Standardizing *any* random variable.

Question

If $X$ is any random variable, and if $Z=\dfrac{X-\mu}{\sigma}$, then are the following true or false:

1. The mean of $Z$ is always $0$, regardless of the distribution of $X$.
2. The variance of $Z$ is always $1$, regardless of the distribution of $X$.
3. $Z$ is always normally distributed, regardless of the distribution of $X$.

I ran some quick calculations, and I'am persuaded that the first two statements are true. For statement #3, my intuition says the answer is false, but I don't really know how to formulate an answer.

Is this a central limit theorem type question?

Answer 1

For (1): Subtracting the mean from X will result in E(x-E(x)) = E(x) - E(x) = 0

For (2): Proof by counter-example (similar to the answer above)

For (3): Proof by counter-example (as comment above), take X as the Uniform distribution for example.

Multiplying X by 1/(sigma) and subtracting (mu)/(sigma) does not make X normal.

Answer 2

$\newcommand{\var}{\operatorname{Var}}$You first need $E[X]$ and $E[X^2]$ to exist (i.e. $E[X^2] < \infty$).

1. Use linearity of expectation to show its true.
2. Use linearity of expectation to show $\var(Z) = E[(Z-0)^2] = E[Z^2] = 1$.
3. Obviously not. Pretty much any non-Gaussian distribution will have this not be Gaussian (e.g. a Bernoulli rv will become $\frac{1}{\sqrt{2}}$ times a rademacher).

Answer 3

A simple counterexample suffices for claim #3: just let $X \sim \mathrm{Bernoulli}(0.5)$. Since $X$ is discrete, $Z = (X - \mu)/\sigma$ is also discrete, which contradicts the claim that $Z$ is normal (hence, continuous).

Of course, even if we restrict the claim to continuous random variables with infinite support, another counterexample is the exponential distribution: $f_X(x) = e^{-x}$, for $x > 0$. And if $X$ is required to be continuous with support over the entire real line, then choose the double exponential distribution $$f_X(x) = \frac{1}{2} e^{-|x|}.$$ The cusp at $f(0)$ cannot be removed by standardization, or indeed, by any nondegenerate location-scale transformation.