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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) – 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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