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given a distribution of $\sigma^2$ ($y \sim LogN(\mu, \sigma^2$) I want to calculate the posterior distribution of the Gini coefficient which is given by:

$G = 2\Phi(\frac{\sigma}{\sqrt{2}})-1$

$\Phi(x)$ is the cumulative density function for the standard normal distribution ($N(0,1$))

If I would have a fixed $\sigma^2$ it would be intuitive to just put it in the equality and print the distribution for G. But how do I proceed when I have a distribution of $\sigma^2$? Can anybody give me a hint?

Thank you already in advance!

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If you could generate a high enough number of values for $\sigma$ using your $LogN$ distribution then you could compute $G$ for all of those values and kind of recreate the distribution of $G$ I think. I would do it that way :

  • Generate $n$ values for $\sigma$ (I don't know what a $LogN$ distribution looks like but random generation would probably be good enough to get a good hang of a 'discretized' form of the distribution)
  • Compute $n$ $G$ values out of those $\sigma$ (as if $\sigma$ was 'fixed') which would in turn allows to get a good hang of $G$ distribution.
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  • $\begingroup$ Thank you for your fast respond. I also thought about this but I am still struggling a little bit. I have already drawn a high number of values for $\sigma$. When I insert every $\sigma$ in the equation for G it will result in many CDF's, won't it? How do I get the information from the CDF's to plot a single PDF? $\endgroup$
    – Rchieve
    Commented Apr 18, 2016 at 11:17
  • $\begingroup$ I don't think you should have several CDF. $G$ is kind of a normal distribution so to simplify let's use the regular standard normal. If you randomly generate a number of $x$ values and compute $\Phi(x_i), i \in [1, n]$ then you would end up with a series of values describing $\Phi$ (more or less acurately depengin on $n$) so just one CDF $\endgroup$
    – Riff
    Commented Apr 18, 2016 at 11:23
  • $\begingroup$ Now I got it! I implemented it in Python and it seems like it works! Thanks a lot for your hint! $\endgroup$
    – Rchieve
    Commented Apr 18, 2016 at 11:40
  • $\begingroup$ As a side note, the example could be viewed as a bayesian problem where the distribution of $x$ simply is a uniform distribution. Swap that with your $LogN$ distribution and you should get the idea. $\endgroup$
    – Riff
    Commented Apr 18, 2016 at 11:42

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