Does adding or/and dividing a random variable by a constant change its probability distribution? Suppose that we have a random variable X with probability distribution $PDF_X(\mu_x,\sigma_x)$
Consider random variable $Y=\frac {X-a}b$ , I
 know that mean and variance of Y would be:
$$\mu_y=\frac {\mu_x-a}b, \sigma_y=\sqrt{ \frac {\sigma_x^2} {b^2}  }$$
Would X and Y have the same type of probability distribution (Of course with different mean and variance)?
For example I know that if X is a Normal random variable, Y would be again a Normal random variable. Is this true for all the other probability distribution?
Thank you.
 A: Informally, if we have a random variable $X$, and $Y=aX+b$, where $a$ and $b$ are constants and $b\ne 0$, then $X$ and $Y$ are close relatives. 
But the distributions of $X$ and $Y$ need not have the same type of name. As you pointed out, if $X$ has normal distribution, then so does $Y$. Similarly, if $X$ has uniform distribution, so does $Y$.
However, if $X$ has binomial distribution, then $aX+b$ only has binomial distribution if $a=1$ and $b=0$.  Similar comments could be made about the hypergeometric, the Poisson, and many others.  
The fact that if $X$ has binomial distribution, then (usually) $aX+b$ does not has no real mathematical significance. It just has to do with the kind of distributions we choose to call binomial.  The very close relationship between $X$ and $aX+b$ remains, even if we do not happen to give their distributions the same name.
A: $c\cdot X$ does not necessarily follows the same family of distributions as $X$; depends on the family of distribution $X$ belongs to (Not true e.g. for the Beta distribution, but true for Normal). 
