(m2 / np.sqrt(v + m2))
    nrm_var = np.log((v + m2) / m2)
    return nrm_mean, nrm_var


lognrm_mean = 1e-5
lognrm_var = 10
norm_mean, norm_var = get_underlying_normal_params(lognrm_mean, \
        lognrm_var)
sample = [lognorm.rvs(scale=np.exp(norm_mean), s=np.sqrt(norm_var)) \
        for _ in range(50000)]
print("Sample mean: ", statistics.mean(sample))
print("Sample variance: ", statistics.variance(sample))

the produced output is:

Sample mean:  1.3404377964314877e-06
Sample variance:  5.009857166696648e-09

We can see then, that the created sample has variance ~5e-9, far from the desired variance, 10.

If however, I run the same code setting lognrm_mean = 1, the output is:

Sample mean:  1.0077730236251317
Sample variance:  9.250922743416222

And now the sample variance is actually close to the desired variance.

Why was I able to pre-determine the variance of the sample on the second case but not on the first case? Am I doing something conceptually wrong?

Edit explaining why this is not a duplicate of this other question: in my question I ask if there is something wrong in the procedure I used to create a lognormal with a pre-determined sample mean and variance. There could be a conceptual problem that I am missing or even a procedural problem, concerning for instance floating point erors; I am specifically interested in the behaviour of the sample variance. The other question asks the relation between parameters of a lognormal distribution (associated to an underlying normal distribution) and its mean and variance. Of course this relates to this question, however that is not the question I am asking and, moreover, I have given "formulas" for this relation in my question.

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  • $\begingroup$ I have indicated a "duplicate" that provides appropriate formulas. However, it's unclear what you are doing or asking. In particular, there is nothing in your code named lognormal_mu, so it's not evident what calculation is being performed by setting that variable's value to 1. It's also puzzling why you are printing out moments of simulated values, because they have little to do with the relationships among the moments of the underlying distributions. $\endgroup$ – whuber Sep 9 '19 at 18:41
  • $\begingroup$ I still don't get it: your output very closely matches the input in both cases (usually 3 - 4 significant figures, which is great for samples of just 50K). Exactly what does the "however" refer to? $\endgroup$ – whuber Sep 9 '19 at 19:01
  • $\begingroup$ It refers to the sample variance in the first case, which is very far from the desired variance (lognormal variance). In both runs, the desired variance is 10, however this is only reasonably fulfilled in the second run, when the desired lognormal mean is 1. $\endgroup$ – Gustavo Estrela Sep 9 '19 at 19:04
  • $\begingroup$ @whuber I have edited the question and I hope I could detail more what is my question. I also believe that this shouldn't be marked as duplicate since I have already derived the relation between the (first and second) moments of a lognormal distribution and its underlying normal; which is exactly the question made in the pointed duplicate. $\endgroup$ – Gustavo Estrela Sep 9 '19 at 19:33
  • $\begingroup$ @whuber I have now stated in my question why this is not a duplicate. $\endgroup$ – Gustavo Estrela Sep 10 '19 at 15:44

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