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While it seems like a rudimentary question one should be able to google, I've tried every related search query I could think of and found nothing. Thus, given the mean and variance of X, how can one find the mean and variance of $\sqrt{X}$

Edit: X is normally distributed.

Dimitriye98
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    One cannot: the mean and variance of $X$ do not determine the mean of $\sqrt{X}$. – Did Apr 06 '16 at 08:11
  • @Did Can it be done if the distribution is normal? – Dimitriye98 Apr 06 '16 at 08:15
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    Square root of a random variable normally distributed? Sure about that? – Did Apr 06 '16 at 08:18
  • @Did I mean if the random variable you're taking the square root of is normally distributed. X is normally distributed, and I need the mean and variance of $\sqrt X$ – Dimitriye98 Apr 06 '16 at 08:24
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    Yeah, and to consider the square root of a random variable, one needs that it is almost surely nonnegative. Do you think a normal random variable fits this condition? – Did Apr 06 '16 at 08:27
  • But √X is defined correctly if and only if supp(X) is subset of R^+_0. Not for N(0,1), for example. – Slepecky Mamut Apr 06 '16 at 08:31
  • @Did I suppose that makes sense. I need it for a geometric mean, so I suppose I'll go ask a separate question on getting that geometric mean. – Dimitriye98 Apr 06 '16 at 08:31

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