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.?)

built into the definitionof the family of normal random variables that the family is closed under translation and (suitably-scaled) dilation. As it stands, I don't think your question is well posed. $\endgroup$ – Yemon Choi Jul 1 '10 at 18:28