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If I have a random variable $X$ which has mean $\mu$ and variance $\sigma^2$, what is the approximate expression of $\log(X)$ and $\sqrt{X}$? Do I assume normal approximation or use Taylor expansion?

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  • $\begingroup$ If this is for homework or self-study, you should add the self-study tag. See stats.stackexchange.com/tags/self-study/info $\endgroup$ – Patrick Coulombe Apr 7 '14 at 16:53
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    $\begingroup$ We can't tell you what you're expected to assume, but my guess is that it's probably Taylor expansion that's is being looked for. $\endgroup$ – Glen_b -Reinstate Monica Apr 8 '14 at 0:48
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    $\begingroup$ Be careful: $\log(X)$ and $\sqrt{X}$ are not even properly defined if $\Pr(X<0)>0$. $\endgroup$ – Zen Apr 9 '15 at 22:09
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without knowing the distribution of $X$ you can't find the mean and variance of $f(X)$ in your cases. Taylor expansion will not help because you only know first two moments, which at best would help you evaluate a couple of terms in the Taylor series.

Normal assumption can't be made arbitrarily. If you think your X is from normal distribution, then you basically defined it by first two moments, i.e. you know the distribution. Computing moments of a transformed variable would have been trivial in this case for some functions.

For $\log(X)$ and $\sqrt{X}$ you have an issue: what are you going to do with negative X? Are you willing to go into complex numbers domain?

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  • $\begingroup$ In place of "$\log(X)$" you must mean $\exp(X)$, because the logarithm has similar problems with negative arguments as the square root. $\endgroup$ – whuber Apr 7 '14 at 21:05
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    $\begingroup$ @whuber, my bad, corrected $\endgroup$ – Aksakal Apr 7 '14 at 21:07
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You mean the mean of $\log(x)$ and $\sqrt{x}$? Yes, it depends on the distribution. Using a normal distribution or log-normal or gamma will give you different values.

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    $\begingroup$ Could you please amplify this answer and provide enough detail to be useful to the O.P.? $\endgroup$ – whuber Apr 7 '14 at 17:08
  • $\begingroup$ I don't have the distribution for X at hand. I was only given that it has mean mu and variance sigma^2. $\endgroup$ – user2350622 Apr 7 '14 at 17:12
  • $\begingroup$ User2350622, please edit your question to reflect that crucial information. Before you do so, though, search our site for delta method because what you find will likely answer your question. $\endgroup$ – whuber Apr 7 '14 at 17:17
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    $\begingroup$ What would be the distribution of $\log(X)$ and/or $\sqrt{X}$ when $X$is a normal random variable? Especially when the normal random variable has significant probability of taking on negative values? $\endgroup$ – Dilip Sarwate Apr 7 '14 at 18:02
  • $\begingroup$ Well first of all you can't take the log() or sqrt() of a negative number so a normal distribution is technically not going to work. If all you know is the mean and sigma then you simply don't have enough information to answer the question. To answer this, you are going to have to make additional assumptions. I'd suggest using the the gamma distribution because you can easily calculate the expectation value of sqrt(x). Note that the expectation value of log(x) is the entropy which is given here. en.wikipedia.org/wiki/Gamma_distribution $\endgroup$ – Dave31415 Apr 7 '14 at 18:06

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