If X is normally distributed, and c is a constant, is cX also normally distributed? If $X\sim N(\mu, \sigma^2)$, and if $c$ is a constant, is $cX$ also normally distributed? How do you show it?
If yes, does this apply to other distributions? So if $Y$ follows some type of distribution, will $cY$ also follow that distribution? 
Thank you very much!
 A: The answer to your first question is yes:  if $c \ne 0$ and $c \in \mathbb R$, then $cX \sim \operatorname{Normal}(c\mu, |c|\sigma)$ if $X$ is normal with mean $\mu$ and standard deviation $\sigma$.  This is because the normal distribution belongs to a location-scale family:  its PDF is $$f_X(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-(x-\mu)^2/(2\sigma^2)}, \quad -\infty < x < \infty.$$  Then $Y = cX$ will have PDF $$f_Y(y) = \frac{1}{|c|} f_X(y/c),$$ and a little algebra will show that this is also a normal PDF.  I have left this second part as an exercise.
The answer to your third question, the answer is no: some distributions do not have a scale parameter.  Obvious examples are:  the binomial distribution and Poisson distribution.  These have discrete support on nonnegative integers.
There are continuous random variables that also do not have scale parameters:  one example is the Beta distribution $$X \sim \operatorname{Beta}(a,b), \quad a > 0, b > 0$$ has the PDF $$f_X(x) = \frac{\Gamma(a + b)}{\Gamma(a)\Gamma(b)} x^{a-1} (1-x)^{b-1}, \quad 0 < x < 1.$$  $cX$ is not itself Beta for obvious reasons similar to the discrete support examples:  the support is no longer $(0,1)$.
