Notation for standardized random variable? I am wondering if there is in use a notation for a standardized random variable? If $X$ is a random variable following some parametric distribution with mean $\mu$ and variance $\sigma^{2}$, then if there is a notation for the standardized random variable $\sigma^{-1}(X-\mu)$?
 A: The standard normal distribution assumptions aren't exactly maintained for standardizing other distributions. For example, the normal distribution is
$$f(x \; | \; \mu, \sigma) = \frac{1}{\sqrt{2\pi\sigma^2} } \; e^{ -\frac{(x-\mu)^2}{2\sigma^2} }\;.$$
The standard normal distribution is a special case when mean $\mu=0$ and standand deviation $\sigma =1$, and it is described by its standard probability density function,
$$\varphi(x) = \frac 1{\sqrt{2\pi}}e^{- \frac 12 x^2}
\;.$$ 
The formula for the probability density function of one method of generalizing the gamma distribution is 
$$f(x) = \frac{(\frac{x-\mu}{\beta})^{\gamma - 1}\exp{(-\frac{x-\mu}
{\beta}})} {\beta\Gamma(\gamma)}  \hspace{.2in}  x \ge \mu; \;\gamma,
\beta > 0\;.$$
The case where $\mu=0$ and $\beta=1$ is called the standard gamma distribution. The equation for the standard gamma distribution reduces to 
$$f(x) = \frac{x^{\gamma - 1}e^{-x}} {\Gamma(\gamma)}  \hspace{.2in}
x \ge 0; \gamma > 0\;,$$
where the mean is $\gamma$ and the variance is also $\gamma$. So, the mean is not set to be zero. Nor can it be, unlike for the symmetric normal distribution, we cannot define a gamma distribution to be $-\infty<X<\infty$. In the standard  case, the gamma distribution is zero for $X<0$. In other words, the support is semi-infinite.
Generalizing from these two distributions, we can write standard forms for distributions, but the rules for what a standard distribution is vary according to the nature of the distribution itself.
