4

More specific: I have two variables, X and Y, they are exponentially distributed with parameter a. I know that the distribution of X + Y follows a Gamma(2,a) distribution IF they are independent (I already know that they are identically distributed). But can I safely assume that they are independent?

A yes or no would suffice :)

Wouter Stultiens
  • 41
  • 1
  • 8
    Let $X=Y$. Do X and Y have the same PDF? Are $X$ and $Y$ independent? – Matthew Gunn Jun 22 '17 at 20:20
  • 1
    @ Matthew Gunn if X = Y and if X (and therefore Y) equals a (particular) constant with probability 1 (i.e., is "deterministic), then X is independent of Y (nevertheless, i upvoted your comment). Of course, this doesn't apply if X 9and Y) are exponentially distributed. – Mark L. Stone Jun 22 '17 at 21:39
  • 1
    Though not your example if you take any distribution symmetric about 0 then X and Y could be identically distributed with Y= -X. – Michael R. Chernick Jun 22 '17 at 23:15

1 Answers1

5

Safety depends on context.

But in general, identicalness of distribution does not imply independence. In particular situations (applications), random variables may happen to be independent, or in some cases, even if not, they may be close enough to being independent to suit certain practical purposes (calculations of adequate accuracy).

There is a commonly stated specification for random variables known as i.i.d, which stands for independent and identically distributed. It is a rather strong condition. In legal documents, you may see such terms as 'cease and desist", both words of which mean pretty much the same thing. Not so for i.i.d, in which each "i" has a different meaning. Random variables can be 1) independent without being identically distributed or 2) identically distributed without being independent or 3) independent and identically distributed or 4) neither independent nor identically distributed

kjetil b halvorsen
  • 63,378
  • 26
  • 142
  • 467
Mark L. Stone
  • 12,546
  • 1
  • 31
  • 51
  • 3
    Let me summarize with respect to OP's ask of "A yes or no would suffice :)".... No. – Eduard Gelman Jun 22 '17 at 21:19
  • 2
    @Eagle The difficulty is that yes-no questions are off limits here. By offering an extended explanation, Mark is attempting to rescue this question from oblivion. – whuber Jun 22 '17 at 21:31
  • 2
    It might help to have a simple example illustrating all four possibilities. Draw a card from a standard 52-card deck. Let $X=1$ if the card is a club (which is black) or a diamond (which is red) and $X=0$ otherwise. Let $Y=1$ if the card is a spade or heart, $Y=0$ otherwise. Let $Z=1$ if the card is red and $Z=0$ otherwise. Let $W$ be the card's value from $1$ through $10$; set it to $10$ for the 12 remaining "court cards". Then $(X,W)$ are independent but not identically distributed, $(X,Y)$ are i.d. but not independent, $(X,Z)$ are i.i.d., and $(Z,W+Z)$ are neither i.d. nor independent. – whuber Jun 22 '17 at 21:40