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if you exponentiate a normal distribution,

Y=exp{X}

where X is a normally distributed random variable (RV), then Y is log-normally distributed.

What is the distribution if you take the logarithm, instead, of a normal distribution? In other words, if

Y=ln(X)

what is the distribution of Y? Is there a closed-form expression, or even an empirical approximation, to that RV?

mdewey
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eSurfsnake
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    Does this answer your question? [Expected value and variance of log(a)](https://stats.stackexchange.com/questions/57715/expected-value-and-variance-of-loga) – Adrian Keister Apr 22 '20 at 18:20
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    I see the math problem, but let's assume I'm doing it numerically (which is why a closed-form would be handy, even a polynomial approximation). For example, a satellite may orbit the earth at 100 miles with a standard deviation of 1 mile...the odds of it being negative are now infinitesimal. I could always check first to insure that P(X<0)<.0001 or="" whatever=""> – eSurfsnake Apr 22 '20 at 18:28
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    In that case the logarithm is extremely well approximated by a linear function, so the distribution is *still* normal. – whuber Apr 22 '20 at 19:06
  • Can you expand on that? For example, if the mean were 100 miles and the stdev were 30 miles, three-sigma lower would be 10. We could still truncate negative numbers and be "at least 99.7% right". I strongly suspect that logarithm would not be well approximated by a normal. – eSurfsnake Apr 22 '20 at 19:19
  • I think you should consider the domain of $\log$ function. Is something wrong with me?! Acording what I understood, you want to find the distribution of $\log X$ when $X$ is normal?!! – Masoud Apr 22 '20 at 20:23
  • If $x$ is normally distributed, then the domain of $x$ is $(-\infty,+\infty)$, while $\log()$ only accept values in $(0,+\infty)$. So you can't define $y=\log(x)$ – Haotian Chen Apr 22 '20 at 20:34
  • You want to find the distribution of $\log X$ when $X$ has the normal distribution?!! The domain of the $\log$ function is $(0,\infty)$. – Masoud Apr 22 '20 at 20:33
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    Surfsnake, you are correct--but you have simply found that intermediate regime where it makes no sense to take the log, because the chance of $X$ being negative is appreciable. Because you are asking for a mathematical impossibility (as amply pointed out by the last several comments), we are struggling to make *some* practical sense of your question. In the case you posited earlier, the mean is 100 and the SD 1. The Normal approximation to $\log X$ in that case is *beautiful.* – whuber Apr 22 '20 at 21:21
  • agreed...N(100,000, 1) is going to look just like a normal curve. I'm trying to find a way to re-pose the question where the distribution gives it sense. Sort of like how $$exp{\frac{1}{1-x^2}$$ over [-1,1] "traps" all of function between -1 and +1. – eSurfsnake Apr 23 '20 at 01:11

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