Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.?)

share|improve this question
1  
Well, the translated variable does not have the same distribution, for the mean, for example, is obviously different... –  Mariano Suárez-Alvarez Jul 1 '10 at 18:27
    
The situation is exactly as you say. It doesn't have the same distribution, but it's just shifted along the line. –  Robin Chapman Jul 1 '10 at 18:28
1  
Surely it depends on what you mean by "the type of a distribution"? My point of view is that it is built into the definition of 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. –  Yemon Choi Jul 1 '10 at 18:28
    
I've changed "distribution" to "distribution type" to address Mariano and Yemon's comments. By "distribution type" I mean, as Yemon says, closure under the translation operation. So, generally speaking, the assertion is whether any random variable univariate exhibits closure under translation (e.g. shifted Uniform remains Uniform - obvious; shifted Beta remains Beta, etc.). The mean should be different, the other moments remain unchanged; I think this is a basic property of the translation operation. –  baudolino Jul 1 '10 at 18:50
1  
Sorry - meant central moments. They most certainly are translation invariant. Alternatively, you can look at cumulants, too. With respect to translation, the first cumulant of $Y$ (the mean) is equivariant, while all the other cumulants of $Y$ are invariant. –  baudolino Jul 1 '10 at 19:11
show 2 more comments

closed as not a real question by Robin Chapman, Mariano Suárez-Alvarez, Yemon Choi, Steve Huntsman, Pete L. Clark Jul 2 '10 at 1:31

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
    
Thank you for the reference! Although the definition says "if $a$ and $b$ exist then...", whereas I was asserting that it doesn't matter what $b$ is (so it always exists), there is a Proposition in the reference a little bit further down on Page 31 that clears it up. –  baudolino Jul 2 '10 at 2:45
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.