This might be a very silly question, but I just wanted to make sure I have all the right steps.


Suppose we have a univariate continuous random variable $X$, with some pdf and cdf ${{f}\_{X}}(x)$ and ${{F}\_{X}}(x)$, respectively. Now look at the transformation $Y = X + k$, with $k\in \mathbb{R}$. Then, the cdf and pdf of $Y$ are ${{F}\_{Y}}(y)={{F}\_{X}}(y-k)$ and ${{f}\_{Y}}(y)={{f}\_{X}}(y-k)$.


Does this imply that $Y$ has the same distribution type as $X$? In other words, does a translation (shift) either to the left or to the right of the random variable preserve its distribution type (e.g. a translated Normal variate obviously remains Normal, but is this true of any distribution? It would seem obvious that the answer is "yes" (basically I'm taking the shape and moving it without distortions), but I've yet to see a reference on this yet. Any suggestions? Maybe it's so trivial that nobody bothered.


On the other hand, if the answer is "not necessarily", is it just because the domain shifts as well (e.g. shift an Exponential distribution to the right some amount $k$, and now the domain changes from $[0, \infty]$ to $[k, \infty]$, therefore the translated $Y$ is not technically "Exponential", even though the pdf ${{f}\_{Y}}(y)=\lambda e^{-\lambda (y-k)}$ is that of an Exponential r.v.?)



Does the translation of a random variable preserve its distribution type?
Traditionally, the distributions of r.v.'s $X$ and $Y$ are said to be of the same type if there are constants $a>0$ and $b\in\mathbb{R}$ such that the distribution of $aX+b$ coincides with that of $Y$, see, e.g., p.31 in "A modern approach to probability theory" by Bert Fristedt,Lawrence F. Gray (look for it at google books). According to this definition your statement is trivially true. Of course, it is precisely meant to say that dilations and translations should not destroy the type of the distribution.


