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can any random variable, not necessarily normally distributed, scaled and shifted in such a way that the new mean is 0 and the new variance is 1?

If not, which can? I remember hearing about location-scale families. Is this connected? Or what about the exponential family distributions or Pearson distribution?

If it is possible (in some cases), can it always be done via a similar transformation as for the normal distribution $$X\sim \mathcal{N}( \mu,\sigma),\ Z=\frac{X-\mu}{\sigma}\ \rightarrow\ Z \sim\mathcal{N}( 0,1)$$?

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  • $\begingroup$ See the answer of Jack. There are very much situations in which it is very very handsome to look at rv $X$ as $X=\sigma Y+\mu$ where $Y$ has mean $0$ and variance $1$. You can work with $Y$ without annoying parameters and - as last stage - the results found on $Y$ can be projected on $X$. Make it a custom to check wether this can be practicized! $\endgroup$ – drhab Sep 30 '14 at 10:06
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If $X$ is a random variable with mean $\mu$ and variance $\sigma^2$, then $$ Y\triangleq \frac{X-\mu}{\sigma}$$ has zero mean and unit variance, hence yes.

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  • $\begingroup$ One small thing: $\sigma>0$ is requested of course. $\endgroup$ – drhab Sep 30 '14 at 10:11
  • $\begingroup$ @drhab: oh, sure. $\endgroup$ – Jack D'Aurizio Sep 30 '14 at 10:14
  • $\begingroup$ One less small thing, $\mu$ and $\sigma$ must be finite (defined). This is not the case for the Cauchy distribution for example. $\endgroup$ – Paul Sep 30 '14 at 10:17
  • $\begingroup$ @Paul: that is right, for sure. But obviously, if $X$ does not have a mean or a variance there is little hope that a scaled version of $X$ has a mean or a variance. $\endgroup$ – Jack D'Aurizio Sep 30 '14 at 10:19

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